Explanation:
Okay, so I need to find z_{2002} given that z_0 = 1/137 + i, and each subsequent term is defined by z_n = F(z_{n-1}) where F(z) = (z + i)/(z - i). Hmm, functional iteration. Let me think about how to approach this.

First, maybe I should try to understand what the function F does. Let me write it down again: F(z) = (z + i)/(z - i). This looks like a Möbius transformation, which is a function of the form f(z) = (az + b)/(cz + d). These transformations can sometimes be analyzed by looking at their fixed points or by conjugating them to simpler transformations, like rotations or translations. Maybe that's the way to go here?

Alternatively, since we're dealing with iterations, perhaps there's a pattern or periodicity in the sequence z_n. If I can find a pattern or a cycle, then I can determine z_{2002} based on the cycle length.

Let me start by computing the first few terms manually to see if I can spot a pattern. Given z_0 = 1/137 + i.

Compute z_1 = F(z_0) = (z_0 + i)/(z_0 - i). Let me compute that.

First, z_0 = 1/137 + i. Let me write that as a complex number: let me denote real part as 1/137 and imaginary part as 1.

So z_0 = (1/137) + i*1. Then z_0 + i is (1/137) + i*(1 + 1) = 1/137 + 2i.

z_0 - i is (1/137) + i*(1 - 1) = 1/137 + 0i = 1/137.

Therefore z_1 = (1/137 + 2i) / (1/137) = [ (1 + 2i*137)/137 ] / (1/137) ) = 1 + 2i*137. Wait, let me check that again.

Wait, numerator is (1/137 + 2i), denominator is 1/137. So dividing both terms by 1/137:

( (1/137)/(1/137) ) + (2i)/(1/137) ) = 1 + 2i * 137. Yes, that's right. So z_1 = 1 + 2*137 i. Let me compute 2*137: 2*100=200, 2*37=74, so total 274. So z_1 = 1 + 274i.

Okay, so z_1 is 1 + 274i. Let me compute z_2 = F(z_1) = (z_1 + i)/(z_1 - i). Let me compute numerator and denominator.

z_1 + i = (1 + 274i) + i = 1 + 275i.

z_1 - i = (1 + 274i) - i = 1 + 273i.

So z_2 = (1 + 275i)/(1 + 273i). Hmm, to simplify this, maybe multiply numerator and denominator by the conjugate of the denominator?

So multiply numerator and denominator by (1 - 273i):

Numerator becomes (1 + 275i)(1 - 273i)

Denominator becomes (1 + 273i)(1 - 273i) = 1^2 + 273^2 = 1 + 273².

Let me compute numerator:

First expand (1)(1) + (1)(-273i) + (275i)(1) + (275i)(-273i)

= 1 - 273i + 275i - 275*273 i²

Simplify:

The real part: 1 + 275*273 (since i² = -1)

Imaginary part: (-273i + 275i) = 2i

So numerator is [1 + 275*273] + 2i.

Let me compute 275 * 273. Hmm, 275*273. Let me compute that:

First, note that 275 = 273 + 2, so 275*273 = 273² + 2*273.

Compute 273²: 273^2. Let me recall that 270² = 72900, 270² + 2*270*3 + 3² = 72900 + 1620 + 9 = 74529. Wait, no, 273 = 270 + 3, so (a + b)^2 = a² + 2ab + b².

a = 270, b = 3: 270² + 2*270*3 + 3² = 72900 + 1620 + 9 = 74529. Then 2*273 = 546. Therefore total numerator real part is 1 + 74529 + 546 = 1 + 74529 is 74530, plus 546 is 75076. Therefore numerator is 75076 + 2i.

Denominator is 1 + 273² = 1 + 74529 = 74530. So z_2 is (75076 + 2i)/74530.

Let me simplify that fraction. Let me check if 75076 and 74530 have a common factor. Let me compute GCD(75076, 74530). Let me compute the difference: 75076 - 74530 = 546. Then GCD(74530, 546). Let me divide 74530 by 546.

First, 546 * 136 = 546*(100 + 30 + 6) = 54600 + 16380 + 3276 = 54600 + 16380 is 70980, plus 3276 is 74256. Then 74530 - 74256 = 274. So GCD(546, 274). Then compute GCD(546,274). 546 ÷ 274 = 2 with a remainder of 546 - 548 = wait, 274*2 = 548, which is larger than 546. So actually 546 ÷ 274 = 1 with remainder 546 - 274 = 272. Then GCD(274, 272). Then 274 - 272 = 2. GCD(272,2). Which is 2. Therefore GCD is 2. Therefore, numerator and denominator can be divided by 2.

So numerator: 75076 ÷ 2 = 37538, 2i ÷ 2 = i. Denominator: 74530 ÷ 2 = 37265. Therefore z_2 = (37538 + i)/37265. Wait, but 37538 divided by 37265? Wait, actually, z_2 = [ (75076 + 2i)/74530 ] simplifies to [ (37538 + i)/37265 ]? Wait, no. Wait, the numerator was 75076 + 2i, denominator 74530. Dividing numerator and denominator by 2 gives (37538 + i)/37265. So z_2 = (37538 + i)/37265. Let me check if that reduces further? Probably not, since GCD of numerator terms? Let me check if 37538 and 37265 have a common factor. Let me compute difference: 37538 - 37265 = 273. Then GCD(37265, 273). Let's see, 273 divides into 37265 how many times?

Divide 37265 by 273. Let me compute 273*100 = 27300, subtract that from 37265: 37265 - 27300 = 9965. Then 273*30 = 8190, subtract from 9965: 9965 - 8190 = 1775. Then 273*6 = 1638, subtract: 1775 - 1638 = 137. Then GCD(273,137). 137 divides into 273 twice, remainder 273 - 274 = -1? Wait 273 ÷ 137 = 1 with remainder 273 - 137 = 136. Then GCD(137,136) = 1. Therefore, the fraction cannot be reduced further. Therefore z_2 = (37538 + i)/37265.

Hmm, this seems complicated. Maybe I need another approach. Computing terms manually is getting messy. Maybe there's a pattern or a cycle here? Let me think.

Since F is a Möbius transformation, perhaps it has a fixed point or can be conjugated to a simpler transformation. Let me recall that Möbius transformations can be represented by matrices, and their iterations can be analyzed by diagonalizing the matrix or finding its eigenvalues. Alternatively, maybe I can find a function that conjugates F to a multiplication by a scalar, which would make iteration straightforward.

Alternatively, perhaps we can use complex analysis. Let me try to find fixed points of F. Fixed points satisfy F(z) = z, so:

(z + i)/(z - i) = z. Multiply both sides by (z - i):

z + i = z(z - i)

Expand RHS: z² - iz

Bring all terms to one side:

z² - iz - z - i = 0 => z² - z(1 + i) - i = 0

Solve quadratic equation: z = [ (1 + i) ± sqrt( (1 + i)^2 + 4i ) ] / 2

Let me compute discriminant D = (1 + i)^2 + 4i

First compute (1 + i)^2 = 1 + 2i + i² = 1 + 2i -1 = 2i

Then D = 2i + 4i = 6i

Therefore z = [ (1 + i) ± sqrt(6i) ] / 2

Hmm, sqrt(6i). Let me compute sqrt(6i). Let me write 6i as 0 + 6i. The square roots of a complex number can be found by converting to polar form. Let me recall that sqrt(a + bi) can be found by solving (x + yi)^2 = a + bi.

But maybe it's easier to note that sqrt(i) is e^{iπ/4} and e^{i5π/4}, so sqrt(6i) would be sqrt(6) * e^{iπ/4} and sqrt(6) * e^{i5π/4}.

Therefore sqrt(6i) = sqrt(6)(cos(π/4) + i sin(π/4)) = sqrt(6)*(√2/2 + i√2/2) = sqrt(3) + i sqrt(3). Similarly the other root is -sqrt(3) - i sqrt(3).

Wait, let me check:

Let me square sqrt(6i). Let me take sqrt(6) e^{iπ/4}. Squaring that gives 6 e^{iπ/2} = 6i. Yes, correct. Similarly, the other square root is sqrt(6) e^{i5π/4} which squares to 6 e^{i5π/2} = 6 e^{iπ/2} = 6i. So yes, sqrt(6i) is sqrt(6)(cos π/4 + i sin π/4) and sqrt(6)(cos 5π/4 + i sin 5π/4). So sqrt(6i) = sqrt(6)( (sqrt(2)/2 + i sqrt(2)/2 )) = sqrt(3) + i sqrt(3) and -sqrt(3) - i sqrt(3).

Therefore, the solutions for z are:

[ (1 + i) ± (sqrt(3) + i sqrt(3)) ] / 2.

Let me compute both possibilities:

First, with the plus sign:

Numerator: (1 + i) + sqrt(3) + i sqrt(3) = (1 + sqrt(3)) + i(1 + sqrt(3))

Therefore z = [ (1 + sqrt(3)) + i(1 + sqrt(3)) ] / 2 = [ (1 + sqrt(3))/2 ] (1 + i )

Similarly, with the minus sign:

Numerator: (1 + i) - sqrt(3) - i sqrt(3) = (1 - sqrt(3)) + i(1 - sqrt(3)) = [1 - sqrt(3)](1 + i)

Therefore z = [1 - sqrt(3)](1 + i)/2.

So the fixed points are z = [ (1 + sqrt(3))/2 ](1 + i) and z = [ (1 - sqrt(3))/2 ](1 + i). Let me note these as α and β, maybe?

Alternatively, perhaps I can use these fixed points to conjugate the transformation F into a simpler form. Let me recall that if we have a Möbius transformation with two fixed points, we can conjugate it to a scaling transformation. Let me try that.

Let me denote the fixed points as α and β. Let me try to find a Möbius transformation that maps α to 0 and β to ∞, which would transform F into a scaling transformation. Let me recall that if we have a Möbius transformation φ such that φ(α) = 0 and φ(β) = ∞, then φ(F(z)) = k φ(z), for some constant k. Then the iteration becomes φ(z_n) = k^n φ(z_0). Therefore, z_n can be obtained by inverting φ.

So let me try to construct φ. Let me take α and β as fixed points. Let me compute them numerically to get a sense? Wait, maybe not. Let me proceed symbolically.

Let me denote φ(z) = (z - α)/(z - β). Then, since φ(α) = 0 and φ(β) = ∞, then φ(F(z)) = k φ(z). Let me verify this.

Alternatively, maybe it's easier to use the formula for conjugating Möbius transformations. Let me recall that if F has fixed points α and β, then the transformation can be written as F(z) = ( (1 - k)z + k α + (k - 1) β ) / ( (k - 1)z + k β - (k - 1) α ) )? Not sure. Maybe I need a better approach.

Alternatively, let me suppose that φ(F(z)) = k φ(z). Let me write φ(z) = (z - α)/(z - β). Then φ(F(z)) = [F(z) - α]/[F(z) - β]. Let me compute that:

F(z) = (z + i)/(z - i). Therefore,

φ(F(z)) = [ ( (z + i)/(z - i) ) - α ] / [ ( (z + i)/(z - i) ) - β ]

Multiply numerator and denominator by (z - i):

Numerator: (z + i) - α(z - i) = z + i - α z + α i = z(1 - α) + i(1 + α)

Denominator: (z + i) - β(z - i) = z + i - β z + β i = z(1 - β) + i(1 + β)

Therefore φ(F(z)) = [ z(1 - α) + i(1 + α) ] / [ z(1 - β) + i(1 + β) ]

If we want this to be equal to k φ(z) = k*(z - α)/(z - β), then we need:

[ z(1 - α) + i(1 + α) ] / [ z(1 - β) + i(1 + β) ] = k*(z - α)/(z - β)

Cross multiplying:

[ z(1 - α) + i(1 + α) ]*(z - β) = k*(z - α)*[ z(1 - β) + i(1 + β) ]

This should hold for all z (except the poles), so coefficients of z², z, and the constants must match. Let me expand both sides.

Left-hand side (LHS):

Multiply [ z(1 - α) + i(1 + α) ] by (z - β):

= z(1 - α)(z - β) + i(1 + α)(z - β)

= (1 - α)z² - (1 - α)β z + i(1 + α) z - i(1 + α)β

= (1 - α)z² + [ - (1 - α)β + i(1 + α) ] z - i(1 + α)β

Right-hand side (RHS):

k*(z - α)*[ z(1 - β) + i(1 + β) ]

First compute [ z(1 - β) + i(1 + β) ] = (1 - β) z + i(1 + β)

Multiply by (z - α):

= (1 - β) z(z - α) + i(1 + β)(z - α)

= (1 - β) z² - (1 - β)α z + i(1 + β) z - i(1 + β)α

= (1 - β) z² + [ - (1 - β)α + i(1 + β) ] z - i(1 + β)α

Therefore RHS is k times that:

k*(1 - β) z² + k*[ - (1 - β)α + i(1 + β) ] z - k*i(1 + β)α

Therefore equating LHS and RHS:

(1 - α)z² + [ - (1 - α)β + i(1 + α) ] z - i(1 + α)β = k*(1 - β) z² + k*[ - (1 - β)α + i(1 + β) ] z - k*i(1 + β)α

Since this holds for all z, coefficients of z², z, and constants must be equal. Therefore:

1. Coefficient of z²: (1 - α) = k*(1 - β). Hence k = (1 - α)/(1 - β)

2. Coefficient of z: [ - (1 - α)β + i(1 + α) ] = k*[ - (1 - β)α + i(1 + β) ]

3. Constant term: -i(1 + α)β = -k i(1 + β) α

Let me check the constant term equation first:

- i(1 + α)β = -k i(1 + β) α => (1 + α)β = k (1 + β) α => k = [ (1 + α)β ] / [ (1 + β) α ]

From coefficient of z², we had k = (1 - α)/(1 - β). Therefore equating these two expressions for k:

(1 - α)/(1 - β) = [ (1 + α)β ] / [ (1 + β) α ]

Cross multiplying:

(1 - α)(1 + β) α = (1 + α)β (1 - β)

Let me expand both sides:

Left: α(1 - α)(1 + β) = α[ (1)(1 + β) - α(1 + β) ] = α(1 + β - α - α β)

Right: β(1 + α)(1 - β) = β[ (1)(1 - β) + α(1 - β) ] = β(1 - β + α - α β)

So left side: α(1 + β - α - α β)

Right side: β(1 - β + α - α β)

Let me expand both:

Left: α + α β - α² - α² β

Right: β - β² + α β - α β²

Set left equal to right:

α + α β - α² - α² β = β - β² + α β - α β²

Simplify both sides by subtracting right from left:

α + α β - α² - α² β - β + β² - α β + α β² = 0

Simplify term by term:

α - β + α β - α β + (- α²) + β² + (- α² β) + α β² = 0

Simplify:

α - β - α² + β² - α² β + α β² = 0

Factor terms:

(α - β) + (-α² + β²) + (- α² β + α β²) = 0

Note that -α² + β² = -(α² - β²) = -(α - β)(α + β)

Similarly, - α² β + α β² = α β ( - α + β ) = α β ( β - α ) = - α β ( α - β )

Therefore:

(α - β) - (α - β)(α + β) - α β ( α - β ) = 0

Factor out (α - β):

(α - β)[1 - (α + β) - α β ] = 0

Therefore, this equation holds if either α - β = 0, which would mean α = β, but α and β are distinct fixed points (since discriminant D = 6i ≠ 0), so α ≠ β. Therefore, the other factor must be zero:

1 - (α + β) - α β = 0 → α + β + α β = 1

So, we have that the fixed points α and β satisfy α + β + α β = 1.

Let me check whether that is true with our found fixed points.

Earlier, we had fixed points:

α = [ (1 + sqrt(3))/2 ](1 + i )

and β = [ (1 - sqrt(3))/2 ](1 + i )

Let me compute α + β:

α + β = [ (1 + sqrt(3))/2 + (1 - sqrt(3))/2 ](1 + i ) = [ (2/2 ) ](1 + i ) = 1*(1 + i ) = 1 + i

Then compute α β:

α β = [ (1 + sqrt(3))/2 * (1 - sqrt(3))/2 ] * (1 + i )^2

First compute the scalar factor: [ (1 + sqrt(3))(1 - sqrt(3)) ] / 4 = (1 - 3)/4 = (-2)/4 = -1/2

Then compute (1 + i )² = 1 + 2i + i² = 1 + 2i - 1 = 2i

Therefore α β = (-1/2)(2i) = -i

So α + β + α β = (1 + i ) + (-i ) = 1 + i - i = 1. Yes! It satisfies the equation. Therefore, the condition is satisfied. Good.

So, going back, we have that k = (1 - α)/(1 - β). Let me compute that.

But maybe instead of computing k explicitly, we can use the relation α + β + α β = 1. Let me think if I can find k in terms of α and β?

Alternatively, maybe I can proceed with φ(z) = (z - α)/(z - β) as the conjugating map, then φ(F(z)) = k φ(z). Therefore, φ(z_n) = k^n φ(z_0). Therefore, once we compute φ(z_0), we can get φ(z_n) = k^n φ(z_0), then invert φ to get z_n.

That seems promising. Let me try this approach.

First, compute φ(z_0) where φ(z) = (z - α)/(z - β). Then z_n = φ^{-1}(k^n φ(z_0)).

Therefore, if I can compute φ(z_0), then compute k, then find z_n.

So, first, need to find φ(z_0). Let me compute that.

But to compute φ(z_0), I need to know α and β. But α and β are complicated expressions. Maybe there's a better way?

Alternatively, since we have α + β + α β = 1, and also from the fixed points equation.

Alternatively, perhaps there's a relation between k and α, β?

We had earlier that k = (1 - α)/(1 - β). Let me compute that.

But maybe I can find k? Let me think. Let me also note that from the earlier equation when equating coefficients for z term, which was:

[ - (1 - α)β + i(1 + α) ] = k*[ - (1 - β)α + i(1 + β) ]

But maybe we can use k = (1 - α)/(1 - β). Let me denote k = (1 - α)/(1 - β). Then:

Let me compute the RHS of the z coefficient equation: k*[ - (1 - β)α + i(1 + β) ] = [ (1 - α)/(1 - β) ] * [ - (1 - β)α + i(1 + β) ] = (1 - α)[ - α + i(1 + β)/(1 - β) ]? Wait, no. Wait:

Wait, expanding:

= [ (1 - α)/(1 - β) ] * [ - (1 - β)α + i(1 + β) ]

= (1 - α)/ (1 - β) * [ - (1 - β)α ] + (1 - α)/ (1 - β) * i(1 + β )

= - (1 - α) α + i (1 - α)(1 + β)/(1 - β)

Hmm, complicated. Maybe not helpful. Let me try to compute k?

Alternatively, since we know α and β, maybe plug them in. But they are complex numbers. Let me recall that:

Earlier, we had α = [ (1 + sqrt(3))/2 ](1 + i ) and β = [ (1 - sqrt(3))/2 ](1 + i ). Let me denote these as:

Let me write α = A(1 + i), where A = (1 + sqrt(3))/2

Similarly β = B(1 + i), where B = (1 - sqrt(3))/2

Note that A and B are real numbers. Let me check: A = (1 + sqrt(3))/2 ≈ (1 + 1.732)/2 ≈ 1.366, B = (1 - sqrt(3))/2 ≈ (1 - 1.732)/2 ≈ -0.366.

So, α = A(1 + i), β = B(1 + i). Therefore, maybe I can express k = (1 - α)/(1 - β). Let me compute 1 - α and 1 - β.

Compute 1 - α = 1 - A(1 + i) = 1 - A - A i

Similarly 1 - β = 1 - B(1 + i) = 1 - B - B i

So k = [1 - A - A i ] / [1 - B - B i ]

But since A and B are real numbers, this is a complex number. However, maybe there's a simplification here. Let me compute k.

Alternatively, maybe there is a relation between A and B? Let me note that A + B = [ (1 + sqrt(3))/2 + (1 - sqrt(3))/2 ] = 1, and A * B = [ (1 + sqrt(3))/2 ] * [ (1 - sqrt(3))/2 ] = (1 - 3)/4 = -2/4 = -1/2, as computed earlier.

Also, since α + β + α β = 1, and α + β = (A + B)(1 + i) = 1*(1 + i) = 1 + i, and α β = -i, so indeed 1 + i - i = 1.

Alternatively, going back to k.

Let me compute numerator and denominator:

Numerator: 1 - α = 1 - A(1 + i) = (1 - A) - A i

Denominator: 1 - β = 1 - B(1 + i) = (1 - B) - B i

But since A + B = 1, then 1 - A = B, and 1 - B = A. Therefore:

Numerator: B - A i

Denominator: A - B i

Therefore k = (B - A i ) / (A - B i )

Multiply numerator and denominator by the conjugate of the denominator, which is (A + B i ):

k = [ (B - A i)(A + B i ) ] / [ (A - B i)(A + B i ) ]

Compute denominator: A² + (B i)^2? Wait, no, denominator is (A - B i)(A + B i ) = A² - (B i )² = A² - ( - B² ) = A² + B². Wait:

Wait, (A - Bi)(A + Bi) = A² - (Bi)^2 = A² - B² i² = A² - B² (-1) = A² + B². Yes.

Numerator: (B - A i)(A + B i ) = B*A + B*(B i ) - A i * A - A i * B i

= AB + B² i - A² i - A B i²

Simplify term by term:

AB + B² i - A² i - AB (-1) because i² = -1

So AB + B² i - A² i + AB

Combine like terms:

AB + AB = 2AB

B² i - A² i = i (B² - A² )

So numerator is 2AB + i (B² - A² )

Therefore, k = [ 2AB + i (B² - A² ) ] / (A² + B² )

Now, let me compute AB, A² + B², and B² - A².

We know that A + B = 1 and A * B = -1/2.

Therefore, A² + B² = (A + B)^2 - 2AB = 1² - 2*(-1/2) = 1 + 1 = 2.

Also, B² - A² = (B - A)(B + A) = (B - A)*1 = B - A.

But A = (1 + sqrt(3))/2, B = (1 - sqrt(3))/2. Therefore B - A = [ (1 - sqrt(3))/2 - (1 + sqrt(3))/2 ] = [ (1 - sqrt(3) - 1 - sqrt(3))/2 ] = [ -2 sqrt(3)/2 ] = -sqrt(3)

Therefore, numerator becomes 2AB + i*(B² - A² ) = 2*(-1/2) + i*(-sqrt(3)) = -1 - i sqrt(3)

Denominator is A² + B² = 2. Therefore, k = (-1 - i sqrt(3))/2

So k = - (1 + i sqrt(3))/2. Let me note that.

So k is a complex number. Let me write it in polar form. The modulus of k: |k| = sqrt( (-1/2)^2 + ( - sqrt(3)/2 )^2 ) = sqrt(1/4 + 3/4 ) = sqrt(1) = 1. The angle: arctangent( (-sqrt(3)/2 ) / (-1/2 )) = arctangent( sqrt(3) ). But since both real and imaginary parts are negative, it's in the third quadrant. The reference angle is π/3, so the angle is π + π/3 = 4π/3. Therefore, k = e^{i 4π/3}.

Alternatively, since k is a complex number of modulus 1 and angle 4π/3. Therefore, k^n = e^{i 4πn /3}.

But this suggests that φ(z_n) = k^n φ(z_0). Therefore, if we can compute φ(z_0), then we can compute φ(z_n) and then invert φ to get z_n.

So let me compute φ(z_0) where φ(z) = (z - α)/(z - β). Remember that α = A(1 + i) and β = B(1 + i), where A = (1 + sqrt(3))/2 and B = (1 - sqrt(3))/2.

Given z_0 = 1/137 + i.

So φ(z_0) = [ z_0 - α ] / [ z_0 - β ].

Let me compute numerator and denominator.

First, compute z_0 - α: (1/137 + i ) - A(1 + i )

Similarly z_0 - β: (1/137 + i ) - B(1 + i )

Let me compute numerator:

= (1/137 + i ) - A(1 + i ) = [ 1/137 - A ] + i [ 1 - A ]

Similarly denominator:

= [ 1/137 - B ] + i [ 1 - B ]

But A = (1 + sqrt(3))/2, B = (1 - sqrt(3))/2.

Let me compute 1/137 - A and 1 - A, same for B.

First, compute 1 - A: 1 - (1 + sqrt(3))/2 = (2 - 1 - sqrt(3))/2 = (1 - sqrt(3))/2 = B.

Similarly, 1 - B = 1 - (1 - sqrt(3))/2 = (2 - 1 + sqrt(3))/2 = (1 + sqrt(3))/2 = A.

Therefore, numerator becomes:

[ 1/137 - A ] + i * B

Denominator becomes:

[ 1/137 - B ] + i * A

So φ(z_0) = [ (1/137 - A) + i B ] / [ (1/137 - B) + i A ]

Now, let me compute 1/137 - A and 1/137 - B.

Compute 1/137 - A = 1/137 - (1 + sqrt(3))/2. Similarly for the other term.

But maybe I can write it as follows:

Let me denote that φ(z_0) = [ (C + i B) ] / [ (D + i A) ], where C = 1/137 - A and D = 1/137 - B.

But this seems messy. However, notice that since A and B are related, perhaps we can find a relation between numerator and denominator?

Alternatively, maybe there's a better way. Let me think. Since φ(z) is (z - α)/(z - β), and we have expressions involving 1/137 - A and similar.

Alternatively, maybe we can use the earlier expression for φ(z_n) and relate it to k^n φ(z_0). But unless we have more info, perhaps I need to compute φ(z_0). Alternatively, since z_0 is given, maybe I can find a relation?

Alternatively, perhaps there's a pattern in the iteration. Earlier, we computed z_0 = 1/137 + i, z_1 = 1 + 274i, z_2 = (37538 + i)/37265. Wait, this seems messy, but maybe if we continue computing terms, we can see a pattern?

Alternatively, perhaps I can use the formula for φ(z_n) = k^n φ(z_0). If I can compute φ(z_0), then maybe φ(z_n) is a complex number which can be inverted.

Alternatively, maybe there's a way to write φ(z_0) in terms of k? Let me think.

Wait, given that k is known, and φ(z_n) = k^n φ(z_0), then if we can write φ(z_0) as something involving k?

Alternatively, maybe there's a relation between z_0 and the fixed points?

Alternatively, perhaps we can find that φ(z_0) is equal to some power of k multiplied by something?

Alternatively, maybe I can compute φ(z_0) numerically?

Let me try that. Let me compute the numerical values.

First, compute A = (1 + sqrt(3))/2 ≈ (1 + 1.732)/2 ≈ 2.732 / 2 ≈ 1.366

Similarly, B = (1 - sqrt(3))/2 ≈ (1 - 1.732)/2 ≈ (-0.732)/2 ≈ -0.366

Then z_0 = 1/137 + i ≈ 0.0073 + i. Let me compute numerator and denominator:

Numerator: 1/137 - A ≈ 0.0073 - 1.366 ≈ -1.3587, and i * B ≈ i * (-0.366) ≈ -0.366i. Therefore numerator ≈ -1.3587 - 0.366i.

Denominator: 1/137 - B ≈ 0.0073 - (-0.366) ≈ 0.0073 + 0.366 ≈ 0.3733, and i * A ≈ i * 1.366 ≈ 1.366i. Therefore denominator ≈ 0.3733 + 1.366i.

So φ(z_0) ≈ (-1.3587 - 0.366i)/(0.3733 + 1.366i). Let me compute this division.

First, multiply numerator and denominator by the conjugate of the denominator: 0.3733 - 1.366i.

Numerator becomes (-1.3587 - 0.366i)(0.3733 - 1.366i)

Denominator becomes (0.3733)^2 + (1.366)^2 ≈ let me compute that.

First, approximate denominator:

0.3733² ≈ 0.139, 1.366² ≈ 1.866, so total ≈ 0.139 + 1.866 ≈ 2.005 ≈ 2. Let me check with exact values? Well, since earlier we had A² + B² = 2, but here the denominator is (0.3733)^2 + (1.366)^2. Wait, note that 0.3733 is approximately 1/137 - B? Wait no, denominator was [1/137 - B] + i*A, but when I calculated the denominator as approx 0.3733 + 1.366i, then the denominator squared magnitude is (1/137 - B)^2 + A^2. Wait, but earlier I had A² + B² = 2, but here it's different.

But regardless, let me proceed with approximate calculation.

So denominator ≈ 2. Then compute numerator:

Let me compute (-1.3587)(0.3733) + (-1.3587)(-1.366i) + (-0.366i)(0.3733) + (-0.366i)(-1.366i)

First compute real parts:

-1.3587 * 0.3733 ≈ -0.506

Imaginary parts from first term: + (-1.3587)(-1.366i) ≈ + (1.3587 * 1.366) i ≈ approximately, 1.3587 * 1.366 ≈ let me compute 1.35 * 1.36 ≈ 1.35 * 1.36 ≈ 1.836. So approx +1.836i

Then next term: -0.366i * 0.3733 ≈ -0.136i

Last term: -0.366i * -1.366i = + 0.366 * 1.366 * i² = approx 0.5 * (-1) = -0.5. Wait, exact calculation:

0.366 * 1.366 ≈ Let me calculate 0.366 * 1.366 ≈ approx 0.366 * 1 + 0.366 * 0.366 ≈ 0.366 + 0.133 ≈ 0.499. So +0.499 * (-1) = -0.499.

Therefore total real parts: -0.506 - 0.499 ≈ -1.005

Total imaginary parts: 1.836i - 0.136i ≈ 1.7i

Therefore numerator ≈ -1.005 + 1.7i. Divided by denominator ≈ 2, gives φ(z_0) ≈ (-1.005 + 1.7i)/2 ≈ -0.5025 + 0.85i.

So φ(z_0) ≈ -0.5 + 0.85i. Let me check if this is related to k?

Earlier, we had k = (-1 - i sqrt(3))/2 ≈ (-1 - i * 1.732)/2 ≈ (-1 - i1.732)/2 ≈ -0.5 - i0.866. Which is similar to the approximate value of φ(z_0) here: -0.5 + 0.85i. Close, except the imaginary part is positive instead of negative. Wait, but maybe my approximations are off?

Alternatively, maybe φ(z_0) is k^{-1} or something?

Wait, let me think. If φ(z_n) = k^n φ(z_0), then if we can find φ(z_0) = c, then z_n is given by φ^{-1}(k^n c). However, without knowing exactly what φ(z_0) is, it's hard to proceed. However, given the approximate value of φ(z_0) ≈ -0.5 + 0.85i and k ≈ -0.5 - i0.866, which is e^{i 4π/3} as mentioned before. So k is a complex number with angle 4π/3, modulus 1.

Let me note that k is e^{i 4π/3}, so k^n is e^{i 4πn /3}.

But φ(z_0) is approximately some value, but given the exact expressions, maybe there's a relation?

Alternatively, maybe φ(z_0) is equal to k^{-1}? Let me check:

If φ(z_0) = k^{-1}, then φ(z_n) = k^n * k^{-1} = k^{n - 1}, but this is just a guess. Alternatively, maybe φ(z_0) = something else.

Alternatively, maybe I can use the exact expressions. Recall that φ(z_0) = [ (1/137 - A ) + i B ] / [ (1/137 - B ) + i A ].

But since 1 - A = B, then 1/137 - A = 1/137 - A = 1/137 - (1 - B) = 1/137 - 1 + B = B - (1 - 1/137 ) = B - 136/137. Similarly, but maybe this isn't helpful.

Wait, let me compute 1/137 - A. A = (1 + sqrt(3))/2, so 1/137 - A = 1/137 - 1/2 - sqrt(3)/2 = (2 - 137)/274 - sqrt(3)/2 = ( - 135 / 274 ) - sqrt(3)/2. Wait, no:

Wait, 1/137 - (1 + sqrt(3))/2 = 1/137 - 1/2 - sqrt(3)/2 = (1/137 - 1/2) - sqrt(3)/2. Let me compute 1/137 - 1/2 = (2 - 137)/274 = (-135)/274. So, 1/137 - A = -135/274 - sqrt(3)/2. Similarly for the other term: 1/137 - B = 1/137 - (1 - sqrt(3))/2 = 1/137 - 1/2 + sqrt(3)/2 = (2 - 137)/274 + sqrt(3)/2 = -135/274 + sqrt(3)/2.

So the numerator is [ -135/274 - sqrt(3)/2 ] + i * B. Wait, B = (1 - sqrt(3))/2.

So numerator is:

Real part: -135/274 - sqrt(3)/2

Imaginary part: i * (1 - sqrt(3))/2

Similarly denominator is [ -135/274 + sqrt(3)/2 ] + i * A, where A = (1 + sqrt(3))/2.

This seems complicated, but maybe we can write this as:

Numerator: C + i D

Denominator: E + i F

Where C = -135/274 - sqrt(3)/2, D = (1 - sqrt(3))/2, E = -135/274 + sqrt(3)/2, F = (1 + sqrt(3))/2.

Then φ(z_0) = (C + i D)/(E + i F). Let me multiply numerator and denominator by the conjugate of the denominator (E - i F):

Numerator becomes (C + i D)(E - i F)

Denominator becomes |E + i F|².

Let me compute denominator first. |E + i F|² = E² + F².

E = -135/274 + sqrt(3)/2, F = (1 + sqrt(3))/2.

So E² + F² = ?

Let me compute:

Let me note that sqrt(3)/2 is approximately 0.866, and -135/274 is approximately -0.493. But exact calculation is better.

Alternatively, let me note that F = A, and earlier we had A² + B² = 2, but E is related to 1/137 - B?

Wait, perhaps not. Let me proceed.

Alternatively, maybe there's a way to relate numerator and denominator through previous expressions?

Wait, given that φ(z_0) = [ z_0 - α ] / [ z_0 - β ] = something. But perhaps we can relate this to k?

Recall that earlier, we found that k = (B - A i ) / (A - B i ) = (-1 - i sqrt(3))/2. Let me check:

Wait, we had that k = [ 2AB + i (B² - A² ) ] / (A² + B² ). We computed numerator as -1 - i sqrt(3) and denominator as 2, giving k = (-1 - i sqrt(3))/2.

But given that φ(z_0) is some complex number, and if we could find its relation to k, it would help. However, without more information, it's challenging.

Alternatively, maybe we can consider that the iteration leads to periodic behavior. Given that k is a complex number on the unit circle with angle 4π/3, which is a rotation by 4π/3 radians. Since the angle is 4π/3, the order of k is 3, since 4π/3 * 3 = 4π, which is equivalent to 0 modulo 2π. Wait, 4π/3 * 3 = 4π, but 4π is 2π*2, so the order is 3? Wait, no. The multiplicative order of k is the smallest integer m such that k^m = 1. Since k = e^{i4π/3}, then k^3 = e^{i4π} = e^{i0} = 1. Hence, the order is 3. Therefore, k^3 = 1.

Therefore, φ(z_n) = k^n φ(z_0). So, since k^3 = 1, then φ(z_{n}) = φ(z_0) * k^{n} = φ(z_0) * k^{n mod 3}.

Therefore, φ(z_n) has period 3. Therefore, if we can find φ(z_0), then φ(z_n) cycles every 3 steps. Therefore, if we compute φ(z_0), then φ(z_{2002}) = φ(z_0) * k^{2002} = φ(z_0) * k^{2002 mod 3}.

Therefore, compute 2002 divided by 3. Let me compute 2002 / 3:

3*667 = 2001, so 2002 = 3*667 + 1, so remainder 1. Therefore, 2002 ≡ 1 mod 3. Therefore, φ(z_{2002}) = φ(z_0) * k^1 = φ(z_0) * k.

Therefore, if I can compute φ(z_0) * k, then invert φ to get z_{2002}.

But how?

Alternatively, if I can find that φ(z_0) * k is equal to something that can be inverted?

Alternatively, since φ(z_0) = (z_0 - α)/(z_0 - β), and k = (B - A i ) / (A - B i ). Wait, earlier we had φ(z_0) = [ (1/137 - A ) + i B ] / [ (1/137 - B ) + i A ].

Wait, perhaps instead of trying to compute it directly, there's a relation between φ(z_0) and some term?

Alternatively, since we know that k^3 = 1, and we need to find φ(z_0) * k. But without knowing φ(z_0), this seems difficult.

Wait, let me think again. If k^3 = 1, then the sequence φ(z_n) cycles every 3 terms. So z_0, z_1, z_2, z_3 = z_0, etc. But wait, if φ(z_n) cycles every 3, then z_n cycles every 3 steps?

Wait, no. Because φ(z_n) = k^n φ(z_0). If φ(z_n) cycles every 3, then φ(z_{n + 3}) = φ(z_n). Therefore, z_{n + 3} = φ^{-1}(φ(z_n)) = z_n. Therefore, the sequence z_n is periodic with period 3. But is that true?

Wait, if φ(z_n) = k^n φ(z_0), and k^3 = 1, then φ(z_{n + 3}) = k^{n + 3} φ(z_0) = k^n * k^3 φ(z_0) = k^n * 1 * φ(z_0) = φ(z_n). Therefore, φ(z_{n + 3}) = φ(z_n). Therefore, if φ is invertible (which it is, since it's a Möbius transformation not equal to identity), then z_{n + 3} = z_n. Therefore, the sequence is periodic with period 3. Therefore, z_{n} = z_{n mod 3}.

Wait, but if that's the case, then z_{2002} = z_{2002 mod 3}. Let me compute 2002 mod 3. As above, 2001 is divisible by 3, so 2002 = 3*667 + 1, so 2002 ≡ 1 mod 3. Therefore, z_{2002} = z_1. But earlier we computed z_1 = 1 + 274i. Is this the case? Wait, but earlier when I computed z_1 and z_2, they didn't immediately cycle back to z_0. Let me check.

Wait, if the sequence is periodic with period 3, then z_0, z_1, z_2, z_0, z_1, z_2,... So z_3 should be z_0. Let me compute z_3 using the earlier terms.

We had z_0, z_1 = 1 + 274i, z_2 = (37538 + i)/37265. Let me compute z_3 = F(z_2).

But this would be complicated, but if the period is 3, then z_3 should equal z_0. Let me check with approximate values.

From earlier, z_1 ≈ 1 + 274i, which is a large complex number, z_2 ≈ (37538 + i)/37265 ≈ (approx) 37538/37265 ≈ 1.007 (real part) and i/37265 ≈ 0.000027i. So z_2 ≈ 1.007 + 0.000027i. Then z_3 = F(z_2) = (z_2 + i)/(z_2 - i). Let me approximate this.

Numerator: z_2 + i ≈ 1.007 + 0.000027i + i ≈ 1.007 + 1.000027i

Denominator: z_2 - i ≈ 1.007 + 0.000027i - i ≈ 1.007 - 0.999973i

Divide numerator by denominator. Since denominator is close to 1.007 - 1i, and numerator is approximately 1.007 + 1i. So z_3 ≈ (1.007 + 1i)/(1.007 - 1i). Multiply numerator and denominator by (1.007 + 1i):

Numerator becomes (1.007 + 1i)^2 ≈ (1.007)^2 - (1)^2 + 2*1.007*1i ≈ 1.014 + 2.014i

Denominator becomes (1.007)^2 + (1)^2 ≈ 1.014 + 1 = 2.014

Therefore z_3 ≈ (1.014 + 2.014i)/2.014 ≈ 0.5 + i. Which is approximately 0.5 + i, which is not z_0 = 1/137 + i ≈ 0.0073 + i. So z_3 ≈ 0.5 + i ≠ z_0. Therefore, contradiction. Hence, my assumption that the period is 3 is wrong?

Wait, what's wrong here? Earlier, we said that φ(z_n) = k^n φ(z_0), and since k^3 = 1, φ(z_{n + 3}) = φ(z_n). Hence, z_{n + 3} = φ^{-1}(φ(z_n)) = z_n. But in reality, according to approximate calculation, z_3 is not equal to z_0. So there must be a mistake in my reasoning.

Wait, perhaps the assumption that φ(z_n) = k^n φ(z_0) is incorrect? Or maybe my calculation for φ(z_0) is incorrect?

Wait, let me recall that φ(z_n) = k^n φ(z_0). So if I have φ(z_3) = k^3 φ(z_0) = 1 * φ(z_0) = φ(z_0). Therefore, φ(z_3) = φ(z_0) => z_3 = z_0, but our approximate calculation suggests otherwise. So this implies that either my calculation of z_3 is wrong, or my reasoning about φ is wrong.

Wait, let me recheck the calculation of φ(z_n). Earlier, I had φ(F(z)) = k φ(z). So this should mean that φ(z_1) = k φ(z_0). Then φ(z_2) = k φ(z_1) = k^2 φ(z_0), and φ(z_3) = k^3 φ(z_0) = φ(z_0). Hence z_3 = φ^{-1}(φ(z_0)) = z_0. Therefore, if my approximate calculation shows that z_3 is not equal to z_0, there must be an error in either the calculation of z_3 or the logic.

Let me recalculate z_3 more carefully. Given z_2 is approximately 1.007 + 0.000027i. Let me write z_2 = x + yi, where x ≈ 1.007, y ≈ 0.000027.

Then F(z_2) = (z_2 + i)/(z_2 - i) = (x + (y + 1)i)/(x - (1 - y)i). Let me compute this.

Let me compute numerator: x + (y + 1)i

Denominator: x - (1 - y)i

Let me multiply numerator and denominator by the conjugate of the denominator: x + (1 - y)i.

Wait, no, the conjugate of the denominator x - (1 - y)i is x + (1 - y)i.

So numerator becomes [x + (y + 1)i][x + (1 - y)i]

Denominator becomes x^2 + (1 - y)^2

Let me expand numerator:

= x*x + x*(1 - y)i + (y + 1)i*x + (y + 1)i*(1 - y)i

= x² + x(1 - y)i + x(y + 1)i + (y + 1)(1 - y)i²

Simplify terms:

First term: x²

Second and third terms: [x(1 - y) + x(y + 1)]i = x[(1 - y) + (y + 1)]i = x[2]i = 2x i

Fourth term: (y + 1)(1 - y) * i² = (1 - y²)(-1) = -1 + y²

So overall numerator is x² - 1 + y² + 2x i

Therefore, z_3 = [x² - 1 + y² + 2x i] / [x² + (1 - y)^2]

Now, with x ≈ 1.007, y ≈ 0.000027:

Compute x² ≈ (1.007)^2 ≈ 1.014

Compute y² ≈ (0.000027)^2 ≈ 7e-10, negligible.

Compute numerator ≈ 1.014 - 1 + 0 + 2*1.007*i ≈ 0.014 + 2.014i

Denominator ≈ x² + (1 - y)^2 ≈ 1.014 + 1^2 ≈ 2.014

Thus, z_3 ≈ (0.014 + 2.014i)/2.014 ≈ 0.007 + i.

Which is approximately z_0 = 1/137 + i ≈ 0.0073 + i. This matches!

Ah, I see! Earlier when I approximated, I thought z_3 was ≈ 0.5 + i, but actually, the real part is ≈ 0.014 / 2.014 ≈ 0.007, which is approximately z_0. Therefore, the sequence is indeed periodic with period 3. My previous approximation was wrong because I miscalculated the numerator. So z_3 ≈ z_0, which confirms the periodicity. Great!

Therefore, the sequence is periodic with period 3. Hence, z_{n+3} = z_n. Therefore, to find z_{2002}, we can find 2002 modulo 3.

As before, 2002 divided by 3: 3*667 = 2001, so 2002 = 3*667 + 1, hence 2002 ≡ 1 mod 3. Therefore, z_{2002} = z_1.

Earlier, we computed z_1 = 1 + 274i. However, wait, earlier when I computed z_1, I had z_0 = 1/137 + i. Let me confirm:

Yes, z_1 = (z_0 + i)/(z_0 - i). Given z_0 = 1/137 + i, then z_0 + i = 1/137 + 2i, and z_0 - i = 1/137. So z_1 = (1/137 + 2i) / (1/137) = 1 + 2i*137 = 1 + 274i. Yes, correct.

But according to the periodicity, z_3 = z_0, z_4 = z_1, etc. Hence, z_{3k + 1} = z_1. Therefore, since 2002 = 3*667 + 1, z_{2002} = z_1 = 1 + 274i.

But the problem says "find z_{2002}." So is this the answer? Wait, but let me check if the period is indeed 3. Since z_3 = z_0, z_4 = z_1, z_5 = z_2, z_6 = z_0, etc. Yes, so the period is 3. Hence, z_{2002} = z_1. However, let me check with n = 0,1,2,3,...

z_0 = 1/137 + i

z_1 = 1 + 274i

z_2 = (37538 + i)/37265 ≈ 1.007 + 0.000027i

z_3 = z_0

So yes, the period is 3. Therefore, z_{2002} = z_{1} = 1 + 274i. But let me check with n=3k+1, so for example n=1: z_1, n=4: z_4 = z_1, yes. So 2002 = 3*667 + 1, so z_{2002} = z_1 = 1 + 274i.

Wait, but the problem says z_0 = 1/137 + i, and asks for z_{2002}. So is it possible that the answer is 1 + 274i? But that seems too straightforward. However, given the periodicity, it should be correct.

Wait, but let me double-check with another term. Let me compute z_3 = F(z_2). Let me try to compute it exactly. Earlier, we had z_2 = (37538 + i)/37265. Let me compute z_3 = F(z_2) = (z_2 + i)/(z_2 - i). Let me compute this exactly.

Let me denote z_2 = (37538 + i)/37265.

Then z_2 + i = (37538 + i)/37265 + i = (37538 + i + 37265i)/37265 = [37538 + 37266i]/37265.

Similarly, z_2 - i = (37538 + i)/37265 - i = (37538 + i - 37265i)/37265 = [37538 - 37264i]/37265.

Therefore, z_3 = [37538 + 37266i]/[37538 - 37264i]. Let me simplify this fraction.

Let me note that 37538 and 37264 might have common factors. Let me check GCD(37538, 37264). Compute 37538 - 37264 = 274. Then GCD(37264, 274). Let me divide 37264 by 274.

But this might take time. Alternatively, perhaps there is a pattern here.

Alternatively, let me multiply numerator and denominator by the conjugate of the denominator, which is 37538 + 37264i.

So numerator becomes (37538 + 37266i)(37538 + 37264i)

Denominator becomes (37538)^2 + (37264)^2.

This seems very complicated. However, if the periodicity is correct, z_3 should equal z_0 = 1/137 + i. Let me check if this is the case.

Let me assume z_3 = z_0. Then:

[37538 + 37266i]/[37538 - 37264i] = 1/137 + i

Let me cross multiply:

(37538 + 37266i) = (1/137 + i)(37538 - 37264i)

Compute RHS:

= (1/137)*37538 - (1/137)*37264i + i*37538 - i^2 * 37264

= (37538 / 137) - (37264 / 137)i + 37538i + 37264

First compute 37538 / 137:

Since z_0 = 1/137 + i, and earlier we had z_1 = (z_0 + i)/(z_0 - i) = 1 + 274i. Let me check 37538 / 137.

Wait, earlier when computing z_2, the numerator was 75076 + 2i over 74530, which reduced to 37538 + i over 37265. So 37538 = 75076 / 2, and 37265 = 74530 / 2. Now, 75076 / 137: let me compute 137*547 = ? Wait, maybe there's a connection here. Let me note that z_0 = 1/137 + i, so maybe there's a relation between 37538 and 137?

Wait, 137 * 274 = let me compute 137*200 = 27400, 137*74 = let's see, 137*70 = 9590, 137*4 = 548, total 9590 + 548 = 10138. Therefore total 27400 + 10138 = 37538. Yes! So 137 * 274 = 37538. Similarly, 37264: let me check if 37264 is related. Let me compute 137*272. 137*272: Let me compute 137*270 = 137*200 + 137*70 = 27400 + 9590 = 36990, then 137*2 = 274, so total 36990 + 274 = 37264. Yes! Therefore, 37264 = 137*272.

Therefore, RHS:

= (137 * 274)/137 - (137*272)/137 * i + 37538i + 37264

Simplify:

= 274 - 272i + 37538i + 37264

Combine real parts: 274 + 37264 = 37538

Combine imaginary parts: -272i + 37538i = (37538 - 272)i = 37266i

Therefore RHS = 37538 + 37266i, which matches the numerator. Hence, z_3 = z_0. Hence, verified!

Therefore, the sequence is indeed periodic with period 3. Hence, z_{2002} = z_{3*667 + 1} = z_1 = 1 + 274i. Therefore, the answer is 1 + 274i.

But let me confirm once again with the initial terms:

z_0 = 1/137 + i

z_1 = 1 + 274i

z_2 = (37538 + i)/37265

z_3 = z_0

z_4 = z_1, etc.

Hence, n mod 3 = 0: z_0, 1: z_1, 2: z_2. Since 2002 mod 3 is 1, so z_{2002} = z_1 = 1 + 274i.

Therefore, the final answer is \boxed{1 + 274i}.

**Final Answer**
\boxed{1 + 274i}

To solve the given problem, we consider the recursive sequence defined by the Möbius transformation:

$$
z_n = F(z_{n-1}) = \frac{z_{n-1} + i}{z_{n-1} - i}
$$
with initial value $ z_0 = \frac{1}{137} + i $.

---

### **Step 1: Understanding the Function**

The mapping $ F(z) = \frac{z + i}{z - i} $ is a Möbius transformation. Such transformations are invertible and often exhibit periodic or repeating behavior under iteration, especially when analyzed via fixed points or conjugation methods.

We begin by noting that this sequence is generated by a Möbius transformation with two fixed points. These are the solutions to the fixed point equation:
$$
F(z) = z \quad \Rightarrow \quad \frac{z + i}{z - i} = z
$$
Multiply both sides by $ z - i $ and simplify:
$$
z + i = z(z - i) \Rightarrow z + i = z^2 - iz
\Rightarrow z^2 - iz - z - i = 0
\Rightarrow z^2 - z(1+i) - i = 0
$$
Solving this quadratic yields the fixed points:
$$
z = \frac{(1 + i) \pm \sqrt{(1 + i)^2 + 4i}}{2}
$$
Computing the discriminant:
$$
(1 + i)^2 + 4i = 1 + 2i - 1 + 4i = 6i
\Rightarrow \sqrt{6i} = \sqrt{6} \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = \sqrt{3} (1 + i)
$$
Thus, the fixed points are:
$$
z = \frac{(1 + i) \pm \sqrt{3}(1 + i)}{2} = \left( \frac{1 \pm \sqrt{3}}{2} \right)(1 + i)
$$
Let these be $ \alpha = \frac{(1+\sqrt{3})}{2}(1 + i) $, $ \beta = \frac{(1 - \sqrt{3})}{2}(1 + i) $. These fixed points are complex conjugates in the form $ a(1 + i) $.

---

### **Step 2: Conjugating the Transformation**

Let $ \phi(z) = \frac{z - \alpha}{z - \beta} $, a Möbius transformation mapping $ \alpha \mapsto 0 $, $ \beta \mapsto \infty $. This transformation conjugates the function $ F $ to a scaling transformation:
$$
\phi(F(z)) = k \phi(z), \quad \text{where } |k| = 1
$$

It can be shown that the transformation $ k = \frac{(1 - \alpha)}{(1 - \beta)} $, and the value of $ k $ ends up being:
$$
k = \frac{-1 - i \sqrt{3}}{2}
$$
This complex number lies on the unit circle and satisfies $ k^3 = 1 $, meaning it has **order 3**.

---

### **Step 3: Applying the Conjugation**

Under this mapping, the iteration satisfies:
$$
\phi(z_n) = k^n \phi(z_0)
$$
Since $ k^3 = 1 $, the sequence $ \phi(z_n) $ is periodic with period 3:
$$
\phi(z_{n+3}) = \phi(z_n) \Rightarrow z_{n+3} = z_n
$$
That is, the original sequence $ \{ z_n \} $ is **periodic with period 3**.

---

### **Step 4: Computing $ z_{2002} $**

We compute:
$$
2002 \mod 3 = 1 \Rightarrow z_{2002} = z_1
$$

We already computed:
$$
z_0 = \frac{1}{137} + i, \quad z_1 = \frac{z_0 + i}{z_0 - i}
$$
Plug in $ z_0 = \frac{1}{137} + i $:
- Numerator: $ z_0 + i = \frac{1}{137} + i + i = \frac{1}{137} + 2i $
- Denominator: $ z_0 - i = \frac{1}{137} + i - i = \frac{1}{137} $

$$
z_1 = \frac{\frac{1}{137} + 2i}{\frac{1}{137}} = 1 + 2(137)i = 1 + 274i
$$

---

### **Final Answer**

$$
\boxed{1 + 274i}
$$

[Assistant]
Explanation:
Okay, so I need to find z_{2002} given that z_0 = 1/137 + i, and each subsequent term is defined by z_n = F(z_{n-1}) where F(z) = (z + i)/(z - i). Hmm, functional iteration. Let me think about how to approach this.

First, maybe I should try to understand what the function F does. Let me write it down again: F(z) = (z + i)/(z - i). This looks like a Möbius transformation, which is a function of the form f(z) = (az + b)/(cz + d). These transformations can sometimes be analyzed by looking at their fixed points or by conjugating them to simpler transformations, like rotations or translations. Maybe that's the way to go here?

Alternatively, since we're dealing with iterations, perhaps there's a pattern or periodicity in the sequence z_n. If I can find a pattern or a cycle, then I can determine z_{2002} based on the cycle length.

Let me start by computing the first few terms manually to see if I can spot a pattern. Given z_0 = 1/137 + i.

Compute z_1 = F(z_0) = (z_0 + i)/(z_0 - i). Let me compute that.

First, z_0 = 1/137 + i. Let me write that as a complex number: let me denote real part as 1/137 and imaginary part as 1.

So z_0 = (1/137) + i*1. Then z_0 + i is (1/137) + i*(1 + 1) = 1/137 + 2i.

z_0 - i is (1/137) + i*(1 - 1) = 1/137 + 0i = 1/137.

Therefore z_1 = (1/137 + 2i) / (1/137) = [ (1 + 2i*137)/137 ] / (1/137) ) = 1 + 2i*137. Wait, let me check that again.

Wait, numerator is (1/137 + 2i), denominator is 1/137. So dividing both terms by 1/137:

( (1/137)/(1/137) ) + (2i)/(1/137) ) = 1 + 2i * 137. Yes, that's right. So z_1 = 1 + 2*137 i. Let me compute 2*137: 2*100=200, 2*37=74, so total 274. So z_1 = 1 + 274i.

Okay, so z_1 is 1 + 274i. Let me compute z_2 = F(z_1) = (z_1 + i)/(z_1 - i). Let me compute numerator and denominator.

z_1 + i = (1 + 274i) + i = 1 + 275i.

z_1 - i = (1 + 274i) - i = 1 + 273i.

So z_2 = (1 + 275i)/(1 + 273i). Hmm, to simplify this, maybe multiply numerator and denominator by the conjugate of the denominator?

So multiply numerator and denominator by (1 - 273i):

Numerator becomes (1 + 275i)(1 - 273i)

Denominator becomes (1 + 273i)(1 - 273i) = 1^2 + 273^2 = 1 + 273².

Let me compute numerator:

First expand (1)(1) + (1)(-273i) + (275i)(1) + (275i)(-273i)

= 1 - 273i + 275i - 275*273 i²

Simplify:

The real part: 1 + 275*273 (since i² = -1)

Imaginary part: (-273i + 275i) = 2i

So numerator is [1 + 275*273] + 2i.

Let me compute 275 * 273. Hmm, 275*273. Let me compute that:

First, note that 275 = 273 + 2, so 275*273 = 273² + 2*273.

Compute 273²: 273^2. Let me recall that 270² = 72900, 270² + 2*270*3 + 3² = 72900 + 1620 + 9 = 74529. Wait, no, 273 = 270 + 3, so (a + b)^2 = a² + 2ab + b².

a = 270, b = 3: 270² + 2*270*3 + 3² = 72900 + 1620 + 9 = 74529. Then 2*273 = 546. Therefore total numerator real part is 1 + 74529 + 546 = 1 + 74529 is 74530, plus 546 is 75076. Therefore numerator is 75076 + 2i.

Denominator is 1 + 273² = 1 + 74529 = 74530. So z_2 is (75076 + 2i)/74530.

Let me simplify that fraction. Let me check if 75076 and 74530 have a common factor. Let me compute GCD(75076, 74530). Let me compute the difference: 75076 - 74530 = 546. Then GCD(74530, 546). Let me divide 74530 by 546.

First, 546 * 136 = 546*(100 + 30 + 6) = 54600 + 16380 + 3276 = 54600 + 16380 is 70980, plus 3276 is 74256. Then 74530 - 74256 = 274. So GCD(546, 274). Then compute GCD(546,274). 546 ÷ 274 = 2 with a remainder of 546 - 548 = wait, 274*2 = 548, which is larger than 546. So actually 546 ÷ 274 = 1 with remainder 546 - 274 = 272. Then GCD(274, 272). Then 274 - 272 = 2. GCD(272,2). Which is 2. Therefore GCD is 2. Therefore, numerator and denominator can be divided by 2.

So numerator: 75076 ÷ 2 = 37538, 2i ÷ 2 = i. Denominator: 74530 ÷ 2 = 37265. Therefore z_2 = (37538 + i)/37265. Wait, but 37538 divided by 37265? Wait, actually, z_2 = [ (75076 + 2i)/74530 ] simplifies to [ (37538 + i)/37265 ]? Wait, no. Wait, the numerator was 75076 + 2i, denominator 74530. Dividing numerator and denominator by 2 gives (37538 + i)/37265. So z_2 = (37538 + i)/37265. Let me check if that reduces further? Probably not, since GCD of numerator terms? Let me check if 37538 and 37265 have a common factor. Let me compute difference: 37538 - 37265 = 273. Then GCD(37265, 273). Let's see, 273 divides into 37265 how many times?

Divide 37265 by 273. Let me compute 273*100 = 27300, subtract that from 37265: 37265 - 27300 = 9965. Then 273*30 = 8190, subtract from 9965: 9965 - 8190 = 1775. Then 273*6 = 1638, subtract: 1775 - 1638 = 137. Then GCD(273,137). 137 divides into 273 twice, remainder 273 - 274 = -1? Wait 273 ÷ 137 = 1 with remainder 273 - 137 = 136. Then GCD(137,136) = 1. Therefore, the fraction cannot be reduced further. Therefore z_2 = (37538 + i)/37265.

Hmm, this seems complicated. Maybe I need another approach. Computing terms manually is getting messy. Maybe there's a pattern or a cycle here? Let me think.

Since F is a Möbius transformation, perhaps it has a fixed point or can be conjugated to a simpler transformation. Let me recall that Möbius transformations can be represented by matrices, and their iterations can be analyzed by diagonalizing the matrix or finding its eigenvalues. Alternatively, maybe I can find a function that conjugates F to a multiplication by a scalar, which would make iteration straightforward.

Alternatively, perhaps we can use complex analysis. Let me try to find fixed points of F. Fixed points satisfy F(z) = z, so:

(z + i)/(z - i) = z. Multiply both sides by (z - i):

z + i = z(z - i)

Expand RHS: z² - iz

Bring all terms to one side:

z² - iz - z - i = 0 => z² - z(1 + i) - i = 0

Solve quadratic equation: z = [ (1 + i) ± sqrt( (1 + i)^2 + 4i ) ] / 2

Let me compute discriminant D = (1 + i)^2 + 4i

First compute (1 + i)^2 = 1 + 2i + i² = 1 + 2i -1 = 2i

Then D = 2i + 4i = 6i

Therefore z = [ (1 + i) ± sqrt(6i) ] / 2

Hmm, sqrt(6i). Let me compute sqrt(6i). Let me write 6i as 0 + 6i. The square roots of a complex number can be found by converting to polar form. Let me recall that sqrt(a + bi) can be found by solving (x + yi)^2 = a + bi.

But maybe it's easier to note that sqrt(i) is e^{iπ/4} and e^{i5π/4}, so sqrt(6i) would be sqrt(6) * e^{iπ/4} and sqrt(6) * e^{i5π/4}.

Therefore sqrt(6i) = sqrt(6)(cos(π/4) + i sin(π/4)) = sqrt(6)*(√2/2 + i√2/2) = sqrt(3) + i sqrt(3). Similarly the other root is -sqrt(3) - i sqrt(3).

Wait, let me check:

Let me square sqrt(6i). Let me take sqrt(6) e^{iπ/4}. Squaring that gives 6 e^{iπ/2} = 6i. Yes, correct. Similarly, the other square root is sqrt(6) e^{i5π/4} which squares to 6 e^{i5π/2} = 6 e^{iπ/2} = 6i. So yes, sqrt(6i) is sqrt(6)(cos π/4 + i sin π/4) and sqrt(6)(cos 5π/4 + i sin 5π/4). So sqrt(6i) = sqrt(6)( (sqrt(2)/2 + i sqrt(2)/2 )) = sqrt(3) + i sqrt(3) and -sqrt(3) - i sqrt(3).

Therefore, the solutions for z are:

[ (1 + i) ± (sqrt(3) + i sqrt(3)) ] / 2.

Let me compute both possibilities:

First, with the plus sign:

Numerator: (1 + i) + sqrt(3) + i sqrt(3) = (1 + sqrt(3)) + i(1 + sqrt(3))

Therefore z = [ (1 + sqrt(3)) + i(1 + sqrt(3)) ] / 2 = [ (1 + sqrt(3))/2 ] (1 + i )

Similarly, with the minus sign:

Numerator: (1 + i) - sqrt(3) - i sqrt(3) = (1 - sqrt(3)) + i(1 - sqrt(3)) = [1 - sqrt(3)](1 + i)

Therefore z = [1 - sqrt(3)](1 + i)/2.

So the fixed points are z = [ (1 + sqrt(3))/2 ](1 + i) and z = [ (1 - sqrt(3))/2 ](1 + i). Let me note these as α and β, maybe?

Alternatively, perhaps I can use these fixed points to conjugate the transformation F into a simpler form. Let me recall that if we have a Möbius transformation with two fixed points, we can conjugate it to a scaling transformation. Let me try that.

Let me denote the fixed points as α and β. Let me try to find a Möbius transformation that maps α to 0 and β to ∞, which would transform F into a scaling transformation. Let me recall that if we have a Möbius transformation φ such that φ(α) = 0 and φ(β) = ∞, then φ(F(z)) = k φ(z), for some constant k. Then the iteration becomes φ(z_n) = k^n φ(z_0). Therefore, z_n can be obtained by inverting φ.

So let me try to construct φ. Let me take α and β as fixed points. Let me compute them numerically to get a sense? Wait, maybe not. Let me proceed symbolically.

Let me denote φ(z) = (z - α)/(z - β). Then, since φ(α) = 0 and φ(β) = ∞, then φ(F(z)) = k φ(z). Let me verify this.

Alternatively, maybe it's easier to use the formula for conjugating Möbius transformations. Let me recall that if F has fixed points α and β, then the transformation can be written as F(z) = ( (1 - k)z + k α + (k - 1) β ) / ( (k - 1)z + k β - (k - 1) α ) )? Not sure. Maybe I need a better approach.

Alternatively, let me suppose that φ(F(z)) = k φ(z). Let me write φ(z) = (z - α)/(z - β). Then φ(F(z)) = [F(z) - α]/[F(z) - β]. Let me compute that:

F(z) = (z + i)/(z - i). Therefore,

φ(F(z)) = [ ( (z + i)/(z - i) ) - α ] / [ ( (z + i)/(z - i) ) - β ]

Multiply numerator and denominator by (z - i):

Numerator: (z + i) - α(z - i) = z + i - α z + α i = z(1 - α) + i(1 + α)

Denominator: (z + i) - β(z - i) = z + i - β z + β i = z(1 - β) + i(1 + β)

Therefore φ(F(z)) = [ z(1 - α) + i(1 + α) ] / [ z(1 - β) + i(1 + β) ]

If we want this to be equal to k φ(z) = k*(z - α)/(z - β), then we need:

[ z(1 - α) + i(1 + α) ] / [ z(1 - β) + i(1 + β) ] = k*(z - α)/(z - β)

Cross multiplying:

[ z(1 - α) + i(1 + α) ]*(z - β) = k*(z - α)*[ z(1 - β) + i(1 + β) ]

This should hold for all z (except the poles), so coefficients of z², z, and the constants must match. Let me expand both sides.

Left-hand side (LHS):

Multiply [ z(1 - α) + i(1 + α) ] by (z - β):

= z(1 - α)(z - β) + i(1 + α)(z - β)

= (1 - α)z² - (1 - α)β z + i(1 + α) z - i(1 + α)β

= (1 - α)z² + [ - (1 - α)β + i(1 + α) ] z - i(1 + α)β

Right-hand side (RHS):

k*(z - α)*[ z(1 - β) + i(1 + β) ]

First compute [ z(1 - β) + i(1 + β) ] = (1 - β) z + i(1 + β)

Multiply by (z - α):

= (1 - β) z(z - α) + i(1 + β)(z - α)

= (1 - β) z² - (1 - β)α z + i(1 + β) z - i(1 + β)α

= (1 - β) z² + [ - (1 - β)α + i(1 + β) ] z - i(1 + β)α

Therefore RHS is k times that:

k*(1 - β) z² + k*[ - (1 - β)α + i(1 + β) ] z - k*i(1 + β)α

Therefore equating LHS and RHS:

(1 - α)z² + [ - (1 - α)β + i(1 + α) ] z - i(1 + α)β = k*(1 - β) z² + k*[ - (1 - β)α + i(1 + β) ] z - k*i(1 + β)α

Since this holds for all z, coefficients of z², z, and constants must be equal. Therefore:

1. Coefficient of z²: (1 - α) = k*(1 - β). Hence k = (1 - α)/(1 - β)

2. Coefficient of z: [ - (1 - α)β + i(1 + α) ] = k*[ - (1 - β)α + i(1 + β) ]

3. Constant term: -i(1 + α)β = -k i(1 + β) α

Let me check the constant term equation first:

- i(1 + α)β = -k i(1 + β) α => (1 + α)β = k (1 + β) α => k = [ (1 + α)β ] / [ (1 + β) α ]

From coefficient of z², we had k = (1 - α)/(1 - β). Therefore equating these two expressions for k:

(1 - α)/(1 - β) = [ (1 + α)β ] / [ (1 + β) α ]

Cross multiplying:

(1 - α)(1 + β) α = (1 + α)β (1 - β)

Let me expand both sides:

Left: α(1 - α)(1 + β) = α[ (1)(1 + β) - α(1 + β) ] = α(1 + β - α - α β)

Right: β(1 + α)(1 - β) = β[ (1)(1 - β) + α(1 - β) ] = β(1 - β + α - α β)

So left side: α(1 + β - α - α β)

Right side: β(1 - β + α - α β)

Let me expand both:

Left: α + α β - α² - α² β

Right: β - β² + α β - α β²

Set left equal to right:

α + α β - α² - α² β = β - β² + α β - α β²

Simplify both sides by subtracting right from left:

α + α β - α² - α² β - β + β² - α β + α β² = 0

Simplify term by term:

α - β + α β - α β + (- α²) + β² + (- α² β) + α β² = 0

Simplify:

α - β - α² + β² - α² β + α β² = 0

Factor terms:

(α - β) + (-α² + β²) + (- α² β + α β²) = 0

Note that -α² + β² = -(α² - β²) = -(α - β)(α + β)

Similarly, - α² β + α β² = α β ( - α + β ) = α β ( β - α ) = - α β ( α - β )

Therefore:

(α - β) - (α - β)(α + β) - α β ( α - β ) = 0

Factor out (α - β):

(α - β)[1 - (α + β) - α β ] = 0

Therefore, this equation holds if either α - β = 0, which would mean α = β, but α and β are distinct fixed points (since discriminant D = 6i ≠ 0), so α ≠ β. Therefore, the other factor must be zero:

1 - (α + β) - α β = 0 → α + β + α β = 1

So, we have that the fixed points α and β satisfy α + β + α β = 1.

Let me check whether that is true with our found fixed points.

Earlier, we had fixed points:

α = [ (1 + sqrt(3))/2 ](1 + i )

and β = [ (1 - sqrt(3))/2 ](1 + i )

Let me compute α + β:

α + β = [ (1 + sqrt(3))/2 + (1 - sqrt(3))/2 ](1 + i ) = [ (2/2 ) ](1 + i ) = 1*(1 + i ) = 1 + i

Then compute α β:

α β = [ (1 + sqrt(3))/2 * (1 - sqrt(3))/2 ] * (1 + i )^2

First compute the scalar factor: [ (1 + sqrt(3))(1 - sqrt(3)) ] / 4 = (1 - 3)/4 = (-2)/4 = -1/2

Then compute (1 + i )² = 1 + 2i + i² = 1 + 2i - 1 = 2i

Therefore α β = (-1/2)(2i) = -i

So α + β + α β = (1 + i ) + (-i ) = 1 + i - i = 1. Yes! It satisfies the equation. Therefore, the condition is satisfied. Good.

So, going back, we have that k = (1 - α)/(1 - β). Let me compute that.

But maybe instead of computing k explicitly, we can use the relation α + β + α β = 1. Let me think if I can find k in terms of α and β?

Alternatively, maybe I can proceed with φ(z) = (z - α)/(z - β) as the conjugating map, then φ(F(z)) = k φ(z). Therefore, φ(z_n) = k^n φ(z_0). Therefore, once we compute φ(z_0), we can get φ(z_n) = k^n φ(z_0), then invert φ to get z_n.

That seems promising. Let me try this approach.

First, compute φ(z_0) where φ(z) = (z - α)/(z - β). Then z_n = φ^{-1}(k^n φ(z_0)).

Therefore, if I can compute φ(z_0), then compute k, then find z_n.

So, first, need to find φ(z_0). Let me compute that.

But to compute φ(z_0), I need to know α and β. But α and β are complicated expressions. Maybe there's a better way?

Alternatively, since we have α + β + α β = 1, and also from the fixed points equation.

Alternatively, perhaps there's a relation between k and α, β?

We had earlier that k = (1 - α)/(1 - β). Let me compute that.

But maybe I can find k? Let me think. Let me also note that from the earlier equation when equating coefficients for z term, which was:

[ - (1 - α)β + i(1 + α) ] = k*[ - (1 - β)α + i(1 + β) ]

But maybe we can use k = (1 - α)/(1 - β). Let me denote k = (1 - α)/(1 - β). Then:

Let me compute the RHS of the z coefficient equation: k*[ - (1 - β)α + i(1 + β) ] = [ (1 - α)/(1 - β) ] * [ - (1 - β)α + i(1 + β) ] = (1 - α)[ - α + i(1 + β)/(1 - β) ]? Wait, no. Wait:

Wait, expanding:

= [ (1 - α)/(1 - β) ] * [ - (1 - β)α + i(1 + β) ]

= (1 - α)/ (1 - β) * [ - (1 - β)α ] + (1 - α)/ (1 - β) * i(1 + β )

= - (1 - α) α + i (1 - α)(1 + β)/(1 - β)

Hmm, complicated. Maybe not helpful. Let me try to compute k?

Alternatively, since we know α and β, maybe plug them in. But they are complex numbers. Let me recall that:

Earlier, we had α = [ (1 + sqrt(3))/2 ](1 + i ) and β = [ (1 - sqrt(3))/2 ](1 + i ). Let me denote these as:

Let me write α = A(1 + i), where A = (1 + sqrt(3))/2

Similarly β = B(1 + i), where B = (1 - sqrt(3))/2

Note that A and B are real numbers. Let me check: A = (1 + sqrt(3))/2 ≈ (1 + 1.732)/2 ≈ 1.366, B = (1 - sqrt(3))/2 ≈ (1 - 1.732)/2 ≈ -0.366.

So, α = A(1 + i), β = B(1 + i). Therefore, maybe I can express k = (1 - α)/(1 - β). Let me compute 1 - α and 1 - β.

Compute 1 - α = 1 - A(1 + i) = 1 - A - A i

Similarly 1 - β = 1 - B(1 + i) = 1 - B - B i

So k = [1 - A - A i ] / [1 - B - B i ]

But since A and B are real numbers, this is a complex number. However, maybe there's a simplification here. Let me compute k.

Alternatively, maybe there is a relation between A and B? Let me note that A + B = [ (1 + sqrt(3))/2 + (1 - sqrt(3))/2 ] = 1, and A * B = [ (1 + sqrt(3))/2 ] * [ (1 - sqrt(3))/2 ] = (1 - 3)/4 = -2/4 = -1/2, as computed earlier.

Also, since α + β + α β = 1, and α + β = (A + B)(1 + i) = 1*(1 + i) = 1 + i, and α β = -i, so indeed 1 + i - i = 1.

Alternatively, going back to k.

Let me compute numerator and denominator:

Numerator: 1 - α = 1 - A(1 + i) = (1 - A) - A i

Denominator: 1 - β = 1 - B(1 + i) = (1 - B) - B i

But since A + B = 1, then 1 - A = B, and 1 - B = A. Therefore:

Numerator: B - A i

Denominator: A - B i

Therefore k = (B - A i ) / (A - B i )

Multiply numerator and denominator by the conjugate of the denominator, which is (A + B i ):

k = [ (B - A i)(A + B i ) ] / [ (A - B i)(A + B i ) ]

Compute denominator: A² + (B i)^2? Wait, no, denominator is (A - B i)(A + B i ) = A² - (B i )² = A² - ( - B² ) = A² + B². Wait:

Wait, (A - Bi)(A + Bi) = A² - (Bi)^2 = A² - B² i² = A² - B² (-1) = A² + B². Yes.

Numerator: (B - A i)(A + B i ) = B*A + B*(B i ) - A i * A - A i * B i

= AB + B² i - A² i - A B i²

Simplify term by term:

AB + B² i - A² i - AB (-1) because i² = -1

So AB + B² i - A² i + AB

Combine like terms:

AB + AB = 2AB

B² i - A² i = i (B² - A² )

So numerator is 2AB + i (B² - A² )

Therefore, k = [ 2AB + i (B² - A² ) ] / (A² + B² )

Now, let me compute AB, A² + B², and B² - A².

We know that A + B = 1 and A * B = -1/2.

Therefore, A² + B² = (A + B)^2 - 2AB = 1² - 2*(-1/2) = 1 + 1 = 2.

Also, B² - A² = (B - A)(B + A) = (B - A)*1 = B - A.

But A = (1 + sqrt(3))/2, B = (1 - sqrt(3))/2. Therefore B - A = [ (1 - sqrt(3))/2 - (1 + sqrt(3))/2 ] = [ (1 - sqrt(3) - 1 - sqrt(3))/2 ] = [ -2 sqrt(3)/2 ] = -sqrt(3)

Therefore, numerator becomes 2AB + i*(B² - A² ) = 2*(-1/2) + i*(-sqrt(3)) = -1 - i sqrt(3)

Denominator is A² + B² = 2. Therefore, k = (-1 - i sqrt(3))/2

So k = - (1 + i sqrt(3))/2. Let me note that.

So k is a complex number. Let me write it in polar form. The modulus of k: |k| = sqrt( (-1/2)^2 + ( - sqrt(3)/2 )^2 ) = sqrt(1/4 + 3/4 ) = sqrt(1) = 1. The angle: arctangent( (-sqrt(3)/2 ) / (-1/2 )) = arctangent( sqrt(3) ). But since both real and imaginary parts are negative, it's in the third quadrant. The reference angle is π/3, so the angle is π + π/3 = 4π/3. Therefore, k = e^{i 4π/3}.

Alternatively, since k is a complex number of modulus 1 and angle 4π/3. Therefore, k^n = e^{i 4πn /3}.

But this suggests that φ(z_n) = k^n φ(z_0). Therefore, if we can compute φ(z_0), then we can compute φ(z_n) and then invert φ to get z_n.

So let me compute φ(z_0) where φ(z) = (z - α)/(z - β). Remember that α = A(1 + i) and β = B(1 + i), where A = (1 + sqrt(3))/2 and B = (1 - sqrt(3))/2.

Given z_0 = 1/137 + i.

So φ(z_0) = [ z_0 - α ] / [ z_0 - β ].

Let me compute numerator and denominator.

First, compute z_0 - α: (1/137 + i ) - A(1 + i )

Similarly z_0 - β: (1/137 + i ) - B(1 + i )

Let me compute numerator:

= (1/137 + i ) - A(1 + i ) = [ 1/137 - A ] + i [ 1 - A ]

Similarly denominator:

= [ 1/137 - B ] + i [ 1 - B ]

But A = (1 + sqrt(3))/2, B = (1 - sqrt(3))/2.

Let me compute 1/137 - A and 1 - A, same for B.

First, compute 1 - A: 1 - (1 + sqrt(3))/2 = (2 - 1 - sqrt(3))/2 = (1 - sqrt(3))/2 = B.

Similarly, 1 - B = 1 - (1 - sqrt(3))/2 = (2 - 1 + sqrt(3))/2 = (1 + sqrt(3))/2 = A.

Therefore, numerator becomes:

[ 1/137 - A ] + i * B

Denominator becomes:

[ 1/137 - B ] + i * A

So φ(z_0) = [ (1/137 - A) + i B ] / [ (1/137 - B) + i A ]

Now, let me compute 1/137 - A and 1/137 - B.

Compute 1/137 - A = 1/137 - (1 + sqrt(3))/2. Similarly for the other term.

But maybe I can write it as follows:

Let me denote that φ(z_0) = [ (C + i B) ] / [ (D + i A) ], where C = 1/137 - A and D = 1/137 - B.

But this seems messy. However, notice that since A and B are related, perhaps we can find a relation between numerator and denominator?

Alternatively, maybe there's a better way. Let me think. Since φ(z) is (z - α)/(z - β), and we have expressions involving 1/137 - A and similar.

Alternatively, maybe we can use the earlier expression for φ(z_n) and relate it to k^n φ(z_0). But unless we have more info, perhaps I need to compute φ(z_0). Alternatively, since z_0 is given, maybe I can find a relation?

Alternatively, perhaps there's a pattern in the iteration. Earlier, we computed z_0 = 1/137 + i, z_1 = 1 + 274i, z_2 = (37538 + i)/37265. Wait, this seems messy, but maybe if we continue computing terms, we can see a pattern?

Alternatively, perhaps I can use the formula for φ(z_n) = k^n φ(z_0). If I can compute φ(z_0), then maybe φ(z_n) is a complex number which can be inverted.

Alternatively, maybe there's a way to write φ(z_0) in terms of k? Let me think.

Wait, given that k is known, and φ(z_n) = k^n φ(z_0), then if we can write φ(z_0) as something involving k?

Alternatively, maybe there's a relation between z_0 and the fixed points?

Alternatively, perhaps we can find that φ(z_0) is equal to some power of k multiplied by something?

Alternatively, maybe I can compute φ(z_0) numerically?

Let me try that. Let me compute the numerical values.

First, compute A = (1 + sqrt(3))/2 ≈ (1 + 1.732)/2 ≈ 2.732 / 2 ≈ 1.366

Similarly, B = (1 - sqrt(3))/2 ≈ (1 - 1.732)/2 ≈ (-0.732)/2 ≈ -0.366

Then z_0 = 1/137 + i ≈ 0.0073 + i. Let me compute numerator and denominator:

Numerator: 1/137 - A ≈ 0.0073 - 1.366 ≈ -1.3587, and i * B ≈ i * (-0.366) ≈ -0.366i. Therefore numerator ≈ -1.3587 - 0.366i.

Denominator: 1/137 - B ≈ 0.0073 - (-0.366) ≈ 0.0073 + 0.366 ≈ 0.3733, and i * A ≈ i * 1.366 ≈ 1.366i. Therefore denominator ≈ 0.3733 + 1.366i.

So φ(z_0) ≈ (-1.3587 - 0.366i)/(0.3733 + 1.366i). Let me compute this division.

First, multiply numerator and denominator by the conjugate of the denominator: 0.3733 - 1.366i.

Numerator becomes (-1.3587 - 0.366i)(0.3733 - 1.366i)

Denominator becomes (0.3733)^2 + (1.366)^2 ≈ let me compute that.

First, approximate denominator:

0.3733² ≈ 0.139, 1.366² ≈ 1.866, so total ≈ 0.139 + 1.866 ≈ 2.005 ≈ 2. Let me check with exact values? Well, since earlier we had A² + B² = 2, but here the denominator is (0.3733)^2 + (1.366)^2. Wait, note that 0.3733 is approximately 1/137 - B? Wait no, denominator was [1/137 - B] + i*A, but when I calculated the denominator as approx 0.3733 + 1.366i, then the denominator squared magnitude is (1/137 - B)^2 + A^2. Wait, but earlier I had A² + B² = 2, but here it's different.

But regardless, let me proceed with approximate calculation.

So denominator ≈ 2. Then compute numerator:

Let me compute (-1.3587)(0.3733) + (-1.3587)(-1.366i) + (-0.366i)(0.3733) + (-0.366i)(-1.366i)

First compute real parts:

-1.3587 * 0.3733 ≈ -0.506

Imaginary parts from first term: + (-1.3587)(-1.366i) ≈ + (1.3587 * 1.366) i ≈ approximately, 1.3587 * 1.366 ≈ let me compute 1.35 * 1.36 ≈ 1.35 * 1.36 ≈ 1.836. So approx +1.836i

Then next term: -0.366i * 0.3733 ≈ -0.136i

Last term: -0.366i * -1.366i = + 0.366 * 1.366 * i² = approx 0.5 * (-1) = -0.5. Wait, exact calculation:

0.366 * 1.366 ≈ Let me calculate 0.366 * 1.366 ≈ approx 0.366 * 1 + 0.366 * 0.366 ≈ 0.366 + 0.133 ≈ 0.499. So +0.499 * (-1) = -0.499.

Therefore total real parts: -0.506 - 0.499 ≈ -1.005

Total imaginary parts: 1.836i - 0.136i ≈ 1.7i

Therefore numerator ≈ -1.005 + 1.7i. Divided by denominator ≈ 2, gives φ(z_0) ≈ (-1.005 + 1.7i)/2 ≈ -0.5025 + 0.85i.

So φ(z_0) ≈ -0.5 + 0.85i. Let me check if this is related to k?

Earlier, we had k = (-1 - i sqrt(3))/2 ≈ (-1 - i * 1.732)/2 ≈ (-1 - i1.732)/2 ≈ -0.5 - i0.866. Which is similar to the approximate value of φ(z_0) here: -0.5 + 0.85i. Close, except the imaginary part is positive instead of negative. Wait, but maybe my approximations are off?

Alternatively, maybe φ(z_0) is k^{-1} or something?

Wait, let me think. If φ(z_n) = k^n φ(z_0), then if we can find φ(z_0) = c, then z_n is given by φ^{-1}(k^n c). However, without knowing exactly what φ(z_0) is, it's hard to proceed. However, given the approximate value of φ(z_0) ≈ -0.5 + 0.85i and k ≈ -0.5 - i0.866, which is e^{i 4π/3} as mentioned before. So k is a complex number with angle 4π/3, modulus 1.

Let me note that k is e^{i 4π/3}, so k^n is e^{i 4πn /3}.

But φ(z_0) is approximately some value, but given the exact expressions, maybe there's a relation?

Alternatively, maybe φ(z_0) is equal to k^{-1}? Let me check:

If φ(z_0) = k^{-1}, then φ(z_n) = k^n * k^{-1} = k^{n - 1}, but this is just a guess. Alternatively, maybe φ(z_0) = something else.

Alternatively, maybe I can use the exact expressions. Recall that φ(z_0) = [ (1/137 - A ) + i B ] / [ (1/137 - B ) + i A ].

But since 1 - A = B, then 1/137 - A = 1/137 - A = 1/137 - (1 - B) = 1/137 - 1 + B = B - (1 - 1/137 ) = B - 136/137. Similarly, but maybe this isn't helpful.

Wait, let me compute 1/137 - A. A = (1 + sqrt(3))/2, so 1/137 - A = 1/137 - 1/2 - sqrt(3)/2 = (2 - 137)/274 - sqrt(3)/2 = ( - 135 / 274 ) - sqrt(3)/2. Wait, no:

Wait, 1/137 - (1 + sqrt(3))/2 = 1/137 - 1/2 - sqrt(3)/2 = (1/137 - 1/2) - sqrt(3)/2. Let me compute 1/137 - 1/2 = (2 - 137)/274 = (-135)/274. So, 1/137 - A = -135/274 - sqrt(3)/2. Similarly for the other term: 1/137 - B = 1/137 - (1 - sqrt(3))/2 = 1/137 - 1/2 + sqrt(3)/2 = (2 - 137)/274 + sqrt(3)/2 = -135/274 + sqrt(3)/2.

So the numerator is [ -135/274 - sqrt(3)/2 ] + i * B. Wait, B = (1 - sqrt(3))/2.

So numerator is:

Real part: -135/274 - sqrt(3)/2

Imaginary part: i * (1 - sqrt(3))/2

Similarly denominator is [ -135/274 + sqrt(3)/2 ] + i * A, where A = (1 + sqrt(3))/2.

This seems complicated, but maybe we can write this as:

Numerator: C + i D

Denominator: E + i F

Where C = -135/274 - sqrt(3)/2, D = (1 - sqrt(3))/2, E = -135/274 + sqrt(3)/2, F = (1 + sqrt(3))/2.

Then φ(z_0) = (C + i D)/(E + i F). Let me multiply numerator and denominator by the conjugate of the denominator (E - i F):

Numerator becomes (C + i D)(E - i F)

Denominator becomes |E + i F|².

Let me compute denominator first. |E + i F|² = E² + F².

E = -135/274 + sqrt(3)/2, F = (1 + sqrt(3))/2.

So E² + F² = ?

Let me compute:

Let me note that sqrt(3)/2 is approximately 0.866, and -135/274 is approximately -0.493. But exact calculation is better.

Alternatively, let me note that F = A, and earlier we had A² + B² = 2, but E is related to 1/137 - B?

Wait, perhaps not. Let me proceed.

Alternatively, maybe there's a way to relate numerator and denominator through previous expressions?

Wait, given that φ(z_0) = [ z_0 - α ] / [ z_0 - β ] = something. But perhaps we can relate this to k?

Recall that earlier, we found that k = (B - A i ) / (A - B i ) = (-1 - i sqrt(3))/2. Let me check:

Wait, we had that k = [ 2AB + i (B² - A² ) ] / (A² + B² ). We computed numerator as -1 - i sqrt(3) and denominator as 2, giving k = (-1 - i sqrt(3))/2.

But given that φ(z_0) is some complex number, and if we could find its relation to k, it would help. However, without more information, it's challenging.

Alternatively, maybe we can consider that the iteration leads to periodic behavior. Given that k is a complex number on the unit circle with angle 4π/3, which is a rotation by 4π/3 radians. Since the angle is 4π/3, the order of k is 3, since 4π/3 * 3 = 4π, which is equivalent to 0 modulo 2π. Wait, 4π/3 * 3 = 4π, but 4π is 2π*2, so the order is 3? Wait, no. The multiplicative order of k is the smallest integer m such that k^m = 1. Since k = e^{i4π/3}, then k^3 = e^{i4π} = e^{i0} = 1. Hence, the order is 3. Therefore, k^3 = 1.

Therefore, φ(z_n) = k^n φ(z_0). So, since k^3 = 1, then φ(z_{n}) = φ(z_0) * k^{n} = φ(z_0) * k^{n mod 3}.

Therefore, φ(z_n) has period 3. Therefore, if we can find φ(z_0), then φ(z_n) cycles every 3 steps. Therefore, if we compute φ(z_0), then φ(z_{2002}) = φ(z_0) * k^{2002} = φ(z_0) * k^{2002 mod 3}.

Therefore, compute 2002 divided by 3. Let me compute 2002 / 3:

3*667 = 2001, so 2002 = 3*667 + 1, so remainder 1. Therefore, 2002 ≡ 1 mod 3. Therefore, φ(z_{2002}) = φ(z_0) * k^1 = φ(z_0) * k.

Therefore, if I can compute φ(z_0) * k, then invert φ to get z_{2002}.

But how?

Alternatively, if I can find that φ(z_0) * k is equal to something that can be inverted?

Alternatively, since φ(z_0) = (z_0 - α)/(z_0 - β), and k = (B - A i ) / (A - B i ). Wait, earlier we had φ(z_0) = [ (1/137 - A ) + i B ] / [ (1/137 - B ) + i A ].

Wait, perhaps instead of trying to compute it directly, there's a relation between φ(z_0) and some term?

Alternatively, since we know that k^3 = 1, and we need to find φ(z_0) * k. But without knowing φ(z_0), this seems difficult.

Wait, let me think again. If k^3 = 1, then the sequence φ(z_n) cycles every 3 terms. So z_0, z_1, z_2, z_3 = z_0, etc. But wait, if φ(z_n) cycles every 3, then z_n cycles every 3 steps?

Wait, no. Because φ(z_n) = k^n φ(z_0). If φ(z_n) cycles every 3, then φ(z_{n + 3}) = φ(z_n). Therefore, z_{n + 3} = φ^{-1}(φ(z_n)) = z_n. Therefore, the sequence z_n is periodic with period 3. But is that true?

Wait, if φ(z_n) = k^n φ(z_0), and k^3 = 1, then φ(z_{n + 3}) = k^{n + 3} φ(z_0) = k^n * k^3 φ(z_0) = k^n * 1 * φ(z_0) = φ(z_n). Therefore, φ(z_{n + 3}) = φ(z_n). Therefore, if φ is invertible (which it is, since it's a Möbius transformation not equal to identity), then z_{n + 3} = z_n. Therefore, the sequence is periodic with period 3. Therefore, z_{n} = z_{n mod 3}.

Wait, but if that's the case, then z_{2002} = z_{2002 mod 3}. Let me compute 2002 mod 3. As above, 2001 is divisible by 3, so 2002 = 3*667 + 1, so 2002 ≡ 1 mod 3. Therefore, z_{2002} = z_1. But earlier we computed z_1 = 1 + 274i. Is this the case? Wait, but earlier when I computed z_1 and z_2, they didn't immediately cycle back to z_0. Let me check.

Wait, if the sequence is periodic with period 3, then z_0, z_1, z_2, z_0, z_1, z_2,... So z_3 should be z_0. Let me compute z_3 using the earlier terms.

We had z_0, z_1 = 1 + 274i, z_2 = (37538 + i)/37265. Let me compute z_3 = F(z_2).

But this would be complicated, but if the period is 3, then z_3 should equal z_0. Let me check with approximate values.

From earlier, z_1 ≈ 1 + 274i, which is a large complex number, z_2 ≈ (37538 + i)/37265 ≈ (approx) 37538/37265 ≈ 1.007 (real part) and i/37265 ≈ 0.000027i. So z_2 ≈ 1.007 + 0.000027i. Then z_3 = F(z_2) = (z_2 + i)/(z_2 - i). Let me approximate this.

Numerator: z_2 + i ≈ 1.007 + 0.000027i + i ≈ 1.007 + 1.000027i

Denominator: z_2 - i ≈ 1.007 + 0.000027i - i ≈ 1.007 - 0.999973i

Divide numerator by denominator. Since denominator is close to 1.007 - 1i, and numerator is approximately 1.007 + 1i. So z_3 ≈ (1.007 + 1i)/(1.007 - 1i). Multiply numerator and denominator by (1.007 + 1i):

Numerator becomes (1.007 + 1i)^2 ≈ (1.007)^2 - (1)^2 + 2*1.007*1i ≈ 1.014 + 2.014i

Denominator becomes (1.007)^2 + (1)^2 ≈ 1.014 + 1 = 2.014

Therefore z_3 ≈ (1.014 + 2.014i)/2.014 ≈ 0.5 + i. Which is approximately 0.5 + i, which is not z_0 = 1/137 + i ≈ 0.0073 + i. So z_3 ≈ 0.5 + i ≠ z_0. Therefore, contradiction. Hence, my assumption that the period is 3 is wrong?

Wait, what's wrong here? Earlier, we said that φ(z_n) = k^n φ(z_0), and since k^3 = 1, φ(z_{n + 3}) = φ(z_n). Hence, z_{n + 3} = φ^{-1}(φ(z_n)) = z_n. But in reality, according to approximate calculation, z_3 is not equal to z_0. So there must be a mistake in my reasoning.

Wait, perhaps the assumption that φ(z_n) = k^n φ(z_0) is incorrect? Or maybe my calculation for φ(z_0) is incorrect?

Wait, let me recall that φ(z_n) = k^n φ(z_0). So if I have φ(z_3) = k^3 φ(z_0) = 1 * φ(z_0) = φ(z_0). Therefore, φ(z_3) = φ(z_0) => z_3 = z_0, but our approximate calculation suggests otherwise. So this implies that either my calculation of z_3 is wrong, or my reasoning about φ is wrong.

Wait, let me recheck the calculation of φ(z_n). Earlier, I had φ(F(z)) = k φ(z). So this should mean that φ(z_1) = k φ(z_0). Then φ(z_2) = k φ(z_1) = k^2 φ(z_0), and φ(z_3) = k^3 φ(z_0) = φ(z_0). Hence z_3 = φ^{-1}(φ(z_0)) = z_0. Therefore, if my approximate calculation shows that z_3 is not equal to z_0, there must be an error in either the calculation of z_3 or the logic.

Let me recalculate z_3 more carefully. Given z_2 is approximately 1.007 + 0.000027i. Let me write z_2 = x + yi, where x ≈ 1.007, y ≈ 0.000027.

Then F(z_2) = (z_2 + i)/(z_2 - i) = (x + (y + 1)i)/(x - (1 - y)i). Let me compute this.

Let me compute numerator: x + (y + 1)i

Denominator: x - (1 - y)i

Let me multiply numerator and denominator by the conjugate of the denominator: x + (1 - y)i.

Wait, no, the conjugate of the denominator x - (1 - y)i is x + (1 - y)i.

So numerator becomes [x + (y + 1)i][x + (1 - y)i]

Denominator becomes x^2 + (1 - y)^2

Let me expand numerator:

= x*x + x*(1 - y)i + (y + 1)i*x + (y + 1)i*(1 - y)i

= x² + x(1 - y)i + x(y + 1)i + (y + 1)(1 - y)i²

Simplify terms:

First term: x²

Second and third terms: [x(1 - y) + x(y + 1)]i = x[(1 - y) + (y + 1)]i = x[2]i = 2x i

Fourth term: (y + 1)(1 - y) * i² = (1 - y²)(-1) = -1 + y²

So overall numerator is x² - 1 + y² + 2x i

Therefore, z_3 = [x² - 1 + y² + 2x i] / [x² + (1 - y)^2]

Now, with x ≈ 1.007, y ≈ 0.000027:

Compute x² ≈ (1.007)^2 ≈ 1.014

Compute y² ≈ (0.000027)^2 ≈ 7e-10, negligible.

Compute numerator ≈ 1.014 - 1 + 0 + 2*1.007*i ≈ 0.014 + 2.014i

Denominator ≈ x² + (1 - y)^2 ≈ 1.014 + 1^2 ≈ 2.014

Thus, z_3 ≈ (0.014 + 2.014i)/2.014 ≈ 0.007 + i.

Which is approximately z_0 = 1/137 + i ≈ 0.0073 + i. This matches!

Ah, I see! Earlier when I approximated, I thought z_3 was ≈ 0.5 + i, but actually, the real part is ≈ 0.014 / 2.014 ≈ 0.007, which is approximately z_0. Therefore, the sequence is indeed periodic with period 3. My previous approximation was wrong because I miscalculated the numerator. So z_3 ≈ z_0, which confirms the periodicity. Great!

Therefore, the sequence is periodic with period 3. Hence, z_{n+3} = z_n. Therefore, to find z_{2002}, we can find 2002 modulo 3.

As before, 2002 divided by 3: 3*667 = 2001, so 2002 = 3*667 + 1, hence 2002 ≡ 1 mod 3. Therefore, z_{2002} = z_1.

Earlier, we computed z_1 = 1 + 274i. However, wait, earlier when I computed z_1, I had z_0 = 1/137 + i. Let me confirm:

Yes, z_1 = (z_0 + i)/(z_0 - i). Given z_0 = 1/137 + i, then z_0 + i = 1/137 + 2i, and z_0 - i = 1/137. So z_1 = (1/137 + 2i) / (1/137) = 1 + 2i*137 = 1 + 274i. Yes, correct.

But according to the periodicity, z_3 = z_0, z_4 = z_1, etc. Hence, z_{3k + 1} = z_1. Therefore, since 2002 = 3*667 + 1, z_{2002} = z_1 = 1 + 274i.

But the problem says "find z_{2002}." So is this the answer? Wait, but let me check if the period is indeed 3. Since z_3 = z_0, z_4 = z_1, z_5 = z_2, z_6 = z_0, etc. Yes, so the period is 3. Hence, z_{2002} = z_1. However, let me check with n = 0,1,2,3,...

z_0 = 1/137 + i

z_1 = 1 + 274i

z_2 = (37538 + i)/37265 ≈ 1.007 + 0.000027i

z_3 = z_0

So yes, the period is 3. Therefore, z_{2002} = z_{1} = 1 + 274i. But let me check with n=3k+1, so for example n=1: z_1, n=4: z_4 = z_1, yes. So 2002 = 3*667 + 1, so z_{2002} = z_1 = 1 + 274i.

Wait, but the problem says z_0 = 1/137 + i, and asks for z_{2002}. So is it possible that the answer is 1 + 274i? But that seems too straightforward. However, given the periodicity, it should be correct.

Wait, but let me double-check with another term. Let me compute z_3 = F(z_2). Let me try to compute it exactly. Earlier, we had z_2 = (37538 + i)/37265. Let me compute z_3 = F(z_2) = (z_2 + i)/(z_2 - i). Let me compute this exactly.

Let me denote z_2 = (37538 + i)/37265.

Then z_2 + i = (37538 + i)/37265 + i = (37538 + i + 37265i)/37265 = [37538 + 37266i]/37265.

Similarly, z_2 - i = (37538 + i)/37265 - i = (37538 + i - 37265i)/37265 = [37538 - 37264i]/37265.

Therefore, z_3 = [37538 + 37266i]/[37538 - 37264i]. Let me simplify this fraction.

Let me note that 37538 and 37264 might have common factors. Let me check GCD(37538, 37264). Compute 37538 - 37264 = 274. Then GCD(37264, 274). Let me divide 37264 by 274.

But this might take time. Alternatively, perhaps there is a pattern here.

Alternatively, let me multiply numerator and denominator by the conjugate of the denominator, which is 37538 + 37264i.

So numerator becomes (37538 + 37266i)(37538 + 37264i)

Denominator becomes (37538)^2 + (37264)^2.

This seems very complicated. However, if the periodicity is correct, z_3 should equal z_0 = 1/137 + i. Let me check if this is the case.

Let me assume z_3 = z_0. Then:

[37538 + 37266i]/[37538 - 37264i] = 1/137 + i

Let me cross multiply:

(37538 + 37266i) = (1/137 + i)(37538 - 37264i)

Compute RHS:

= (1/137)*37538 - (1/137)*37264i + i*37538 - i^2 * 37264

= (37538 / 137) - (37264 / 137)i + 37538i + 37264

First compute 37538 / 137:

Since z_0 = 1/137 + i, and earlier we had z_1 = (z_0 + i)/(z_0 - i) = 1 + 274i. Let me check 37538 / 137.

Wait, earlier when computing z_2, the numerator was 75076 + 2i over 74530, which reduced to 37538 + i over 37265. So 37538 = 75076 / 2, and 37265 = 74530 / 2. Now, 75076 / 137: let me compute 137*547 = ? Wait, maybe there's a connection here. Let me note that z_0 = 1/137 + i, so maybe there's a relation between 37538 and 137?

Wait, 137 * 274 = let me compute 137*200 = 27400, 137*74 = let's see, 137*70 = 9590, 137*4 = 548, total 9590 + 548 = 10138. Therefore total 27400 + 10138 = 37538. Yes! So 137 * 274 = 37538. Similarly, 37264: let me check if 37264 is related. Let me compute 137*272. 137*272: Let me compute 137*270 = 137*200 + 137*70 = 27400 + 9590 = 36990, then 137*2 = 274, so total 36990 + 274 = 37264. Yes! Therefore, 37264 = 137*272.

Therefore, RHS:

= (137 * 274)/137 - (137*272)/137 * i + 37538i + 37264

Simplify:

= 274 - 272i + 37538i + 37264

Combine real parts: 274 + 37264 = 37538

Combine imaginary parts: -272i + 37538i = (37538 - 272)i = 37266i

Therefore RHS = 37538 + 37266i, which matches the numerator. Hence, z_3 = z_0. Hence, verified!

Therefore, the sequence is indeed periodic with period 3. Hence, z_{2002} = z_{3*667 + 1} = z_1 = 1 + 274i. Therefore, the answer is 1 + 274i.

But let me confirm once again with the initial terms:

z_0 = 1/137 + i

z_1 = 1 + 274i

z_2 = (37538 + i)/37265

z_3 = z_0

z_4 = z_1, etc.

Hence, n mod 3 = 0: z_0, 1: z_1, 2: z_2. Since 2002 mod 3 is 1, so z_{2002} = z_1 = 1 + 274i.

Therefore, the final answer is \boxed{1 + 274i}.

**Final Answer**
\boxed{1 + 274i}

To solve the given problem, we consider the recursive sequence defined by the Möbius transformation:

$$
z_n = F(z_{n-1}) = \frac{z_{n-1} + i}{z_{n-1} - i}
$$
with initial value $ z_0 = \frac{1}{137} + i $.

---

### **Step 1: Understanding the Function**

The mapping $ F(z) = \frac{z + i}{z - i} $ is a Möbius transformation. Such transformations are invertible and often exhibit periodic or repeating behavior under iteration, especially when analyzed via fixed points or conjugation methods.

We begin by noting that this sequence is generated by a Möbius transformation with two fixed points. These are the solutions to the fixed point equation:
$$
F(z) = z \quad \Rightarrow \quad \frac{z + i}{z - i} = z
$$
Multiply both sides by $ z - i $ and simplify:
$$
z + i = z(z - i) \Rightarrow z + i = z^2 - iz
\Rightarrow z^2 - iz - z - i = 0
\Rightarrow z^2 - z(1+i) - i = 0
$$
Solving this quadratic yields the fixed points:
$$
z = \frac{(1 + i) \pm \sqrt{(1 + i)^2 + 4i}}{2}
$$
Computing the discriminant:
$$
(1 + i)^2 + 4i = 1 + 2i - 1 + 4i = 6i
\Rightarrow \sqrt{6i} = \sqrt{6} \left( \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = \sqrt{3} (1 + i)
$$
Thus, the fixed points are:
$$
z = \frac{(1 + i) \pm \sqrt{3}(1 + i)}{2} = \left( \frac{1 \pm \sqrt{3}}{2} \right)(1 + i)
$$
Let these be $ \alpha = \frac{(1+\sqrt{3})}{2}(1 + i) $, $ \beta = \frac{(1 - \sqrt{3})}{2}(1 + i) $. These fixed points are complex conjugates in the form $ a(1 + i) $.

---

### **Step 2: Conjugating the Transformation**

Let $ \phi(z) = \frac{z - \alpha}{z - \beta} $, a Möbius transformation mapping $ \alpha \mapsto 0 $, $ \beta \mapsto \infty $. This transformation conjugates the function $ F $ to a scaling transformation:
$$
\phi(F(z)) = k \phi(z), \quad \text{where } |k| = 1
$$

It can be shown that the transformation $ k = \frac{(1 - \alpha)}{(1 - \beta)} $, and the value of $ k $ ends up being:
$$
k = \frac{-1 - i \sqrt{3}}{2}
$$
This complex number lies on the unit circle and satisfies $ k^3 = 1 $, meaning it has **order 3**.

---

### **Step 3: Applying the Conjugation**

Under this mapping, the iteration satisfies:
$$
\phi(z_n) = k^n \phi(z_0)
$$
Since $ k^3 = 1 $, the sequence $ \phi(z_n) $ is periodic with period 3:
$$
\phi(z_{n+3}) = \phi(z_n) \Rightarrow z_{n+3} = z_n
$$
That is, the original sequence $ \{ z_n \} $ is **periodic with period 3**.

---

### **Step 4: Computing $ z_{2002} $**

We compute:
$$
2002 \mod 3 = 1 \Rightarrow z_{2002} = z_1
$$

We already computed:
$$
z_0 = \frac{1}{137} + i, \quad z_1 = \frac{z_0 + i}{z_0 - i}
$$
Plug in $ z_0 = \frac{1}{137} + i $:
- Numerator: $ z_0 + i = \frac{1}{137} + i + i = \frac{1}{137} + 2i $
- Denominator: $ z_0 - i = \frac{1}{137} + i - i = \frac{1}{137} $

$$
z_1 = \frac{\frac{1}{137} + 2i}{\frac{1}{137}} = 1 + 2(137)i = 1 + 274i
$$

---

### **Final Answer**

$$
\boxed{1 + 274i}
$$

Solution:
Iterating $F$ a few times, we get \[\begin{aligned} F(F(z)) &= \frac{\frac{z+i}{z-i}+i}{\frac{z+i}{z-i}-i} = \frac{(z+i)+i(z-i)}{(z+i)-i(z-i)}= \frac{z+i+zi+1}{z+i-zi-1}= \frac{(z+1)(i+1)}{(z-1)(1-i)}\\
&= \frac{(z+1)(i+1)^2}{(z-1) \cdot 2}= \frac{(z+1)(2i)}{(z-1) \cdot 2} = \frac{z+1}{z-1}i,\\
F(F(F(z))) &= \frac{\frac{z+1}{z-1}i+i}{\frac{z+1}{z-1}i-i} = \frac{\frac{z+1}{z-1}+1}{\frac{z+1}{z-1}-1} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)}= z. \end{aligned}\]Thus, $z_{k+3} = z_k$ for all $k.$ Since $2002 \equiv 1 \pmod{3},$ we then have \[z_{2002} = z_1 = \frac{z_0+i}{z_0-i} = \frac{1/137 + 2i}{1/137} = \boxed{1+274i}.\]