Explanation:
Okay, so I need to evaluate the sum $ S = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}} $, where $ \omega $ is a 1997th root of unity, not equal to 1. Hmm, let me think about how to approach this.

First, I remember that roots of unity have some nice properties, especially when dealing with sums. Since $ \omega^{1997} = 1 $, $ \omega $ is a primitive 1997th root of unity or a root of a lower order, but since it's not equal to 1, it's definitely a primitive root or something else. But since 1997 is a prime number (I think so, maybe I should check that... Well, 1997 divided by... Let me check. 1997 is a prime. Yeah, I remember that 1997 is a prime number, so all the roots except 1 are primitive. So, $ \omega $ is a primitive 1997th root of unity.

But how does that help me with the sum? Maybe I can use some properties of roots of unity, or perhaps manipulate the sum into something else. Let me recall that sums involving roots of unity sometimes can be simplified by multiplying numerator and denominator by something, or using symmetry.

Wait, the sum is from $ k=1 $ to $ k=1997 $ of $ \frac{1}{1 + \omega^k} $, but wait, hold on, the original question says "from 1 to 1997", but since $ \omega^{1997} = 1 $, so $ \omega^{1997} $ is 1, which is excluded because $ \omega \neq 1 $. Wait, the problem says "1 + ω^1997" but since ω^1997 is 1, that term would be 1/(1 + 1) = 1/2. But hold on, the original sum is written as starting from 1 to 1997, so does that include the term with k=1997? Wait, maybe that's a typo? Wait, no, let me check the problem again.

Wait, the problem says: "If $ \omega^{1997} = 1 $ and $ \omega \neq 1 $, then evaluate $ \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}} $." So, the sum is from k=1 to k=1997. However, since ω^1997 = 1, the last term is 1/(1 + 1) = 1/2. But the problem says ω ≠ 1, but that term is 1/2 regardless. So maybe that's okay.

But maybe there's a way to compute this sum. Let me think. I remember that sometimes with roots of unity, you can pair terms or use substitution. Also, maybe consider the fact that the sum of all roots of unity is zero. Hmm. Let me think. Let me recall that if we have a root of unity ω, then 1 + ω + ω² + ... + ω^{n-1} = 0 for n-th roots of unity. But here, our sum is different. It's sum of reciprocals of 1 + ω^k.

Alternatively, maybe consider that if I let each term be 1/(1 + ω^k). Let me think if there's a way to relate this to a geometric series or something else.

Alternatively, maybe consider multiplying numerator and denominator by (1 - ω^k). Wait, let me try that. If I take 1/(1 + ω^k) and multiply numerator and denominator by (1 - ω^k), then I get (1 - ω^k)/[(1 + ω^k)(1 - ω^k)] = (1 - ω^k)/(1 - ω^{2k}).

But then, if I do that, maybe the denominator becomes 1 - ω^{2k}, but I don't know if that helps. But maybe if I consider the sum over k from 1 to 1997 of (1 - ω^k)/(1 - ω^{2k}).

Alternatively, maybe there's another approach. Let me think. Let me denote S as the sum:

$ S = \sum_{k=1}^{1997} \frac{1}{1 + \omega^k} $.

But since ω^{1997} = 1, the exponents can be considered modulo 1997. Also, note that ω^k for k from 1 to 1996 are all the roots of unity except 1, and then ω^{1997} is 1. However, in the sum, we have k=1 to 1997, which includes ω^{1997} = 1. But maybe I can separate the term when k=1997?

But let me note that the sum S is the sum over all 1997th roots of unity (including ω=1) except that when k=1997, ω^k = 1, so the term is 1/(1 + 1) = 1/2. However, in the problem statement, ω ≠ 1, but ω is still a root of unity. Wait, but actually, the problem says ω is a 1997th root of unity, ω ≠ 1. So maybe the sum is over k from 1 to 1997 of 1/(1 + ω^k). But since ω^k cycles through all the 1997th roots of unity as k goes from 1 to 1997. Wait, but when k=1997, ω^k = 1. So the sum is over all the roots of unity, including 1. But if I consider that, perhaps there's a formula for the sum of 1/(1 + ζ) where ζ ranges over all n-th roots of unity.

Alternatively, maybe I can consider that if I let ζ range over all n-th roots of unity, then sum_{ζ} 1/(1 + ζ). However, since 1 is included, but in our case, the sum is over ζ = ω^k for k=1 to 1997, which includes all the roots of unity. Let me think. Wait, if ω is a primitive root, then as k goes from 1 to 1997, ω^k cycles through all the roots of unity, because the multiplicative order is 1997. So the sum S is sum_{ζ} 1/(1 + ζ) where ζ ranges over all 1997th roots of unity, including ζ=1. However, the problem says ω ≠ 1, but ω is a root of unity. So maybe the sum includes ζ=1. Let me check, when k=1997, ζ=ω^{1997}=1. Therefore, S is the sum over all 1997th roots of unity of 1/(1 + ζ).

But then, perhaps I can compute this sum. Let me recall that for roots of unity, sometimes we can use the formula for the sum over roots of some function. Let me think. Let me consider that ζ^n = 1, so for each root ζ ≠ 1, we can use properties of roots.

Alternatively, consider that:

Let me think about the sum over all roots ζ of unity (including ζ=1) of 1/(1 + ζ). Let me denote this sum as T. Then, since we have T = sum_{ζ} 1/(1 + ζ).

But if we can compute T, then our desired sum S is T. Wait, but in our case, the original problem's sum is exactly T. But maybe we can compute T?

Alternatively, consider that for each root ζ ≠ -1, we can pair terms? Or maybe relate this to derivatives? Hmm, or perhaps consider that 1/(1 + ζ) can be written in terms of something else.

Alternatively, note that the sum S can be related to the sum over ζ of 1/(1 + ζ) = sum_{ζ} [ (1 - ζ)/(1 - ζ^2) ] if ζ ≠ -1? Wait, similar to before. Wait, if ζ ≠ -1, then 1/(1 + ζ) = (1 - ζ)/(1 - ζ^2). But if ζ = -1, then that would be undefined. However, since our roots of unity are ζ^n = 1, so ζ = -1 would be a root only if n is even. Since 1997 is odd, -1 is not a root of unity here. Therefore, since 1997 is odd, none of the roots ζ satisfy ζ = -1. Therefore, for all ζ in our sum, ζ ≠ -1, so we can safely write 1/(1 + ζ) = (1 - ζ)/(1 - ζ^2).

So, then, the sum T becomes sum_{ζ} (1 - ζ)/(1 - ζ^2). But 1 - ζ^2 factors as (1 - ζ)(1 + ζ). Therefore, we can write:

(1 - ζ)/(1 - ζ^2) = [ (1 - ζ) ] / [ (1 - ζ)(1 + ζ) ) ] = 1/(1 + ζ). Wait, that's just circular. Maybe this isn't helpful.

Wait, but perhaps we can think of T = sum_{ζ} 1/(1 + ζ) = sum_{ζ} [1/(1 + ζ)], and since ζ^n = 1. Let me consider multiplying numerator and denominator by something. Let me think about the following idea: For each ζ ≠ -1, if we write 1/(1 + ζ) as something. Alternatively, consider that the sum over all ζ ≠ -1 of 1/(1 + ζ) is equal to T minus 1/(1 + (-1)) if -1 were in the sum, but since it's not, maybe not.

Alternatively, consider the following approach. Let me consider that if we have a function f(ζ) = 1/(1 + ζ), and we want to compute the sum over all roots ζ of unity. Maybe there's a formula for such sums.

Alternatively, consider that for each root ζ of unity, we can consider the sum over all ζ of f(ζ) where f is a rational function. To compute such sums, sometimes we can use residues or other complex analysis, but maybe there's a more elementary approach.

Alternatively, consider using the fact that if we have ζ ≠ -1, then 1/(1 + ζ) can be written as ∫ something... Hmm, maybe not.

Wait, another idea: Let me consider that the sum T = sum_{ζ} 1/(1 + ζ). Let me consider multiplying numerator and denominator by (1 - ζ), but since ζ is a root of unity, maybe I can relate this to the derivative of the polynomial? Let me think.

Let me recall that the sum over all roots ζ of the polynomial x^n - 1 = 0 is 0. Also, if I have a polynomial P(x) = x^n - 1, then the roots are ζ_k = e^{2πik/n} for k = 0, 1, ..., n-1. But in our case, n = 1997, and the roots are ζ = ω^k for k=0,...,1996. However, our sum T includes k=0 to k=1996 (if we consider all roots of unity) but the problem's sum is from k=1 to 1997, which is k=1 to 1997, but ω^{1997} = 1, so that's ζ=1 again. Wait, so in our problem, the sum is from k=1 to 1997 of 1/(1 + ω^k). Which is ζ=ω^1 to ω^{1997} = 1. So that's the same as ζ=ω^0 to ω^{1996} plus the term ζ=ω^{1997}=ω^0=1. Wait, but the sum from k=1 to 1997 is the same as the sum over all roots of unity (since for each root except ζ=ω^0=1, there is a k from 1 to 1996, and then k=1997 gives ζ=1 again). Wait, but that would mean that the sum S is equal to sum_{ζ} 1/(1 + ζ) where ζ ranges over all roots of unity (including ζ=1). Therefore, T = S. So, how do we compute T?

Let me consider the following approach: Let me denote T = sum_{ζ} 1/(1 + ζ), where ζ runs over all n-th roots of unity (n=1997). Let me consider that 1/(1 + ζ) can be written as something.

Alternatively, note that for each ζ ≠ -1, 1/(1 + ζ) = -1/(ζ - 1) + something? Wait, maybe partial fractions? Alternatively, think about the function f(ζ) = 1/(1 + ζ) and consider the sum over all ζ^n = 1.

Let me recall that for a polynomial P(x) with roots ζ_1, ζ_2, ..., ζ_n, then sum_{k} 1/(a - ζ_k) = P'(a)/P(a). Is that right? Yes, that's a formula from complex analysis: if P(x) = product_{k=1}^n (x - ζ_k), then P'(x)/P(x) = sum_{k=1}^n 1/(x - ζ_k). Therefore, sum_{k=1}^n 1/(x - ζ_k) = P'(x)/P(x).

In our case, our polynomial is P(x) = x^n - 1. Then, P'(x) = n x^{n - 1}. Therefore, sum_{k=1}^n 1/(x - ζ_k) = n x^{n - 1}/(x^n - 1).

But we want sum_{ζ} 1/(1 + ζ). Which is equal to sum_{ζ} 1/( - (-1 - ζ) ). Wait, maybe let me set x = -1. Then, sum_{ζ} 1/( -1 - ζ ) = P'(-1)/P(-1).

But P(-1) = (-1)^n - 1. Since n = 1997 is odd, (-1)^n = -1. Therefore, P(-1) = -1 - 1 = -2. And P'(-1) = n (-1)^{n - 1} = 1997 * (-1)^{1996} = 1997 * 1 = 1997. Therefore, sum_{ζ} 1/( -1 - ζ ) = P'(-1)/P(-1) = 1997 / (-2) = -1997 / 2.

But sum_{ζ} 1/( -1 - ζ ) = - sum_{ζ} 1/(1 + ζ) = -T. Therefore, -T = -1997 / 2 => T = 1997 / 2.

Wait, but hold on. So according to this, T = sum_{ζ} 1/(1 + ζ) = 1997 / 2?

But wait, our original sum S is equal to T? Wait, no. Wait, in our problem, the sum S is from k=1 to 1997 of 1/(1 + ω^k). Which, as we discussed, includes all the roots ζ of unity including ζ = 1. But the formula above gives sum_{ζ} 1/(1 + ζ) = T = 1997 / 2. But then S is equal to T? So does that mean S = 1997 / 2? But that can't be, because when we have ζ = 1, the term is 1/(1 + 1) = 1/2. So the sum T includes that term. But let me check if the formula is correct.

Wait, according to the formula, T = sum_{ζ} 1/(1 + ζ) where ζ ranges over all roots of unity. But according to the calculation, that sum equals 1997 / 2. However, if I check with n=1, which is trivial? Wait, n=1: the only root is ζ=1. Then sum is 1/(1 + 1) = 1/2. According to the formula, n=1, so T = 1/2, which matches.

For n=2, the roots are 1 and -1. Then sum is 1/(1 + 1) + 1/(1 + (-1)) = 1/2 + 1/0, which is undefined. But according to the formula, n=2, sum_{ζ} 1/(1 + ζ) = P'(-1)/P(-1). Let's check. P(x) = x^2 -1. P(-1) = (-1)^2 - 1 = 0, which is division by zero. But that makes sense because ζ=-1 would make denominator zero. So in our problem, n=1997 is odd, so -1 is not a root, so we are okay.

So for n odd, P(-1) = (-1)^n -1 = -1 -1 = -2 ≠ 0. So in our case, the formula holds and T = n / 2. Wait, according to our previous calculation, T = 1997 / 2. But wait, in our case, n=1997, and we got T = n / 2.

So according to this, the sum over all roots ζ of 1/(1 + ζ) is n / 2, so S = n / 2? Wait, but in the case of n=1, S=1/2. For n=3, let me check manually.

Take n=3, roots are 1, ω, ω² where ω is a primitive third root of unity. Then sum is 1/(1+1) + 1/(1 + ω) + 1/(1 + ω²). Let's compute that.

First term: 1/2.

Second term: 1/(1 + ω). Let me compute 1 + ω + ω² = 0 => ω² = -1 - ω. Therefore, 1 + ω² = 1 -1 - ω = -ω. Therefore, 1/(1 + ω²) = 1/(-ω).

Similarly, 1/(1 + ω) can be calculated. Let me note that 1/(1 + ω) + 1/(1 + ω²) = 1/(1 + ω) + 1/(1 + (-1 - ω)) = 1/(1 + ω) + 1/(-ω).

Let me compute 1/(1 + ω). Let me multiply numerator and denominator by (1 - ω):

[1*(1 - ω)] / [(1 + ω)(1 - ω)] = (1 - ω)/(1 - ω²). But ω² = -1 - ω, so 1 - ω² = 1 - (-1 - ω) = 2 + ω. Therefore, the expression is (1 - ω)/(2 + ω). Hmm, maybe not helpful. Alternatively, using the fact that 1 + ω + ω² = 0, so 1 + ω = -ω². Therefore, 1/(1 + ω) = -1/ω². Similarly, 1/(1 + ω²) = -1/ω. Therefore, the sum of these two terms is -1/ω² -1/ω.

But ω + ω² = -1, so - (1/ω + 1/ω²). Let me compute 1/ω + 1/ω². Since ω³ =1, ω² = ω^{-1}, so 1/ω = ω², 1/ω² = ω. Therefore, 1/ω + 1/ω² = ω² + ω = -1. Therefore, the sum is -(-1) = 1. Therefore, adding the first term 1/2, total sum is 1 + 1/2 = 3/2. Which is equal to n/2 = 3/2. So yes! That works.

Therefore, the formula seems to hold. Hence, for general n, sum_{ζ} 1/(1 + ζ) = n / 2. Therefore, in our problem, since n=1997, the sum S = 1997 / 2. However, let me check if the problem's sum includes all roots.

Wait, in the problem, the sum is from k=1 to 1997 of 1/(1 + ω^k). Since ω is a primitive root, ω^1 to ω^{1997} includes each root of unity once? Wait, no. Wait, for n=1997, ω^k for k=1 to 1997 includes ω^{1997} = ω^0 = 1. So, the sum is over all roots of unity, including ζ=1. Therefore, T = sum_{ζ} 1/(1 + ζ) = 1997 / 2. Therefore, the answer would be 1997 / 2.

But let me check again with n=1, which gives 1/2, and n=3 gives 3/2. So that seems to hold. Therefore, the answer is 1997/2. However, wait, but the problem says ω ≠ 1, but in the sum, the term when k=1997 is 1/(1 + 1) = 1/2, which is included. Therefore, the answer is 1997/2.

Wait, but let me think again. The problem says "If ω^{1997} = 1 and ω ≠ 1", but the sum includes 1/(1 + ω^{1997}) = 1/(1 + 1) = 1/2. However, if the problem had said ω is a primitive root, then ω^k for k=1,...,1997 would be all roots including 1, so the same thing. But even if ω is not primitive, but any root, say, a primitive m-th root where m divides 1997. Since 1997 is prime, m is either 1 or 1997. Since ω ≠ 1, m=1997. Therefore, ω is a primitive 1997th root of unity. Therefore, the sum over k=1 to 1997 of 1/(1 + ω^k) is equal to sum over all roots of unity.

Therefore, according to the general formula, the answer is 1997 / 2. Therefore, the answer is \boxed{\dfrac{1997}{2}}.

But wait, but wait. Let me think again. When I computed for n=3, the sum over all roots (including ζ=1) gives 3/2. But in that case, the sum was 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 3/2. However, in our problem, the sum is from k=1 to 1997. But when k=1 to 1997, since ω^{1997} = 1, the term for k=1997 is 1/(1 + 1) = 1/2. So if we have 1997 terms, including the term for k=1997. But when I computed the sum over all roots of unity (ζ=1, ζ=ω, ..., ζ=ω^{1996}), which is 1997 terms (from k=0 to 1996), but the problem's sum is from k=1 to 1997, which is the same as sum from k=0 to 1996 plus the term for k=1997, which is same as term k=0 (since ω^{1997} = ω^0 = 1). Therefore, the sum S in the problem is sum_{k=0}^{1996} 1/(1 + ω^k) + 1/(1 + ω^{1997}) = sum_{ζ} 1/(1 + ζ) + 1/(1 + 1) - 1/(1 + 1) ? Wait, no. Because when k=0 to 1996, that's all the roots, including ζ=1. Then, adding k=1997 which is ζ=1 again. So S is equal to sum_{ζ} 1/(1 + ζ) + 1/(1 + 1). Wait, no. The original problem's sum is from k=1 to 1997, which is:

sum_{k=1}^{1997} 1/(1 + ω^k) = sum_{k=1}^{1996} 1/(1 + ω^k) + 1/(1 + ω^{1997}) = sum_{ζ ≠ 1} 1/(1 + ζ) + 1/(1 + 1).

But sum_{ζ ≠ 1} 1/(1 + ζ) = T - 1/(1 + 1) = (1997 / 2) - 1/2 = (1996 / 2) = 998. Then adding the last term 1/2: total sum is 998 + 1/2 = 998.5 = 1997/2. So actually, same result. Therefore, regardless of how we consider it, the total sum is 1997 / 2.

Therefore, the answer is 1997 over 2, boxed.

Wait, but I need to make sure that my formula is correct. Let me think again. The key step was using the derivative formula. Let me go through that again.

Let me denote P(x) = x^n - 1. Then, the roots are ζ_0, ζ_1, ..., ζ_{n-1} where ζ_k = e^{2πik/n}. Then, sum_{k=0}^{n-1} 1/(a - ζ_k) = P'(a)/P(a). Therefore, if I set a = -1, then sum_{k=0}^{n-1} 1/(-1 - ζ_k) = P'(-1)/P(-1). Therefore, sum_{k=0}^{n-1} 1/(1 + ζ_k) = -P'(-1)/P(-1).

Wait, earlier I thought that sum_{ζ} 1/(1 + ζ) = P'(-1)/P(-1) but that seems to be incorrect. Let me redo this.

Let me write:

sum_{k=0}^{n-1} 1/(a - ζ_k) = P'(a)/P(a).

If I set a = -1, then:

sum_{k=0}^{n-1} 1/(-1 - ζ_k) = P'(-1)/P(-1).

But 1/(-1 - ζ_k) = -1/(1 + ζ_k). Therefore, the left-hand side is -sum_{k=0}^{n-1} 1/(1 + ζ_k). Therefore:

- sum_{k=0}^{n-1} 1/(1 + ζ_k) = P'(-1)/P(-1) => sum_{k=0}^{n-1} 1/(1 + ζ_k) = -P'(-1)/P(-1).

Earlier, I had written that sum_{ζ} 1/(1 + ζ) = P'(-1)/P(-1), but actually it's negative of that. Therefore, correcting that:

For our case, n=1997, P(x) = x^{1997} - 1. Then P(-1) = (-1)^{1997} - 1 = -1 -1 = -2. P'(-1) = 1997 * (-1)^{1996} = 1997 * 1 = 1997. Therefore:

sum_{k=0}^{1996} 1/(1 + ζ_k) = -P'(-1)/P(-1) = - (1997)/(-2) = 1997 / 2.

Therefore, that confirms that the sum over all roots ζ (including ζ=1) of 1/(1 + ζ) is 1997 / 2. Therefore, the problem's sum S is exactly that sum, because as we discussed, the problem's sum is from k=1 to 1997, which includes all the roots. Therefore, the answer is indeed 1997 / 2.

But just to confirm, let me check with n=3 again. For n=3, sum_{k=0}^2 1/(1 + ζ_k) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + [ -1/ω -1/ω² ] (from before) which we said was 1/2 + 1 = 3/2, which is equal to n/2 = 3/2. Therefore, the formula holds.

Therefore, my conclusion is that the answer is 1997 divided by 2.

But just to make sure, let me check if there's another way to think about this. Suppose I consider pairing terms. For example, for each root ζ ≠ 1, there is a root ζ^{-1} since the roots come in conjugate pairs? Wait, since ζ is a root of unity, ζ^{-1} is also a root. For example, ζ and ζ^{-1} are distinct unless ζ is self-inverse. Since n is odd, ζ^{-1} is different from ζ, except for ζ=1 and ζ=-1. But since n is odd, ζ=-1 is not a root. Therefore, each root ζ can be paired with ζ^{-1}.

Therefore, let me consider pairing terms 1/(1 + ζ) and 1/(1 + ζ^{-1}). Let me compute their sum:

1/(1 + ζ) + 1/(1 + ζ^{-1}) = [ (1 + ζ^{-1}) + (1 + ζ) ] / [ (1 + ζ)(1 + ζ^{-1}) ] = [2 + ζ + ζ^{-1} ] / [ 1 + ζ + ζ^{-1} + 1 ].

Wait, wait. Let me compute numerator and denominator step by step:

Numerator: (1 + ζ^{-1}) + (1 + ζ) = 2 + ζ + ζ^{-1}

Denominator: (1 + ζ)(1 + ζ^{-1}) = 1 + ζ^{-1} + ζ + ζ * ζ^{-1} = 1 + ζ^{-1} + ζ + 1 = 2 + ζ + ζ^{-1}

Therefore, the sum is [2 + ζ + ζ^{-1}]/[2 + ζ + ζ^{-1}] = 1.

Wow, that's interesting! Therefore, each pair ζ and ζ^{-1} contributes 1 to the sum. Since the roots come in such pairs, and there are (n - 1)/2 such pairs (excluding ζ=1, which is self-inverse). But wait, when n is odd, the only root that is self-inverse is ζ=1. Because ζ^{-1} = ζ implies ζ^2 = 1, but since n is odd, the only solution is ζ=1. Therefore, the roots can be grouped into (n - 1)/2 pairs (ζ, ζ^{-1}), each contributing 1, and then the term for ζ=1, which is 1/(1 + 1) = 1/2. Therefore, total sum is (n - 1)/2 * 1 + 1/2 = (n - 1)/2 + 1/2 = n / 2.

Therefore, this confirms the same result! For example, with n=3, we have (3 - 1)/2 = 1 pair contributing 1, plus 1/2, total 3/2. Perfect.

Therefore, this method gives the same answer. Therefore, regardless of the approach, the sum is n/2. Therefore, for our problem with n=1997, the answer is 1997/2.

Therefore, the answer is \boxed{\dfrac{1997}{2}}.

**Final Answer**
\boxed{\dfrac{1997}{2}}

To evaluate the sum

$$
S = \sum_{k=1}^{1997} \frac{1}{1 + \omega^k}
$$

where $\omega$ is a **primitive 1997th root of unity** (i.e., $\omega^{1997} = 1$, $\omega \neq 1$), we proceed by utilizing the algebraic properties of complex roots of unity.

---

### Key Insight: Roots of Unity

Let $\omega$ be a **primitive 1997th root of unity**, so all $\omega^k$ for $k = 1, 2, \ldots, 1997$ are **distinct** 1997th roots of unity (including $\omega^{1997} = \omega^0 = 1$). Therefore, the sum

$$
\sum_{k=1}^{1997} \frac{1}{1 + \omega^k}
$$

is nothing more than the **sum over all 1997th roots of unity** of the function $f(\zeta) = \frac{1}{1 + \zeta}$, with $\zeta = \omega^k$.

Let us denote this total sum as:

$$
S = \sum_{\zeta} \frac{1}{1 + \zeta} \quad \text{where } \zeta^n = 1 \text{ and } \zeta \neq -1 \text{ (since } n = 1997 \text{ is odd)}
$$

---

### General Approach via Root Sum Formula

Let $P(x) = x^{1997} - 1$, whose roots are all the 1997th roots of unity (including 1). The derivative is:

$$
P'(x) = 1997 x^{1996}
$$

The following identity holds:

$$
\sum_{k=0}^{1996} \frac{1}{a - \zeta_k} = \frac{P'(a)}{P(a)}
$$

where $\zeta_k = \omega^k$ are the roots of $P(x)$.

Now choose $a = -1$, so we can write:

$$
\sum_{k=0}^{1996} \frac{1}{-1 - \zeta_k} = \frac{P'(-1)}{P(-1)}
$$

Therefore:

$$
\sum_{k=0}^{1996} \frac{1}{1 + \zeta_k} = -\frac{P'(-1)}{P(-1)}
$$

Now compute the right-hand side:

- $P(-1) = (-1)^{1997} - 1 = -1 - 1 = -2$
- $P'(-1) = 1997 \cdot (-1)^{1996} = 1997 \cdot 1 = 1997$

So:

$$
\sum_{k=0}^{1996} \frac{1}{1 + \zeta_k} = -\frac{1997}{-2} = \frac{1997}{2}
$$

This is **exactly** the sum $S$, since the terms in $S$ (from $k=1$ to $1997$) include all the 1997th roots of unity.

---

### Conclusion

Thus, the value of the sum is:

$$
\boxed{\dfrac{1997}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to evaluate the sum $ S = \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}} $, where $ \omega $ is a 1997th root of unity, not equal to 1. Hmm, let me think about how to approach this.

First, I remember that roots of unity have some nice properties, especially when dealing with sums. Since $ \omega^{1997} = 1 $, $ \omega $ is a primitive 1997th root of unity or a root of a lower order, but since it's not equal to 1, it's definitely a primitive root or something else. But since 1997 is a prime number (I think so, maybe I should check that... Well, 1997 divided by... Let me check. 1997 is a prime. Yeah, I remember that 1997 is a prime number, so all the roots except 1 are primitive. So, $ \omega $ is a primitive 1997th root of unity.

But how does that help me with the sum? Maybe I can use some properties of roots of unity, or perhaps manipulate the sum into something else. Let me recall that sums involving roots of unity sometimes can be simplified by multiplying numerator and denominator by something, or using symmetry.

Wait, the sum is from $ k=1 $ to $ k=1997 $ of $ \frac{1}{1 + \omega^k} $, but wait, hold on, the original question says "from 1 to 1997", but since $ \omega^{1997} = 1 $, so $ \omega^{1997} $ is 1, which is excluded because $ \omega \neq 1 $. Wait, the problem says "1 + ω^1997" but since ω^1997 is 1, that term would be 1/(1 + 1) = 1/2. But hold on, the original sum is written as starting from 1 to 1997, so does that include the term with k=1997? Wait, maybe that's a typo? Wait, no, let me check the problem again.

Wait, the problem says: "If $ \omega^{1997} = 1 $ and $ \omega \neq 1 $, then evaluate $ \frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}} $." So, the sum is from k=1 to k=1997. However, since ω^1997 = 1, the last term is 1/(1 + 1) = 1/2. But the problem says ω ≠ 1, but that term is 1/2 regardless. So maybe that's okay.

But maybe there's a way to compute this sum. Let me think. I remember that sometimes with roots of unity, you can pair terms or use substitution. Also, maybe consider the fact that the sum of all roots of unity is zero. Hmm. Let me think. Let me recall that if we have a root of unity ω, then 1 + ω + ω² + ... + ω^{n-1} = 0 for n-th roots of unity. But here, our sum is different. It's sum of reciprocals of 1 + ω^k.

Alternatively, maybe consider that if I let each term be 1/(1 + ω^k). Let me think if there's a way to relate this to a geometric series or something else.

Alternatively, maybe consider multiplying numerator and denominator by (1 - ω^k). Wait, let me try that. If I take 1/(1 + ω^k) and multiply numerator and denominator by (1 - ω^k), then I get (1 - ω^k)/[(1 + ω^k)(1 - ω^k)] = (1 - ω^k)/(1 - ω^{2k}).

But then, if I do that, maybe the denominator becomes 1 - ω^{2k}, but I don't know if that helps. But maybe if I consider the sum over k from 1 to 1997 of (1 - ω^k)/(1 - ω^{2k}).

Alternatively, maybe there's another approach. Let me think. Let me denote S as the sum:

$ S = \sum_{k=1}^{1997} \frac{1}{1 + \omega^k} $.

But since ω^{1997} = 1, the exponents can be considered modulo 1997. Also, note that ω^k for k from 1 to 1996 are all the roots of unity except 1, and then ω^{1997} is 1. However, in the sum, we have k=1 to 1997, which includes ω^{1997} = 1. But maybe I can separate the term when k=1997?

But let me note that the sum S is the sum over all 1997th roots of unity (including ω=1) except that when k=1997, ω^k = 1, so the term is 1/(1 + 1) = 1/2. However, in the problem statement, ω ≠ 1, but ω is still a root of unity. Wait, but actually, the problem says ω is a 1997th root of unity, ω ≠ 1. So maybe the sum is over k from 1 to 1997 of 1/(1 + ω^k). But since ω^k cycles through all the 1997th roots of unity as k goes from 1 to 1997. Wait, but when k=1997, ω^k = 1. So the sum is over all the roots of unity, including 1. But if I consider that, perhaps there's a formula for the sum of 1/(1 + ζ) where ζ ranges over all n-th roots of unity.

Alternatively, maybe I can consider that if I let ζ range over all n-th roots of unity, then sum_{ζ} 1/(1 + ζ). However, since 1 is included, but in our case, the sum is over ζ = ω^k for k=1 to 1997, which includes all the roots of unity. Let me think. Wait, if ω is a primitive root, then as k goes from 1 to 1997, ω^k cycles through all the roots of unity, because the multiplicative order is 1997. So the sum S is sum_{ζ} 1/(1 + ζ) where ζ ranges over all 1997th roots of unity, including ζ=1. However, the problem says ω ≠ 1, but ω is a root of unity. So maybe the sum includes ζ=1. Let me check, when k=1997, ζ=ω^{1997}=1. Therefore, S is the sum over all 1997th roots of unity of 1/(1 + ζ).

But then, perhaps I can compute this sum. Let me recall that for roots of unity, sometimes we can use the formula for the sum over roots of some function. Let me think. Let me consider that ζ^n = 1, so for each root ζ ≠ 1, we can use properties of roots.

Alternatively, consider that:

Let me think about the sum over all roots ζ of unity (including ζ=1) of 1/(1 + ζ). Let me denote this sum as T. Then, since we have T = sum_{ζ} 1/(1 + ζ).

But if we can compute T, then our desired sum S is T. Wait, but in our case, the original problem's sum is exactly T. But maybe we can compute T?

Alternatively, consider that for each root ζ ≠ -1, we can pair terms? Or maybe relate this to derivatives? Hmm, or perhaps consider that 1/(1 + ζ) can be written in terms of something else.

Alternatively, note that the sum S can be related to the sum over ζ of 1/(1 + ζ) = sum_{ζ} [ (1 - ζ)/(1 - ζ^2) ] if ζ ≠ -1? Wait, similar to before. Wait, if ζ ≠ -1, then 1/(1 + ζ) = (1 - ζ)/(1 - ζ^2). But if ζ = -1, then that would be undefined. However, since our roots of unity are ζ^n = 1, so ζ = -1 would be a root only if n is even. Since 1997 is odd, -1 is not a root of unity here. Therefore, since 1997 is odd, none of the roots ζ satisfy ζ = -1. Therefore, for all ζ in our sum, ζ ≠ -1, so we can safely write 1/(1 + ζ) = (1 - ζ)/(1 - ζ^2).

So, then, the sum T becomes sum_{ζ} (1 - ζ)/(1 - ζ^2). But 1 - ζ^2 factors as (1 - ζ)(1 + ζ). Therefore, we can write:

(1 - ζ)/(1 - ζ^2) = [ (1 - ζ) ] / [ (1 - ζ)(1 + ζ) ) ] = 1/(1 + ζ). Wait, that's just circular. Maybe this isn't helpful.

Wait, but perhaps we can think of T = sum_{ζ} 1/(1 + ζ) = sum_{ζ} [1/(1 + ζ)], and since ζ^n = 1. Let me consider multiplying numerator and denominator by something. Let me think about the following idea: For each ζ ≠ -1, if we write 1/(1 + ζ) as something. Alternatively, consider that the sum over all ζ ≠ -1 of 1/(1 + ζ) is equal to T minus 1/(1 + (-1)) if -1 were in the sum, but since it's not, maybe not.

Alternatively, consider the following approach. Let me consider that if we have a function f(ζ) = 1/(1 + ζ), and we want to compute the sum over all roots ζ of unity. Maybe there's a formula for such sums.

Alternatively, consider that for each root ζ of unity, we can consider the sum over all ζ of f(ζ) where f is a rational function. To compute such sums, sometimes we can use residues or other complex analysis, but maybe there's a more elementary approach.

Alternatively, consider using the fact that if we have ζ ≠ -1, then 1/(1 + ζ) can be written as ∫ something... Hmm, maybe not.

Wait, another idea: Let me consider that the sum T = sum_{ζ} 1/(1 + ζ). Let me consider multiplying numerator and denominator by (1 - ζ), but since ζ is a root of unity, maybe I can relate this to the derivative of the polynomial? Let me think.

Let me recall that the sum over all roots ζ of the polynomial x^n - 1 = 0 is 0. Also, if I have a polynomial P(x) = x^n - 1, then the roots are ζ_k = e^{2πik/n} for k = 0, 1, ..., n-1. But in our case, n = 1997, and the roots are ζ = ω^k for k=0,...,1996. However, our sum T includes k=0 to k=1996 (if we consider all roots of unity) but the problem's sum is from k=1 to 1997, which is k=1 to 1997, but ω^{1997} = 1, so that's ζ=1 again. Wait, so in our problem, the sum is from k=1 to 1997 of 1/(1 + ω^k). Which is ζ=ω^1 to ω^{1997} = 1. So that's the same as ζ=ω^0 to ω^{1996} plus the term ζ=ω^{1997}=ω^0=1. Wait, but the sum from k=1 to 1997 is the same as the sum over all roots of unity (since for each root except ζ=ω^0=1, there is a k from 1 to 1996, and then k=1997 gives ζ=1 again). Wait, but that would mean that the sum S is equal to sum_{ζ} 1/(1 + ζ) where ζ ranges over all roots of unity (including ζ=1). Therefore, T = S. So, how do we compute T?

Let me consider the following approach: Let me denote T = sum_{ζ} 1/(1 + ζ), where ζ runs over all n-th roots of unity (n=1997). Let me consider that 1/(1 + ζ) can be written as something.

Alternatively, note that for each ζ ≠ -1, 1/(1 + ζ) = -1/(ζ - 1) + something? Wait, maybe partial fractions? Alternatively, think about the function f(ζ) = 1/(1 + ζ) and consider the sum over all ζ^n = 1.

Let me recall that for a polynomial P(x) with roots ζ_1, ζ_2, ..., ζ_n, then sum_{k} 1/(a - ζ_k) = P'(a)/P(a). Is that right? Yes, that's a formula from complex analysis: if P(x) = product_{k=1}^n (x - ζ_k), then P'(x)/P(x) = sum_{k=1}^n 1/(x - ζ_k). Therefore, sum_{k=1}^n 1/(x - ζ_k) = P'(x)/P(x).

In our case, our polynomial is P(x) = x^n - 1. Then, P'(x) = n x^{n - 1}. Therefore, sum_{k=1}^n 1/(x - ζ_k) = n x^{n - 1}/(x^n - 1).

But we want sum_{ζ} 1/(1 + ζ). Which is equal to sum_{ζ} 1/( - (-1 - ζ) ). Wait, maybe let me set x = -1. Then, sum_{ζ} 1/( -1 - ζ ) = P'(-1)/P(-1).

But P(-1) = (-1)^n - 1. Since n = 1997 is odd, (-1)^n = -1. Therefore, P(-1) = -1 - 1 = -2. And P'(-1) = n (-1)^{n - 1} = 1997 * (-1)^{1996} = 1997 * 1 = 1997. Therefore, sum_{ζ} 1/( -1 - ζ ) = P'(-1)/P(-1) = 1997 / (-2) = -1997 / 2.

But sum_{ζ} 1/( -1 - ζ ) = - sum_{ζ} 1/(1 + ζ) = -T. Therefore, -T = -1997 / 2 => T = 1997 / 2.

Wait, but hold on. So according to this, T = sum_{ζ} 1/(1 + ζ) = 1997 / 2?

But wait, our original sum S is equal to T? Wait, no. Wait, in our problem, the sum S is from k=1 to 1997 of 1/(1 + ω^k). Which, as we discussed, includes all the roots ζ of unity including ζ = 1. But the formula above gives sum_{ζ} 1/(1 + ζ) = T = 1997 / 2. But then S is equal to T? So does that mean S = 1997 / 2? But that can't be, because when we have ζ = 1, the term is 1/(1 + 1) = 1/2. So the sum T includes that term. But let me check if the formula is correct.

Wait, according to the formula, T = sum_{ζ} 1/(1 + ζ) where ζ ranges over all roots of unity. But according to the calculation, that sum equals 1997 / 2. However, if I check with n=1, which is trivial? Wait, n=1: the only root is ζ=1. Then sum is 1/(1 + 1) = 1/2. According to the formula, n=1, so T = 1/2, which matches.

For n=2, the roots are 1 and -1. Then sum is 1/(1 + 1) + 1/(1 + (-1)) = 1/2 + 1/0, which is undefined. But according to the formula, n=2, sum_{ζ} 1/(1 + ζ) = P'(-1)/P(-1). Let's check. P(x) = x^2 -1. P(-1) = (-1)^2 - 1 = 0, which is division by zero. But that makes sense because ζ=-1 would make denominator zero. So in our problem, n=1997 is odd, so -1 is not a root, so we are okay.

So for n odd, P(-1) = (-1)^n -1 = -1 -1 = -2 ≠ 0. So in our case, the formula holds and T = n / 2. Wait, according to our previous calculation, T = 1997 / 2. But wait, in our case, n=1997, and we got T = n / 2.

So according to this, the sum over all roots ζ of 1/(1 + ζ) is n / 2, so S = n / 2? Wait, but in the case of n=1, S=1/2. For n=3, let me check manually.

Take n=3, roots are 1, ω, ω² where ω is a primitive third root of unity. Then sum is 1/(1+1) + 1/(1 + ω) + 1/(1 + ω²). Let's compute that.

First term: 1/2.

Second term: 1/(1 + ω). Let me compute 1 + ω + ω² = 0 => ω² = -1 - ω. Therefore, 1 + ω² = 1 -1 - ω = -ω. Therefore, 1/(1 + ω²) = 1/(-ω).

Similarly, 1/(1 + ω) can be calculated. Let me note that 1/(1 + ω) + 1/(1 + ω²) = 1/(1 + ω) + 1/(1 + (-1 - ω)) = 1/(1 + ω) + 1/(-ω).

Let me compute 1/(1 + ω). Let me multiply numerator and denominator by (1 - ω):

[1*(1 - ω)] / [(1 + ω)(1 - ω)] = (1 - ω)/(1 - ω²). But ω² = -1 - ω, so 1 - ω² = 1 - (-1 - ω) = 2 + ω. Therefore, the expression is (1 - ω)/(2 + ω). Hmm, maybe not helpful. Alternatively, using the fact that 1 + ω + ω² = 0, so 1 + ω = -ω². Therefore, 1/(1 + ω) = -1/ω². Similarly, 1/(1 + ω²) = -1/ω. Therefore, the sum of these two terms is -1/ω² -1/ω.

But ω + ω² = -1, so - (1/ω + 1/ω²). Let me compute 1/ω + 1/ω². Since ω³ =1, ω² = ω^{-1}, so 1/ω = ω², 1/ω² = ω. Therefore, 1/ω + 1/ω² = ω² + ω = -1. Therefore, the sum is -(-1) = 1. Therefore, adding the first term 1/2, total sum is 1 + 1/2 = 3/2. Which is equal to n/2 = 3/2. So yes! That works.

Therefore, the formula seems to hold. Hence, for general n, sum_{ζ} 1/(1 + ζ) = n / 2. Therefore, in our problem, since n=1997, the sum S = 1997 / 2. However, let me check if the problem's sum includes all roots.

Wait, in the problem, the sum is from k=1 to 1997 of 1/(1 + ω^k). Since ω is a primitive root, ω^1 to ω^{1997} includes each root of unity once? Wait, no. Wait, for n=1997, ω^k for k=1 to 1997 includes ω^{1997} = ω^0 = 1. So, the sum is over all roots of unity, including ζ=1. Therefore, T = sum_{ζ} 1/(1 + ζ) = 1997 / 2. Therefore, the answer would be 1997 / 2.

But let me check again with n=1, which gives 1/2, and n=3 gives 3/2. So that seems to hold. Therefore, the answer is 1997/2. However, wait, but the problem says ω ≠ 1, but in the sum, the term when k=1997 is 1/(1 + 1) = 1/2, which is included. Therefore, the answer is 1997/2.

Wait, but let me think again. The problem says "If ω^{1997} = 1 and ω ≠ 1", but the sum includes 1/(1 + ω^{1997}) = 1/(1 + 1) = 1/2. However, if the problem had said ω is a primitive root, then ω^k for k=1,...,1997 would be all roots including 1, so the same thing. But even if ω is not primitive, but any root, say, a primitive m-th root where m divides 1997. Since 1997 is prime, m is either 1 or 1997. Since ω ≠ 1, m=1997. Therefore, ω is a primitive 1997th root of unity. Therefore, the sum over k=1 to 1997 of 1/(1 + ω^k) is equal to sum over all roots of unity.

Therefore, according to the general formula, the answer is 1997 / 2. Therefore, the answer is \boxed{\dfrac{1997}{2}}.

But wait, but wait. Let me think again. When I computed for n=3, the sum over all roots (including ζ=1) gives 3/2. But in that case, the sum was 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + 1/(1 + ω) + 1/(1 + ω²) = 3/2. However, in our problem, the sum is from k=1 to 1997. But when k=1 to 1997, since ω^{1997} = 1, the term for k=1997 is 1/(1 + 1) = 1/2. So if we have 1997 terms, including the term for k=1997. But when I computed the sum over all roots of unity (ζ=1, ζ=ω, ..., ζ=ω^{1996}), which is 1997 terms (from k=0 to 1996), but the problem's sum is from k=1 to 1997, which is the same as sum from k=0 to 1996 plus the term for k=1997, which is same as term k=0 (since ω^{1997} = ω^0 = 1). Therefore, the sum S in the problem is sum_{k=0}^{1996} 1/(1 + ω^k) + 1/(1 + ω^{1997}) = sum_{ζ} 1/(1 + ζ) + 1/(1 + 1) - 1/(1 + 1) ? Wait, no. Because when k=0 to 1996, that's all the roots, including ζ=1. Then, adding k=1997 which is ζ=1 again. So S is equal to sum_{ζ} 1/(1 + ζ) + 1/(1 + 1). Wait, no. The original problem's sum is from k=1 to 1997, which is:

sum_{k=1}^{1997} 1/(1 + ω^k) = sum_{k=1}^{1996} 1/(1 + ω^k) + 1/(1 + ω^{1997}) = sum_{ζ ≠ 1} 1/(1 + ζ) + 1/(1 + 1).

But sum_{ζ ≠ 1} 1/(1 + ζ) = T - 1/(1 + 1) = (1997 / 2) - 1/2 = (1996 / 2) = 998. Then adding the last term 1/2: total sum is 998 + 1/2 = 998.5 = 1997/2. So actually, same result. Therefore, regardless of how we consider it, the total sum is 1997 / 2.

Therefore, the answer is 1997 over 2, boxed.

Wait, but I need to make sure that my formula is correct. Let me think again. The key step was using the derivative formula. Let me go through that again.

Let me denote P(x) = x^n - 1. Then, the roots are ζ_0, ζ_1, ..., ζ_{n-1} where ζ_k = e^{2πik/n}. Then, sum_{k=0}^{n-1} 1/(a - ζ_k) = P'(a)/P(a). Therefore, if I set a = -1, then sum_{k=0}^{n-1} 1/(-1 - ζ_k) = P'(-1)/P(-1). Therefore, sum_{k=0}^{n-1} 1/(1 + ζ_k) = -P'(-1)/P(-1).

Wait, earlier I thought that sum_{ζ} 1/(1 + ζ) = P'(-1)/P(-1) but that seems to be incorrect. Let me redo this.

Let me write:

sum_{k=0}^{n-1} 1/(a - ζ_k) = P'(a)/P(a).

If I set a = -1, then:

sum_{k=0}^{n-1} 1/(-1 - ζ_k) = P'(-1)/P(-1).

But 1/(-1 - ζ_k) = -1/(1 + ζ_k). Therefore, the left-hand side is -sum_{k=0}^{n-1} 1/(1 + ζ_k). Therefore:

- sum_{k=0}^{n-1} 1/(1 + ζ_k) = P'(-1)/P(-1) => sum_{k=0}^{n-1} 1/(1 + ζ_k) = -P'(-1)/P(-1).

Earlier, I had written that sum_{ζ} 1/(1 + ζ) = P'(-1)/P(-1), but actually it's negative of that. Therefore, correcting that:

For our case, n=1997, P(x) = x^{1997} - 1. Then P(-1) = (-1)^{1997} - 1 = -1 -1 = -2. P'(-1) = 1997 * (-1)^{1996} = 1997 * 1 = 1997. Therefore:

sum_{k=0}^{1996} 1/(1 + ζ_k) = -P'(-1)/P(-1) = - (1997)/(-2) = 1997 / 2.

Therefore, that confirms that the sum over all roots ζ (including ζ=1) of 1/(1 + ζ) is 1997 / 2. Therefore, the problem's sum S is exactly that sum, because as we discussed, the problem's sum is from k=1 to 1997, which includes all the roots. Therefore, the answer is indeed 1997 / 2.

But just to confirm, let me check with n=3 again. For n=3, sum_{k=0}^2 1/(1 + ζ_k) = 1/(1 + 1) + 1/(1 + ω) + 1/(1 + ω²) = 1/2 + [ -1/ω -1/ω² ] (from before) which we said was 1/2 + 1 = 3/2, which is equal to n/2 = 3/2. Therefore, the formula holds.

Therefore, my conclusion is that the answer is 1997 divided by 2.

But just to make sure, let me check if there's another way to think about this. Suppose I consider pairing terms. For example, for each root ζ ≠ 1, there is a root ζ^{-1} since the roots come in conjugate pairs? Wait, since ζ is a root of unity, ζ^{-1} is also a root. For example, ζ and ζ^{-1} are distinct unless ζ is self-inverse. Since n is odd, ζ^{-1} is different from ζ, except for ζ=1 and ζ=-1. But since n is odd, ζ=-1 is not a root. Therefore, each root ζ can be paired with ζ^{-1}.

Therefore, let me consider pairing terms 1/(1 + ζ) and 1/(1 + ζ^{-1}). Let me compute their sum:

1/(1 + ζ) + 1/(1 + ζ^{-1}) = [ (1 + ζ^{-1}) + (1 + ζ) ] / [ (1 + ζ)(1 + ζ^{-1}) ] = [2 + ζ + ζ^{-1} ] / [ 1 + ζ + ζ^{-1} + 1 ].

Wait, wait. Let me compute numerator and denominator step by step:

Numerator: (1 + ζ^{-1}) + (1 + ζ) = 2 + ζ + ζ^{-1}

Denominator: (1 + ζ)(1 + ζ^{-1}) = 1 + ζ^{-1} + ζ + ζ * ζ^{-1} = 1 + ζ^{-1} + ζ + 1 = 2 + ζ + ζ^{-1}

Therefore, the sum is [2 + ζ + ζ^{-1}]/[2 + ζ + ζ^{-1}] = 1.

Wow, that's interesting! Therefore, each pair ζ and ζ^{-1} contributes 1 to the sum. Since the roots come in such pairs, and there are (n - 1)/2 such pairs (excluding ζ=1, which is self-inverse). But wait, when n is odd, the only root that is self-inverse is ζ=1. Because ζ^{-1} = ζ implies ζ^2 = 1, but since n is odd, the only solution is ζ=1. Therefore, the roots can be grouped into (n - 1)/2 pairs (ζ, ζ^{-1}), each contributing 1, and then the term for ζ=1, which is 1/(1 + 1) = 1/2. Therefore, total sum is (n - 1)/2 * 1 + 1/2 = (n - 1)/2 + 1/2 = n / 2.

Therefore, this confirms the same result! For example, with n=3, we have (3 - 1)/2 = 1 pair contributing 1, plus 1/2, total 3/2. Perfect.

Therefore, this method gives the same answer. Therefore, regardless of the approach, the sum is n/2. Therefore, for our problem with n=1997, the answer is 1997/2.

Therefore, the answer is \boxed{\dfrac{1997}{2}}.

**Final Answer**
\boxed{\dfrac{1997}{2}}

To evaluate the sum

$$
S = \sum_{k=1}^{1997} \frac{1}{1 + \omega^k}
$$

where $\omega$ is a **primitive 1997th root of unity** (i.e., $\omega^{1997} = 1$, $\omega \neq 1$), we proceed by utilizing the algebraic properties of complex roots of unity.

---

### Key Insight: Roots of Unity

Let $\omega$ be a **primitive 1997th root of unity**, so all $\omega^k$ for $k = 1, 2, \ldots, 1997$ are **distinct** 1997th roots of unity (including $\omega^{1997} = \omega^0 = 1$). Therefore, the sum

$$
\sum_{k=1}^{1997} \frac{1}{1 + \omega^k}
$$

is nothing more than the **sum over all 1997th roots of unity** of the function $f(\zeta) = \frac{1}{1 + \zeta}$, with $\zeta = \omega^k$.

Let us denote this total sum as:

$$
S = \sum_{\zeta} \frac{1}{1 + \zeta} \quad \text{where } \zeta^n = 1 \text{ and } \zeta \neq -1 \text{ (since } n = 1997 \text{ is odd)}
$$

---

### General Approach via Root Sum Formula

Let $P(x) = x^{1997} - 1$, whose roots are all the 1997th roots of unity (including 1). The derivative is:

$$
P'(x) = 1997 x^{1996}
$$

The following identity holds:

$$
\sum_{k=0}^{1996} \frac{1}{a - \zeta_k} = \frac{P'(a)}{P(a)}
$$

where $\zeta_k = \omega^k$ are the roots of $P(x)$.

Now choose $a = -1$, so we can write:

$$
\sum_{k=0}^{1996} \frac{1}{-1 - \zeta_k} = \frac{P'(-1)}{P(-1)}
$$

Therefore:

$$
\sum_{k=0}^{1996} \frac{1}{1 + \zeta_k} = -\frac{P'(-1)}{P(-1)}
$$

Now compute the right-hand side:

- $P(-1) = (-1)^{1997} - 1 = -1 - 1 = -2$
- $P'(-1) = 1997 \cdot (-1)^{1996} = 1997 \cdot 1 = 1997$

So:

$$
\sum_{k=0}^{1996} \frac{1}{1 + \zeta_k} = -\frac{1997}{-2} = \frac{1997}{2}
$$

This is **exactly** the sum $S$, since the terms in $S$ (from $k=1$ to $1997$) include all the 1997th roots of unity.

---

### Conclusion

Thus, the value of the sum is:

$$
\boxed{\dfrac{1997}{2}}
$$

Solution:
Note that
\begin{align*}
\frac{1}{1 + \omega^k} + \frac{1}{1 + \omega^{1997 - k}} &= \frac{1}{1 + \omega^k} + \frac{\omega^k}{\omega^k + \omega^{1997}} \\
&= \frac{1}{1 + \omega^k} + \frac{\omega^k}{\omega^k + 1} \\
&= \frac{1 + \omega^k}{1 + \omega^k} = 1.
\end{align*}Thus, we can pair the terms
\[\frac{1}{1 + \omega}, \ \frac{1}{1 + \omega^2}, \ \dots, \ \frac{1}{1 + \omega^{1995}}, \ \frac{1}{1 + \omega^{1996}}\]into $1996/2 = 998$ pairs, so that the sum of the numbers in each pair is 1.  Also, $\frac{1}{1 + \omega^{1997}} = \frac{1}{2},$ so the sum works out to $998 + \frac{1}{2} = \boxed{\frac{1997}{2}}.$


Explanation:
Okay, so I need to figure out the real values of x that are not in the domain of the function f(x) = 1 divided by the sum of two absolute value expressions: |x² + 3x - 4| + |x² + 9x + 20|. The question is asking for which x values this function is undefined, which would be when the denominator is zero. Since the denominator is a sum of absolute values, which are always non-negative, the only way the denominator can be zero is if both absolute value expressions are zero simultaneously. So, I need to find the x values that make both x² + 3x - 4 = 0 and x² + 9x + 20 = 0. Then, those x values would make the denominator zero, hence not in the domain of f(x).

First, let me solve each quadratic equation separately. Let me start with the first one: x² + 3x - 4 = 0. To solve this quadratic equation, I can factor it or use the quadratic formula. Let me try factoring. Looking for two numbers that multiply to -4 and add to 3. Let me think... 4 and -1? 4 * (-1) = -4, and 4 + (-1) = 3. Yes, that works. So, the equation factors to (x + 4)(x - 1) = 0. Therefore, the solutions are x = -4 and x = 1.

Now, the second quadratic equation: x² + 9x + 20 = 0. Again, let me try factoring. Looking for two numbers that multiply to 20 and add to 9. 4 and 5? 4 * 5 = 20, and 4 + 5 = 9. Perfect. So, this factors to (x + 4)(x + 5) = 0. Therefore, the solutions are x = -4 and x = -5.

So, the first equation has roots at x = -4 and x = 1, and the second equation has roots at x = -4 and x = -5. Now, the denominator is the sum of the absolute values of these quadratics. For the denominator to be zero, both absolute values must be zero. That would require both quadratics to be zero at the same x value. Let me check the solutions.

Looking at the roots of the first equation: x = -4 and x = 1. Let's check if these x-values are also roots of the second equation.

For x = -4: Let me plug into the second equation: (-4)² + 9*(-4) + 20 = 16 - 36 + 20 = 0. Yes, so x = -4 is a common root.

Now, for x = 1: Let me check in the second equation: (1)² + 9*(1) + 20 = 1 + 9 + 20 = 30 ≠ 0. So, x = 1 is not a root of the second equation. Similarly, check x = -5 in the first equation: (-5)^2 + 3*(-5) - 4 = 25 - 15 - 4 = 6 ≠ 0. So x = -5 is not a root of the first equation. Therefore, the only x that makes both quadratics zero is x = -4.

Therefore, the denominator is zero only when x = -4. Hence, the function f(x) is undefined at x = -4. Therefore, the real values of x not in the domain are just x = -4. Wait, but let me make sure I didn't miss anything.

Wait, could there be other x-values where the sum of the absolute values is zero? Well, the absolute value is always non-negative, so the sum of two absolute values is zero only if each absolute value is zero. So, the only way for |A| + |B| = 0 is if A = 0 and B = 0. Therefore, there's no other possibility. Therefore, only x = -4 makes both expressions zero, so only x = -4 is excluded from the domain. So, the answer should be x = -4.

But let me check again. Let me verify that for x = -4, the denominator is indeed zero. Let me compute each absolute value:

First, x² + 3x - 4 at x = -4: (-4)^2 + 3*(-4) - 4 = 16 -12 -4 = 0. Then, the second expression x² + 9x + 20 at x = -4: (-4)^2 + 9*(-4) + 20 = 16 -36 + 20 = 0. So yes, both are zero. Therefore, their sum is zero. Therefore, f(-4) is undefined.

But let me check if there are other x-values where denominator could be zero. Suppose, for example, that one of the absolute values is positive and the other is negative? Wait, no, absolute values are always non-negative. So, their sum can be zero only if both are zero. So, there's no way for the sum to be zero unless both individual terms are zero. Therefore, there are no other x-values.

But let me think again. Maybe there are x-values where even if each absolute value is not zero, their sum can be zero? But since absolute values are non-negative, the only way their sum is zero is if each is zero. So, no. Therefore, only x = -4.

But let me check for example, suppose that one term is positive and the other is negative. But absolute values can't be negative. So, each term is zero or positive. Therefore, the sum is zero only if both terms are zero. Therefore, the only x is -4.

Therefore, the answer is x = -4. Therefore, the real value not in the domain is -4.

Wait, but before I conclude, let me check if maybe the quadratics can be negative, but absolute values make them positive. But even so, their sum can't be zero unless both are zero.

Alternatively, is there a possibility that the denominator is zero for some other x? Let me take some other x-values and see. Let me pick x = 0. Then compute denominator:

First quadratic: 0 + 0 -4 = -4. Absolute value is 4. Second quadratic: 0 + 0 + 20 = 20. Absolute value is 20. Sum is 4 + 20 = 24 ≠ 0.

How about x = -5? Let's check denominator. First quadratic: (-5)^2 + 3*(-5) -4 = 25 -15 -4 = 6. Absolute value is 6. Second quadratic: (-5)^2 + 9*(-5) +20 = 25 -45 +20 = 0. So the sum is 6 + 0 = 6 ≠ 0. Therefore, denominator is non-zero.

Similarly, x = 1: first quadratic is zero, second quadratic is 1 + 9 +20 = 30. Sum is 0 + 30 = 30 ≠ 0.

How about x = -4? As before, both are zero.

What about some other x? Let me take x = -3. First quadratic: (-3)^2 + 3*(-3) -4 = 9 -9 -4 = -4. Absolute value is 4. Second quadratic: (-3)^2 +9*(-3) +20 = 9 -27 +20 = 2. Absolute value is 2. Sum is 4 + 2 = 6 ≠ 0.

How about x = -6? First quadratic: (-6)^2 + 3*(-6) -4 = 36 -18 -4 = 14. Absolute value 14. Second quadratic: (-6)^2 +9*(-6) +20 = 36 -54 +20 = 2. Absolute value 2. Sum is 16 ≠ 0.

How about x = -2? First quadratic: (-2)^2 +3*(-2) -4 = 4 -6 -4 = -6. Absolute value 6. Second quadratic: (-2)^2 +9*(-2) +20 = 4 -18 +20 = 6. Absolute value 6. Sum 12 ≠ 0.

So, in all these test cases, the denominator isn't zero. Therefore, only x = -4.

But let me also check if there are any other common roots. Wait, the first quadratic is x² + 3x -4 = 0, and the second is x² + 9x +20 = 0. If I subtract the two equations, I can find if there's any common roots.

Let me write them:

Equation 1: x² + 3x - 4 = 0

Equation 2: x² + 9x + 20 = 0

Subtract equation 1 from equation 2:

(x² +9x +20) - (x² +3x -4) = 0 - 0 => 6x +24 = 0 => 6x = -24 => x = -4.

So, that gives x = -4, which is the common root. Therefore, that's the only common root. Therefore, the only x that makes both quadratics zero is x = -4. Therefore, only x = -4 is excluded.

Therefore, the answer is x = -4. Hence, the real value of x not in the domain is -4.

Therefore, the final answer is \boxed{-4}

**Final Answer**
The real value of $ x $ not in the domain is \boxed{-4}.

To determine the real values of $ x $ not in the domain of the function
$$
f(x) = \frac{1}{|x^2 + 3x - 4| + |x^2 + 9x + 20|},
$$
we must identify the values of $ x $ for which the **denominator is zero**—this is when the function is undefined.

---

### Step 1: Understand the structure of the denominator

The denominator is the sum of two absolute value expressions:

$$
|x^2 + 3x - 4| + |x^2 + 9x + 20|
$$

Since **absolute values are always non-negative**, the **sum cannot be negative**, and the only way the sum can be zero is if **both terms are zero** simultaneously. This means:

$$
|x^2 + 3x - 4| + |x^2 + 9x + 20| = 0 \iff x^2 + 3x - 4 = 0 \quad \text{and} \quad x^2 + 9x + 20 = 0
$$

So we are looking for **common roots** of these two quadratic equations.

---

### Step 2: Solve the quadratic equations

#### First equation: $ x^2 + 3x - 4 = 0 $

We factor:

$$
x^2 + 3x - 4 = (x + 4)(x - 1) = 0 \Rightarrow x = -4 \text{ or } x = 1
$$

#### Second equation: $ x^2 + 9x + 20 = 0 $

Again, we factor:

$$
x^2 + 9x + 20 = (x + 4)(x + 5) = 0 \Rightarrow x = -4 \text{ or } x = -5
$$

So the roots of the two quadratics are:

- First equation: $ x = -4, 1 $
- Second equation: $ x = -4, -5 $

---

### Step 3: Find common solutions

The values of $ x $ that make **both quadratic expressions zero** are the **common roots**. From above, the **only common root** is:

$$
x = -4
$$

So, only at $ x = -4 $, both absolute value expressions become zero, which makes the **denominator zero**, and the function $ f(x) $ is **undefined**.

---

### Step 4: Confirm that no other values cause the denominator to be zero

Because absolute values are non-negative, the **sum** can only be zero if **both terms are zero**. Therefore, any other value that makes **one of the terms zero** will not make the sum zero. So, apart from $ x = -4 $, these values are **not problematic**—the denominator is nonzero.

---

### Final Answer

The real value of $ x $ that is not in the domain of the function is:

$$
\boxed{-4}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the real values of x that are not in the domain of the function f(x) = 1 divided by the sum of two absolute value expressions: |x² + 3x - 4| + |x² + 9x + 20|. The question is asking for which x values this function is undefined, which would be when the denominator is zero. Since the denominator is a sum of absolute values, which are always non-negative, the only way the denominator can be zero is if both absolute value expressions are zero simultaneously. So, I need to find the x values that make both x² + 3x - 4 = 0 and x² + 9x + 20 = 0. Then, those x values would make the denominator zero, hence not in the domain of f(x).

First, let me solve each quadratic equation separately. Let me start with the first one: x² + 3x - 4 = 0. To solve this quadratic equation, I can factor it or use the quadratic formula. Let me try factoring. Looking for two numbers that multiply to -4 and add to 3. Let me think... 4 and -1? 4 * (-1) = -4, and 4 + (-1) = 3. Yes, that works. So, the equation factors to (x + 4)(x - 1) = 0. Therefore, the solutions are x = -4 and x = 1.

Now, the second quadratic equation: x² + 9x + 20 = 0. Again, let me try factoring. Looking for two numbers that multiply to 20 and add to 9. 4 and 5? 4 * 5 = 20, and 4 + 5 = 9. Perfect. So, this factors to (x + 4)(x + 5) = 0. Therefore, the solutions are x = -4 and x = -5.

So, the first equation has roots at x = -4 and x = 1, and the second equation has roots at x = -4 and x = -5. Now, the denominator is the sum of the absolute values of these quadratics. For the denominator to be zero, both absolute values must be zero. That would require both quadratics to be zero at the same x value. Let me check the solutions.

Looking at the roots of the first equation: x = -4 and x = 1. Let's check if these x-values are also roots of the second equation.

For x = -4: Let me plug into the second equation: (-4)² + 9*(-4) + 20 = 16 - 36 + 20 = 0. Yes, so x = -4 is a common root.

Now, for x = 1: Let me check in the second equation: (1)² + 9*(1) + 20 = 1 + 9 + 20 = 30 ≠ 0. So, x = 1 is not a root of the second equation. Similarly, check x = -5 in the first equation: (-5)^2 + 3*(-5) - 4 = 25 - 15 - 4 = 6 ≠ 0. So x = -5 is not a root of the first equation. Therefore, the only x that makes both quadratics zero is x = -4.

Therefore, the denominator is zero only when x = -4. Hence, the function f(x) is undefined at x = -4. Therefore, the real values of x not in the domain are just x = -4. Wait, but let me make sure I didn't miss anything.

Wait, could there be other x-values where the sum of the absolute values is zero? Well, the absolute value is always non-negative, so the sum of two absolute values is zero only if each absolute value is zero. So, the only way for |A| + |B| = 0 is if A = 0 and B = 0. Therefore, there's no other possibility. Therefore, only x = -4 makes both expressions zero, so only x = -4 is excluded from the domain. So, the answer should be x = -4.

But let me check again. Let me verify that for x = -4, the denominator is indeed zero. Let me compute each absolute value:

First, x² + 3x - 4 at x = -4: (-4)^2 + 3*(-4) - 4 = 16 -12 -4 = 0. Then, the second expression x² + 9x + 20 at x = -4: (-4)^2 + 9*(-4) + 20 = 16 -36 + 20 = 0. So yes, both are zero. Therefore, their sum is zero. Therefore, f(-4) is undefined.

But let me check if there are other x-values where denominator could be zero. Suppose, for example, that one of the absolute values is positive and the other is negative? Wait, no, absolute values are always non-negative. So, their sum can be zero only if both are zero. So, there's no way for the sum to be zero unless both individual terms are zero. Therefore, there are no other x-values.

But let me think again. Maybe there are x-values where even if each absolute value is not zero, their sum can be zero? But since absolute values are non-negative, the only way their sum is zero is if each is zero. So, no. Therefore, only x = -4.

But let me check for example, suppose that one term is positive and the other is negative. But absolute values can't be negative. So, each term is zero or positive. Therefore, the sum is zero only if both terms are zero. Therefore, the only x is -4.

Therefore, the answer is x = -4. Therefore, the real value not in the domain is -4.

Wait, but before I conclude, let me check if maybe the quadratics can be negative, but absolute values make them positive. But even so, their sum can't be zero unless both are zero.

Alternatively, is there a possibility that the denominator is zero for some other x? Let me take some other x-values and see. Let me pick x = 0. Then compute denominator:

First quadratic: 0 + 0 -4 = -4. Absolute value is 4. Second quadratic: 0 + 0 + 20 = 20. Absolute value is 20. Sum is 4 + 20 = 24 ≠ 0.

How about x = -5? Let's check denominator. First quadratic: (-5)^2 + 3*(-5) -4 = 25 -15 -4 = 6. Absolute value is 6. Second quadratic: (-5)^2 + 9*(-5) +20 = 25 -45 +20 = 0. So the sum is 6 + 0 = 6 ≠ 0. Therefore, denominator is non-zero.

Similarly, x = 1: first quadratic is zero, second quadratic is 1 + 9 +20 = 30. Sum is 0 + 30 = 30 ≠ 0.

How about x = -4? As before, both are zero.

What about some other x? Let me take x = -3. First quadratic: (-3)^2 + 3*(-3) -4 = 9 -9 -4 = -4. Absolute value is 4. Second quadratic: (-3)^2 +9*(-3) +20 = 9 -27 +20 = 2. Absolute value is 2. Sum is 4 + 2 = 6 ≠ 0.

How about x = -6? First quadratic: (-6)^2 + 3*(-6) -4 = 36 -18 -4 = 14. Absolute value 14. Second quadratic: (-6)^2 +9*(-6) +20 = 36 -54 +20 = 2. Absolute value 2. Sum is 16 ≠ 0.

How about x = -2? First quadratic: (-2)^2 +3*(-2) -4 = 4 -6 -4 = -6. Absolute value 6. Second quadratic: (-2)^2 +9*(-2) +20 = 4 -18 +20 = 6. Absolute value 6. Sum 12 ≠ 0.

So, in all these test cases, the denominator isn't zero. Therefore, only x = -4.

But let me also check if there are any other common roots. Wait, the first quadratic is x² + 3x -4 = 0, and the second is x² + 9x +20 = 0. If I subtract the two equations, I can find if there's any common roots.

Let me write them:

Equation 1: x² + 3x - 4 = 0

Equation 2: x² + 9x + 20 = 0

Subtract equation 1 from equation 2:

(x² +9x +20) - (x² +3x -4) = 0 - 0 => 6x +24 = 0 => 6x = -24 => x = -4.

So, that gives x = -4, which is the common root. Therefore, that's the only common root. Therefore, the only x that makes both quadratics zero is x = -4. Therefore, only x = -4 is excluded.

Therefore, the answer is x = -4. Hence, the real value of x not in the domain is -4.

Therefore, the final answer is \boxed{-4}

**Final Answer**
The real value of $ x $ not in the domain is \boxed{-4}.

To determine the real values of $ x $ not in the domain of the function
$$
f(x) = \frac{1}{|x^2 + 3x - 4| + |x^2 + 9x + 20|},
$$
we must identify the values of $ x $ for which the **denominator is zero**—this is when the function is undefined.

---

### Step 1: Understand the structure of the denominator

The denominator is the sum of two absolute value expressions:

$$
|x^2 + 3x - 4| + |x^2 + 9x + 20|
$$

Since **absolute values are always non-negative**, the **sum cannot be negative**, and the only way the sum can be zero is if **both terms are zero** simultaneously. This means:

$$
|x^2 + 3x - 4| + |x^2 + 9x + 20| = 0 \iff x^2 + 3x - 4 = 0 \quad \text{and} \quad x^2 + 9x + 20 = 0
$$

So we are looking for **common roots** of these two quadratic equations.

---

### Step 2: Solve the quadratic equations

#### First equation: $ x^2 + 3x - 4 = 0 $

We factor:

$$
x^2 + 3x - 4 = (x + 4)(x - 1) = 0 \Rightarrow x = -4 \text{ or } x = 1
$$

#### Second equation: $ x^2 + 9x + 20 = 0 $

Again, we factor:

$$
x^2 + 9x + 20 = (x + 4)(x + 5) = 0 \Rightarrow x = -4 \text{ or } x = -5
$$

So the roots of the two quadratics are:

- First equation: $ x = -4, 1 $
- Second equation: $ x = -4, -5 $

---

### Step 3: Find common solutions

The values of $ x $ that make **both quadratic expressions zero** are the **common roots**. From above, the **only common root** is:

$$
x = -4
$$

So, only at $ x = -4 $, both absolute value expressions become zero, which makes the **denominator zero**, and the function $ f(x) $ is **undefined**.

---

### Step 4: Confirm that no other values cause the denominator to be zero

Because absolute values are non-negative, the **sum** can only be zero if **both terms are zero**. Therefore, any other value that makes **one of the terms zero** will not make the sum zero. So, apart from $ x = -4 $, these values are **not problematic**—the denominator is nonzero.

---

### Final Answer

The real value of $ x $ that is not in the domain of the function is:

$$
\boxed{-4}
$$

Solution:
$x$ is not in the domain of $f$ if the denominator is zero.  Since both absolute values are nonnegative, both must be zero for the denominator to be zero.  So

\begin{align*}
0=x^2+3x-4=(x+4)(x-1)&\Rightarrow x=-4\text{ or }x=1\\
0=x^2+9x+20=(x+4)(x+5)&\Rightarrow x=-4\text{ or }x=-5
\end{align*}

The only value of $x$ which makes both absolute values zero is $x=\boxed{-4}$.