Explanation:
Okay, so I need to solve this equation: $x^{10} + (13x - 1)^{10} = 0$. The problem says it has 10 complex roots, which are paired as conjugates: $r_1, \overline{r}_1, r_2, \overline{r}_2,$ and so on up to $r_5, \overline{r}_5$. Then I need to find the sum $\frac{1}{r_1\overline{r}_1} + \frac{1}{r_2\overline{r}_2} + \cdots + \frac{1}{r_5\overline{r}_5}$.

First, let me recall that for any complex number $z$, the product $z \overline{z}$ is equal to $|z|^2$, the square of its modulus. So each term in the sum is $1/|r_i|^2$. Therefore, the problem is essentially asking for the sum of the reciprocals of the squared moduli of the roots.

But how do I find that? Maybe I can relate this to the coefficients of the polynomial? Since the polynomial is given, perhaps I can use Vieta's formulas or some other properties of polynomials. However, the polynomial is of degree 10, which is quite high. Maybe there's a way to simplify the equation first?

Let me start by rewriting the equation:

$x^{10} + (13x - 1)^{10} = 0$

This can be rewritten as:

$(13x - 1)^{10} = -x^{10}$

Taking both sides to the 1/10 power, but since we are dealing with complex numbers, there are multiple roots. However, maybe I can write this as:

$\left(\frac{13x - 1}{x}\right)^{10} = -1$

Assuming $x \neq 0$, which I should check later. Let me divide both sides by $x^{10}$:

$\left(13 - \frac{1}{x}\right)^{10} = -1$

Let me set $y = \frac{1}{x}$, so that the equation becomes:

$(13 - y)^{10} = -1$

Therefore, $13 - y$ is a 10th root of -1. Let me write that:

$13 - y = (-1)^{1/10}$

But the 10th roots of -1 are the complex numbers $e^{i(\pi + 2\pi k)/10}$ for $k = 0, 1, ..., 9$. So, each root $y$ is given by:

$y = 13 - e^{i(\pi + 2\pi k)/10}$ for $k = 0, 1, ..., 9$

But since $y = 1/x$, then $x = 1/y = \frac{1}{13 - e^{i(\pi + 2\pi k)/10}}$

But maybe I can find the modulus squared of each root $x$, since we need $1/(r_i \overline{r_i}) = 1/|x|^2$. So if I can find $|x|^2$, then take reciprocal and sum them up.

Alternatively, since $x = 1/y$, then $|x|^2 = 1/|y|^2$, so $1/|x|^2 = |y|^2$. Therefore, the sum we need is the sum of $|y|^2$ over all roots $y$.

So, if I can compute the sum of $|y|^2$ for each root $y$, that would be the answer.

But how do I compute that? Let me recall that for a complex number $y$, $|y|^2 = y \overline{y}$. Therefore, the sum over all roots $y$ of $y \overline{y}$ is equal to the sum over all roots $y$ of $y \overline{y}$. However, since the roots $y$ come in conjugate pairs? Wait, the original equation is $(13 - y)^{10} = -1$. Let me check if the roots $y$ are conjugate pairs.

If $y$ is a root, then $13 - y$ is a 10th root of -1. Let me denote $\omega = e^{i(\pi + 2\pi k)/10}$, so that $13 - y = \omega$. Then, if $y$ is a root, then $\overline{y}$ would correspond to $\overline{13 - \omega} = 13 - \overline{\omega}$, since 13 is real. Therefore, if $\omega$ is a root, then $\overline{\omega}$ is also a root? Wait, but the equation is $(13 - y)^{10} = -1$. Let me check if if $y$ is a root, then $\overline{y}$ is also a root?

Suppose $y$ satisfies $(13 - y)^{10} = -1$. Then taking conjugate of both sides:

$\overline{(13 - y)^{10}} = \overline{-1}$

Which is $(\overline{13 - y})^{10} = -1$, since modulus squared is 1 for roots of unity? Wait, no. Wait, the conjugate of $-1$ is $-1$, since it's real. So:

$(\overline{13 - y})^{10} = -1$

But $\overline{13 - y} = 13 - \overline{y}$, since 13 is real. Therefore, $13 - \overline{y}$ is also a 10th root of -1. Therefore, if $y$ is a root, then $\overline{y}$ is also a root. Therefore, the roots $y$ come in conjugate pairs. Therefore, the roots $y$ are paired as $y$ and $\overline{y}$, so their moduli squared are equal. Therefore, the sum of $|y|^2$ over all roots is equal to twice the sum over each pair. However, since there are 10 roots, which is even, so 5 pairs. Therefore, the sum we need is the sum over each pair of $|y|^2 + |\overline{y}|^2 = 2|y|^2$. Wait, but since $|y|^2 = |\overline{y}|^2$, so each pair contributes twice $|y|^2$. But since we have 5 pairs, the total sum would be 2*(sum over each pair's |y|^2). Wait, but actually, if we have 10 roots, each pair contributes two terms, so the total sum is sum_{i=1}^{10} |y_i|^2 = 2 * sum_{i=1}^5 |y_i|^2. But since each pair is (y, \overline{y}), so each pair contributes |y|^2 + |\overline{y}|^2 = 2|y|^2. Therefore, the total sum is 2*(sum_{i=1}^5 |y_i|^2). Wait, but maybe I need to think differently. Wait, if there are 10 roots, and they come in 5 conjugate pairs, then the total sum is sum_{i=1}^5 (|y_i|^2 + |\overline{y_i}|^2) = sum_{i=1}^5 2|y_i|^2. Therefore, the total sum is 2*(sum_{i=1}^5 |y_i|^2). However, the problem is asking for sum_{i=1}^5 1/(r_i \overline{r_i}) = sum_{i=1}^5 1/|r_i|^2 = sum_{i=1}^5 |y_i|^2. Wait, because earlier we had that 1/|x|^2 = |y|^2. Therefore, the answer is sum_{i=1}^5 |y_i|^2. But according to the above, the total sum over all roots y is 2*(sum_{i=1}^5 |y_i|^2). Therefore, if I can compute the sum over all roots y of |y|^2, then divide by 2, I get the desired answer.

Alternatively, maybe there's a way to compute the sum of |y|^2 for all roots y. How can I compute that?

I remember that for a polynomial, the sum of |y_i|^2 can be related to the coefficients. Let me recall that for a polynomial with roots y_i, the sum of |y_i|^2 is equal to the sum of y_i \overline{y_i} = sum y_i \overline{y_i}. However, since the coefficients of the polynomial are real? Wait, let me check. The original equation was $(13 - y)^{10} = -1$. Let me expand this equation.

Let me write the equation as $(13 - y)^{10} + 1 = 0$. Therefore, the polynomial equation is $P(y) = (13 - y)^{10} + 1 = 0$. Let me expand this polynomial.

But expanding $(13 - y)^{10}$ would be tedious, but maybe I can write it as $(-1)^{10}(y - 13)^{10} + 1 = 0$, which is $(y - 13)^{10} + 1 = 0$. Therefore, the polynomial is $P(y) = (y - 13)^{10} + 1$. Therefore, the roots of P(y) are the solutions to $(y - 13)^{10} = -1$, which is the same as before.

So, the polynomial is $P(y) = (y - 13)^{10} + 1$. Let me denote $z = y - 13$, so that the equation becomes $z^{10} + 1 = 0$, so $z^{10} = -1$. Therefore, the roots z are the 10th roots of -1, which are $z_k = e^{i(\pi + 2\pi k)/10}$ for k = 0, 1, ..., 9. Therefore, the roots y are $y_k = 13 + z_k = 13 + e^{i(\pi + 2\pi k)/10}$.

Therefore, each root y is $13 + e^{i\theta_k}$, where $\theta_k = (\pi + 2\pi k)/10$. Therefore, to compute |y_k|^2, we can compute |13 + e^{i\theta_k}|^2.

So, |y_k|^2 = |13 + e^{i\theta_k}|^2 = (13 + e^{i\theta_k})(13 + e^{-i\theta_k}) = 13^2 + 13 e^{-i\theta_k} + 13 e^{i\theta_k} + 1 = 169 + 13(e^{i\theta_k} + e^{-i\theta_k}) + 1 = 170 + 26 \cos \theta_k.

Therefore, |y_k|^2 = 170 + 26 \cos \theta_k.

Therefore, the sum over all roots y of |y|^2 is sum_{k=0}^9 [170 + 26 \cos \theta_k] = 10*170 + 26 * sum_{k=0}^9 \cos \theta_k.

Therefore, the total sum is 1700 + 26 * sum_{k=0}^9 \cos \theta_k.

But earlier, we saw that the answer we need is sum_{i=1}^5 |y_i|^2, which is half of the total sum over all roots y (since each pair contributes twice). Wait, no. Wait, the total sum over all roots y is sum_{k=0}^9 |y_k|^2. But since the roots come in conjugate pairs, each pair (y_k, y_{9 - k})? Wait, let me check.

The roots z_k are e^{i(\pi + 2\pi k)/10} for k = 0,...,9. Let me note that for each k, the angle theta_k = (\pi + 2\pi k)/10. Let me compute theta_k for k from 0 to 9:

For k=0: theta_0 = pi/10

k=1: (pi + 2pi)/10 = 3pi/10

k=2: (pi + 4pi)/10 = 5pi/10 = pi/2

k=3: (pi + 6pi)/10 = 7pi/10

k=4: (pi + 8pi)/10 = 9pi/10

k=5: (pi + 10pi)/10 = 11pi/10

k=6: (pi + 12pi)/10 = 13pi/10

k=7: (pi + 14pi)/10 = 15pi/10 = 3pi/2

k=8: (pi + 16pi)/10 = 17pi/10

k=9: (pi + 18pi)/10 = 19pi/10

So, the angles theta_k are pi/10, 3pi/10, pi/2, 7pi/10, 9pi/10, 11pi/10, 13pi/10, 3pi/2, 17pi/10, 19pi/10.

Now, note that cos(theta_k) for k and 10 - k - something? Let me check for conjugate pairs. Since the roots z_k are e^{i theta_k}, their conjugates are e^{-i theta_k} = e^{i( - theta_k)}. However, since theta_k is between pi/10 and 19pi/10, the conjugate of z_k is e^{-i theta_k} which would correspond to some other theta_j. Let me check for k and j such that theta_j = -theta_k + 2pi m for some integer m.

Alternatively, since the roots y are 13 + z_k, and their conjugates are 13 + \overline{z_k} = 13 + e^{-i theta_k}. Therefore, if theta_k is one angle, then theta_j = -theta_k + 2pi m would be another angle. Let me check for each k:

For example, take k=0: theta_0 = pi/10. Then -theta_0 = -pi/10. Adding 2pi to get it into the range [0, 2pi): 19pi/10. Which is theta_9. Similarly, for k=1: theta_1 = 3pi/10, then -theta_1 = -3pi/10, adding 2pi gives 17pi/10, which is theta_8. Similarly, k=2: theta_2 = pi/2, -theta_2 = -pi/2, adding 2pi gives 3pi/2, which is theta_7. Similarly, k=3: theta_3 = 7pi/10, -theta_3 = -7pi/10, adding 2pi gives 13pi/10, which is theta_6. Similarly, k=4: theta_4 = 9pi/10, -theta_4 = -9pi/10, adding 2pi gives 11pi/10, which is theta_5.

Therefore, the roots z_k and z_{9 - k} are conjugates. Therefore, the roots y_k and y_{9 - k} are conjugates. Therefore, their |y_k|^2 and |y_{9 - k}|^2 are equal. Therefore, the sum over all roots y of |y|^2 is 2*(sum_{k=0}^4 |y_k|^2). Therefore, the total sum is 2*(sum_{k=0}^4 |y_k|^2). Therefore, the sum we need, which is sum_{i=1}^5 |y_i|^2, is equal to sum_{k=0}^4 |y_k|^2. Therefore, the total sum over all roots is 2*(sum_{i=1}^5 |y_i|^2). Therefore, if I compute the total sum as 1700 + 26 * sum_{k=0}^9 cos(theta_k), then divide by 2, I get the desired answer.

So, let me compute sum_{k=0}^9 cos(theta_k). Let me note that theta_k for k=0,...,9 are pi/10, 3pi/10, pi/2, 7pi/10, 9pi/10, 11pi/10, 13pi/10, 3pi/2, 17pi/10, 19pi/10.

But note that cos(theta) is even, so cos(theta) = cos(-theta) = cos(2pi - theta). Therefore, cos(theta_k) for k=0,...,9 can be paired as follows:

cos(pi/10) and cos(19pi/10) = cos(pi/10) and cos(-pi/10) = cos(pi/10) and cos(pi/10) = 2cos(pi/10)

Similarly, cos(3pi/10) and cos(17pi/10) = cos(3pi/10) and cos(-3pi/10) = 2cos(3pi/10)

cos(pi/2) and cos(3pi/2) = cos(pi/2) and cos(3pi/2) = 0 + 0 = 0

cos(7pi/10) and cos(13pi/10) = cos(7pi/10) and cos(13pi/10) = cos(7pi/10) and cos(13pi/10). But 13pi/10 = pi + 3pi/10, so cos(13pi/10) = cos(pi + 3pi/10) = -cos(3pi/10). Similarly, cos(7pi/10) = cos(pi - 3pi/10) = -cos(3pi/10). Therefore, cos(7pi/10) + cos(13pi/10) = -cos(3pi/10) - cos(3pi/10) = -2cos(3pi/10). Wait, but earlier I thought they would pair to something else. Wait, maybe I need to check again.

Wait, theta_3 = 7pi/10, theta_6 = 13pi/10. So cos(theta_3) + cos(theta_6) = cos(7pi/10) + cos(13pi/10). Let me compute cos(7pi/10) = cos(pi - 3pi/10) = -cos(3pi/10). Similarly, cos(13pi/10) = cos(pi + 3pi/10) = -cos(3pi/10). Therefore, their sum is -cos(3pi/10) - cos(3pi/10) = -2cos(3pi/10). Similarly, theta_4 = 9pi/10, theta_5 = 11pi/10. cos(theta_4) + cos(theta_5) = cos(9pi/10) + cos(11pi/10). cos(9pi/10) = cos(pi - pi/10) = -cos(pi/10). cos(11pi/10) = cos(pi + pi/10) = -cos(pi/10). Therefore, their sum is -cos(pi/10) - cos(pi/10) = -2cos(pi/10).

Therefore, the total sum of cos(theta_k) from k=0 to 9 is:

[cos(pi/10) + cos(19pi/10)] + [cos(3pi/10) + cos(17pi/10)] + [cos(pi/2) + cos(3pi/2)] + [cos(7pi/10) + cos(13pi/10)] + [cos(9pi/10) + cos(11pi/10)]

Which is:

[2cos(pi/10)] + [2cos(3pi/10)] + [0 + 0] + [-2cos(3pi/10)] + [-2cos(pi/10)] = 2cos(pi/10) + 2cos(3pi/10) - 2cos(3pi/10) - 2cos(pi/10) = 0.

Wait, that's zero? So the total sum of cos(theta_k) from k=0 to 9 is zero? That seems surprising, but let me check again.

Yes, because the sum of cos(theta_k) over all roots of z^{10} = -1. Wait, but z^{10} = -1, so the roots are the 10th roots of -1. However, the sum of all roots of z^{10} + 1 = 0 is zero, because the coefficient of z^9 is zero. Wait, but the sum of the roots of z^{10} + 1 = 0 is zero. However, the sum of the roots is sum_{k=0}^9 z_k = 0. But the sum of cos(theta_k) is the real part of sum_{k=0}^9 z_k, which is zero. Therefore, sum_{k=0}^9 cos(theta_k) = 0. Therefore, the total sum is 1700 + 26 * 0 = 1700. Therefore, the total sum over all roots y of |y|^2 is 1700. Therefore, the sum we need, which is sum_{i=1}^5 |y_i|^2, is half of that, which is 1700 / 2 = 850.

Wait, but hold on. Earlier, I thought that the total sum over all roots y is 2*(sum_{i=1}^5 |y_i|^2). Therefore, if the total sum is 1700, then sum_{i=1}^5 |y_i|^2 = 1700 / 2 = 850. Therefore, the answer is 850?

But let me check again. Let me recap:

We have the polynomial P(y) = (y - 13)^{10} + 1 = 0. The roots y are 13 + z_k where z_k^{10} = -1. Then, |y_k|^2 = 170 + 26 cos(theta_k). Therefore, sum_{k=0}^9 |y_k|^2 = 10*170 + 26*sum cos(theta_k) = 1700 + 0 = 1700.

But since the roots y come in conjugate pairs, each pair contributes |y_k|^2 + |y_{9 -k}|^2 = 2|y_k|^2. Therefore, the total sum is 2*(sum_{k=0}^4 |y_k|^2). Therefore, sum_{k=0}^4 |y_k|^2 = 1700 / 2 = 850. Therefore, the answer is 850.

But wait, the problem says that the original equation has roots r_1, \overline{r_1}, ..., r_5, \overline{r_5}, and we need to compute sum_{i=1}^5 1/(r_i \overline{r_i}) = sum_{i=1}^5 1/|r_i|^2. But earlier, we had that 1/|r_i|^2 = |y_i|^2. Therefore, the answer is sum_{i=1}^5 |y_i|^2 = 850. Therefore, the answer is 850?

But let me check with an example. Let me take a simpler case. Suppose I have a quadratic equation. Let me see if my reasoning holds.

Suppose I have equation x^2 + (ax - b)^2 = 0. Then similar steps would lead to something. But maybe that's too time-consuming. Alternatively, let me check if there's another way to compute the sum of 1/|r_i|^2.

Alternatively, since the original equation is x^{10} + (13x - 1)^{10} = 0. Let me consider that equation. Let me write it as x^{10} = - (13x - 1)^{10}. Taking modulus on both sides, |x|^{10} = |13x - 1|^{10}, so |x| = |13x - 1|. Therefore, for each root x, |x| = |13x - 1|. Therefore, |x|^2 = |13x - 1|^2.

But |13x - 1|^2 = (13x - 1)(\overline{13x} - 1) = 169 |x|^2 - 13x - 13 \overline{x} + 1.

Therefore, |x|^2 = 169 |x|^2 - 13(x + \overline{x}) + 1.

Rearranging:

0 = 168 |x|^2 - 13(x + \overline{x}) + 1.

But x + \overline{x} = 2 Re(x). Let me denote |x|^2 = r^2, and Re(x) = a. Then:

0 = 168 r^2 - 26 a + 1.

But also, since |x|^2 = r^2 = a^2 + b^2, where x = a + ib. However, this seems complicated. However, if I can find the sum over all roots of 1/r^2, which is the same as sum over all roots of |y|^2, which we found as 850.

Alternatively, maybe I can use Vieta's formula. Let me think.

But the original equation is x^{10} + (13x - 1)^{10} = 0. Let me expand (13x - 1)^{10} using binomial theorem. However, expanding this would be tedious, but maybe I can write the equation as:

x^{10} + \sum_{k=0}^{10} \binom{10}{k} (13x)^{10 - k} (-1)^k = 0.

But this seems complicated. However, maybe I can write the equation as:

x^{10} + (13x - 1)^{10} = 0 => (13x - 1)^{10} = -x^{10} => [(13x - 1)/x]^{10} = -1 => [13 - 1/x]^{10} = -1.

Which is the same as before. Therefore, leading to y = 1/x, and so on.

Alternatively, since we have that for each root x, |x| = |13x - 1|. Therefore, |x|^2 = |13x - 1|^2. Let me write this equation again:

|x|^2 = |13x - 1|^2 => |x|^2 = 169 |x|^2 - 26 Re(13x) + 1 => 0 = 168 |x|^2 - 26 Re(13x) + 1 => 168 |x|^2 - 26 * 13 Re(x) + 1 = 0 => 168 |x|^2 - 338 Re(x) + 1 = 0.

But Re(x) = (x + \overline{x}) / 2. However, if I denote for each root x, let me denote |x|^2 = s, and Re(x) = a. Then:

168 s - 338 a + 1 = 0 => 168 s = 338 a - 1 => s = (338 a - 1)/168.

But how does this help me? If I can find the sum over all roots of 1/s, which is the sum of 1/|x|^2.

But I need to relate this to Vieta's formula. However, since the equation is of degree 10, and Vieta's formula relates sums of roots, products, etc. However, the equation is not in standard form. Let me write the original equation again:

x^{10} + (13x - 1)^{10} = 0. Let me expand (13x - 1)^{10}:

(13x - 1)^{10} = \sum_{k=0}^{10} \binom{10}{k} (13x)^{k} (-1)^{10 - k} = \sum_{k=0}^{10} \binom{10}{k} (-1)^{10 - k} 13^k x^k.

Therefore, the original equation is:

x^{10} + \sum_{k=0}^{10} \binom{10}{k} (-1)^{10 - k} 13^k x^k = 0.

But note that when k=10, the term is \binom{10}{10} (-1)^{0} 13^{10} x^{10} = 13^{10} x^{10}. Therefore, combining with the x^{10} term:

x^{10} + 13^{10} x^{10} + ... = 0 => (1 + 13^{10})x^{10} + ... = 0.

But this seems messy. However, perhaps I can write the equation as:

(13x - 1)^{10} = -x^{10} => [(13x - 1)/x]^{10} = -1 => [13 - 1/x]^{10} = -1.

Which is the same as before. Therefore, leading to y = 1/x, and the roots y satisfy (13 - y)^{10} = -1. Therefore, the polynomial equation for y is (13 - y)^{10} + 1 = 0.

Therefore, the polynomial is P(y) = (y - 13)^{10} + 1. Therefore, the roots y are 13 + z_k where z_k^{10} = -1.

Therefore, the sum we need is sum_{k=0}^9 |y_k|^2 = 1700, and since the answer is half of that, 850.

But let me check with specific values. Let me take a simple case where I can compute manually. For example, suppose instead of 10th roots, take a quadratic equation. Let me try with degree 2.

Suppose equation is x^2 + (ax - b)^2 = 0. Then similar steps would lead to something. Let me take a=1, b=1. Then equation is x^2 + (x - 1)^2 = 0. Expanding: x^2 + x^2 - 2x + 1 = 0 => 2x^2 - 2x + 1 = 0. The roots are [2 ± sqrt(4 - 8)] / 4 = [2 ± sqrt(-4)] / 4 = [1 ± i]/2. Therefore, roots are (1 + i)/2 and (1 - i)/2. Then |x|^2 for each root is (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2. Therefore, sum of 1/|x|^2 is 2/(1/2) = 4.

Alternatively, using our previous method: Let me write equation as x^2 + (x - 1)^2 = 0 => (x - 1)^2 = -x^2 => [ (x - 1)/x ]^2 = -1 => [1 - 1/x]^2 = -1 => 1 - 2/x + 1/x^2 = -1 => 1/x^2 - 2/x + 2 = 0. Let y = 1/x, then equation is y^2 - 2y + 2 = 0. The roots y are [2 ± sqrt(4 - 8)] / 2 = [2 ± sqrt(-4)] / 2 = 1 ± i. Therefore, |y|^2 = 1^2 + 1^2 = 2. Therefore, sum of |y|^2 is 2 + 2 = 4, which matches the previous result. Therefore, the answer would be 4.

But according to our previous method: For the equation (y - 13)^{10} + 1 = 0, but in this case, the equation is (y - 1)^2 + 1 = 0? Wait, no. Wait, in this case, the equation for y is y^2 - 2y + 2 = 0. Which is different. However, if I follow the same steps as before:

Let me write the equation as (y - a)^n + 1 = 0, where a is some constant. In this case, n=2, a=1. Then the roots y are 1 + z_k where z_k^2 = -1. Therefore, z_k = i, -i. Therefore, y = 1 + i, 1 - i. Then |y|^2 = (1)^2 + (1)^2 = 2 for each, so sum is 4. Which matches.

Therefore, in this case, the total sum over all roots y of |y|^2 is 4, which is equal to 2*(sum_{i=1}^1 |y_i|^2) since there are two roots. Therefore, the answer is 4. Which is correct.

Therefore, returning to our original problem, the answer is 850. Therefore, I think that is correct.

But let me check another way. Let me think about Vieta's formula.

We have the polynomial P(y) = (y - 13)^{10} + 1. Let me expand this polynomial.

Let me denote P(y) = (y - 13)^{10} + 1. Let me write it as y^{10} - 10*13 y^9 + ... + (-13)^{10} + 1. However, expanding this is tedious, but maybe I can find the sum of |y_i|^2 using Vieta's formula.

But how? The sum of |y_i|^2 is sum_{i=1}^{10} |y_i|^2. For complex numbers, this is equal to sum_{i=1}^{10} y_i \overline{y_i}. However, since the coefficients of the polynomial are real, the roots come in conjugate pairs. Therefore, if y is a root, then \overline{y} is also a root. Therefore, the sum becomes sum_{i=1}^{10} y_i \overline{y_i} = sum_{i=1}^{5} (y_i \overline{y_i} + \overline{y_i} y_i) = 2 sum_{i=1}^5 |y_i|^2.

But how to compute sum_{i=1}^{10} y_i \overline{y_i}?

Alternatively, note that for any complex number y, y \overline{y} = |y|^2. Therefore, sum_{i=1}^{10} |y_i|^2 = sum_{i=1}^{10} y_i \overline{y_i}.

But how to compute this sum?

Let me recall that for a polynomial with roots y_i, we have that sum_{i=1}^n y_i = - coefficient of y^{n-1} / coefficient of y^n. Similarly, sum_{i=1}^n y_i y_j = coefficient of y^{n-2} / coefficient of y^n, etc. However, these are for sums of products of roots, not for sums of |y_i|^2.

But perhaps we can relate sum |y_i|^2 to the coefficients.

Note that sum |y_i|^2 = sum y_i \overline{y_i} = sum y_i \overline{y_i}. However, since the polynomial has real coefficients, the conjugate of each root is also a root. Therefore, if we consider the polynomial with roots y_i, then the polynomial with roots \overline{y_i} is the same as the original polynomial. Therefore, the product of the polynomial evaluated at y and its conjugate? Not sure.

Alternatively, consider that if P(y) = \prod_{i=1}^{10} (y - y_i), then the polynomial with roots \overline{y_i} is \prod_{i=1}^{10} (y - \overline{y_i}) = \overline{P(\overline{y})} if P has real coefficients. Wait, maybe not.

Alternatively, consider that sum_{i=1}^{10} y_i \overline{y_i} = sum_{i=1}^{10} y_i \overline{y_i} = sum_{i=1}^{10} y_i \overline{y_i} = sum_{i=1}^{10} y_i \overline{y_i}.

But since the polynomial has real coefficients, then for each root y_i, \overline{y_i} is also a root. Therefore, we can pair them as y_i and \overline{y_i}. Therefore, sum_{i=1}^{10} y_i \overline{y_i} = sum_{j=1}^5 (y_j \overline{y_j} + \overline{y_j} y_j) = 2 sum_{j=1}^5 |y_j|^2.

But how to compute this sum?

Alternatively, consider that for each root y_i, we have that P(y_i) = 0. Therefore, (y_i - 13)^{10} = -1. Therefore, taking modulus squared on both sides: |y_i - 13|^{20} = 1. Therefore, |y_i - 13|^{2} = 1^{1/10} = 1? Wait, no. Wait, modulus squared of both sides: |(y_i - 13)^{10}|^2 = |-1|^2 => |y_i - 13|^{20} = 1. Therefore, |y_i - 13|^{2} = 1^{1/10} = 1? Wait, no. Wait, modulus squared of (y_i - 13)^{10} is |y_i - 13|^{20}, and modulus squared of -1 is 1. Therefore, |y_i - 13|^{20} = 1 => |y_i - 13| = 1^{1/10} = 1. Therefore, |y_i - 13| = 1. Therefore, for each root y_i, the distance from y_i to 13 is 1. Therefore, all roots lie on a circle of radius 1 centered at 13 in the complex plane.

Therefore, |y_i - 13| = 1 => |y_i - 13|^2 = 1 => (y_i - 13)(\overline{y_i} - 13) = 1 => y_i \overline{y_i} - 13 y_i - 13 \overline{y_i} + 169 = 1 => |y_i|^2 - 13(y_i + \overline{y_i}) + 169 = 1 => |y_i|^2 - 13*2 Re(y_i) + 168 = 0 => |y_i|^2 = 26 Re(y_i) - 168.

Therefore, for each root y_i, |y_i|^2 = 26 Re(y_i) - 168. Therefore, the sum over all roots of |y_i|^2 is sum_{i=1}^{10} [26 Re(y_i) - 168] = 26 sum Re(y_i) - 168*10.

But sum Re(y_i) is the real part of sum y_i.

From Vieta's formula, for the polynomial P(y) = (y - 13)^{10} + 1 = y^{10} - 10*13 y^9 + ... + (-13)^{10} + 1. The coefficient of y^9 is -10*13, so sum y_i = 10*13 = 130. Therefore, sum Re(y_i) = Re(sum y_i) = Re(130) = 130.

Therefore, sum |y_i|^2 = 26*130 - 168*10 = 3380 - 1680 = 1700. Which matches our previous result. Therefore, the total sum is 1700, so the answer is 1700 / 2 = 850.

Therefore, this confirms that the answer is 850.

Therefore, the value of the given sum is 850.

**Final Answer**
\boxed{850}

To solve the equation

$$
x^{10} + (13x - 1)^{10} = 0,
$$

we begin by analyzing the structure of the roots. The equation is a polynomial of degree 10, and the problem states that it has 10 complex roots, which are paired as conjugates: $ r_1, \overline{r}_1, r_2, \overline{r}_2, \ldots, r_5, \overline{r}_5 $.

We are asked to compute the sum:

$$
\frac{1}{r_1\overline{r}_1} + \frac{1}{r_2\overline{r}_2} + \cdots + \frac{1}{r_5\overline{r}_5}.
$$

---

### Step 1: Understanding the Expression

For any complex number $ z $, the product $ z\overline{z} = |z|^2 $, the square of its modulus. Therefore, the expression becomes:

$$
\sum_{i=1}^5 \frac{1}{|r_i|^2}.
$$

Let us denote $ x = r_i $, so the expression becomes:

$$
\sum_{i=1}^5 \frac{1}{|x_i|^2}.
$$

---

### Step 2: Transforming the Equation

We rewrite the original equation:

$$
x^{10} + (13x - 1)^{10} = 0 \Rightarrow (13x - 1)^{10} = -x^{10}.
$$

Dividing both sides by $ x^{10} $ (assuming $ x \neq 0 $), we get:

$$
\left(13 - \frac{1}{x}\right)^{10} = -1.
$$

Let $ y = \frac{1}{x} $, so the equation becomes:

$$
(13 - y)^{10} = -1.
$$

This implies that $ 13 - y $ is a 10th root of $ -1 $, i.e.,

$$
13 - y = e^{i(\pi + 2\pi k)/10}, \quad k = 0, 1, \ldots, 9.
$$

Thus, the roots $ y $ are:

$$
y_k = 13 - e^{i(\pi + 2\pi k)/10}, \quad k = 0, 1, \ldots, 9.
$$

Since $ x = \frac{1}{y} $, the roots $ x $ are:

$$
x_k = \frac{1}{13 - e^{i(\pi + 2\pi k)/10}}.
$$

---

### Step 3: Computing the Desired Sum

We are to compute:

$$
\sum_{i=1}^5 \frac{1}{|r_i|^2} = \sum_{i=1}^5 |y_i|^2,
$$

since $ \frac{1}{|x_i|^2} = |y_i|^2 $.

Let us compute $ |y_k|^2 $ for each root $ y_k = 13 - e^{i\theta_k} $, where $ \theta_k = \frac{\pi + 2\pi k}{10} $.

We use the identity:

$$
|y_k|^2 = |13 - e^{i\theta_k}|^2 = (13 - e^{i\theta_k})(13 - e^{-i\theta_k}) = 169 - 13(e^{i\theta_k} + e^{-i\theta_k}) + 1 = 170 - 26\cos\theta_k.
$$

So:

$$
|y_k|^2 = 170 - 26\cos\theta_k.
$$

Now, summing over all 10 roots:

$$
\sum_{k=0}^9 |y_k|^2 = \sum_{k=0}^9 (170 - 26\cos\theta_k) = 10 \cdot 170 - 26 \sum_{k=0}^9 \cos\theta_k = 1700 - 26 \sum_{k=0}^9 \cos\theta_k.
$$

But the roots $ e^{i\theta_k} $ are the 10th roots of $ -1 $, and the sum of all such roots is zero (since the sum of all roots of a polynomial is given by the negative of the coefficient of the second-highest degree term, which is zero in this case). Therefore, the sum of their real parts is also zero:

$$
\sum_{k=0}^9 \cos\theta_k = 0.
$$

Thus:

$$
\sum_{k=0}^9 |y_k|^2 = 1700.
$$

Since the roots $ y_k $ come in conjugate pairs, and each pair contributes $ |y_k|^2 + |\overline{y_k}|^2 = 2|y_k|^2 $, the sum over all 10 roots is twice the sum over the 5 distinct pairs. Therefore, the sum over the 5 distinct $ |y_i|^2 $ is:

$$
\sum_{i=1}^5 |y_i|^2 = \frac{1}{2} \sum_{k=0}^9 |y_k|^2 = \frac{1700}{2} = 850.
$$

---

### Final Answer

$$
\boxed{850}
$$

[Assistant]
Explanation:
Okay, so I need to solve this equation: $x^{10} + (13x - 1)^{10} = 0$. The problem says it has 10 complex roots, which are paired as conjugates: $r_1, \overline{r}_1, r_2, \overline{r}_2,$ and so on up to $r_5, \overline{r}_5$. Then I need to find the sum $\frac{1}{r_1\overline{r}_1} + \frac{1}{r_2\overline{r}_2} + \cdots + \frac{1}{r_5\overline{r}_5}$.

First, let me recall that for any complex number $z$, the product $z \overline{z}$ is equal to $|z|^2$, the square of its modulus. So each term in the sum is $1/|r_i|^2$. Therefore, the problem is essentially asking for the sum of the reciprocals of the squared moduli of the roots.

But how do I find that? Maybe I can relate this to the coefficients of the polynomial? Since the polynomial is given, perhaps I can use Vieta's formulas or some other properties of polynomials. However, the polynomial is of degree 10, which is quite high. Maybe there's a way to simplify the equation first?

Let me start by rewriting the equation:

$x^{10} + (13x - 1)^{10} = 0$

This can be rewritten as:

$(13x - 1)^{10} = -x^{10}$

Taking both sides to the 1/10 power, but since we are dealing with complex numbers, there are multiple roots. However, maybe I can write this as:

$\left(\frac{13x - 1}{x}\right)^{10} = -1$

Assuming $x \neq 0$, which I should check later. Let me divide both sides by $x^{10}$:

$\left(13 - \frac{1}{x}\right)^{10} = -1$

Let me set $y = \frac{1}{x}$, so that the equation becomes:

$(13 - y)^{10} = -1$

Therefore, $13 - y$ is a 10th root of -1. Let me write that:

$13 - y = (-1)^{1/10}$

But the 10th roots of -1 are the complex numbers $e^{i(\pi + 2\pi k)/10}$ for $k = 0, 1, ..., 9$. So, each root $y$ is given by:

$y = 13 - e^{i(\pi + 2\pi k)/10}$ for $k = 0, 1, ..., 9$

But since $y = 1/x$, then $x = 1/y = \frac{1}{13 - e^{i(\pi + 2\pi k)/10}}$

But maybe I can find the modulus squared of each root $x$, since we need $1/(r_i \overline{r_i}) = 1/|x|^2$. So if I can find $|x|^2$, then take reciprocal and sum them up.

Alternatively, since $x = 1/y$, then $|x|^2 = 1/|y|^2$, so $1/|x|^2 = |y|^2$. Therefore, the sum we need is the sum of $|y|^2$ over all roots $y$.

So, if I can compute the sum of $|y|^2$ for each root $y$, that would be the answer.

But how do I compute that? Let me recall that for a complex number $y$, $|y|^2 = y \overline{y}$. Therefore, the sum over all roots $y$ of $y \overline{y}$ is equal to the sum over all roots $y$ of $y \overline{y}$. However, since the roots $y$ come in conjugate pairs? Wait, the original equation is $(13 - y)^{10} = -1$. Let me check if the roots $y$ are conjugate pairs.

If $y$ is a root, then $13 - y$ is a 10th root of -1. Let me denote $\omega = e^{i(\pi + 2\pi k)/10}$, so that $13 - y = \omega$. Then, if $y$ is a root, then $\overline{y}$ would correspond to $\overline{13 - \omega} = 13 - \overline{\omega}$, since 13 is real. Therefore, if $\omega$ is a root, then $\overline{\omega}$ is also a root? Wait, but the equation is $(13 - y)^{10} = -1$. Let me check if if $y$ is a root, then $\overline{y}$ is also a root?

Suppose $y$ satisfies $(13 - y)^{10} = -1$. Then taking conjugate of both sides:

$\overline{(13 - y)^{10}} = \overline{-1}$

Which is $(\overline{13 - y})^{10} = -1$, since modulus squared is 1 for roots of unity? Wait, no. Wait, the conjugate of $-1$ is $-1$, since it's real. So:

$(\overline{13 - y})^{10} = -1$

But $\overline{13 - y} = 13 - \overline{y}$, since 13 is real. Therefore, $13 - \overline{y}$ is also a 10th root of -1. Therefore, if $y$ is a root, then $\overline{y}$ is also a root. Therefore, the roots $y$ come in conjugate pairs. Therefore, the roots $y$ are paired as $y$ and $\overline{y}$, so their moduli squared are equal. Therefore, the sum of $|y|^2$ over all roots is equal to twice the sum over each pair. However, since there are 10 roots, which is even, so 5 pairs. Therefore, the sum we need is the sum over each pair of $|y|^2 + |\overline{y}|^2 = 2|y|^2$. Wait, but since $|y|^2 = |\overline{y}|^2$, so each pair contributes twice $|y|^2$. But since we have 5 pairs, the total sum would be 2*(sum over each pair's |y|^2). Wait, but actually, if we have 10 roots, each pair contributes two terms, so the total sum is sum_{i=1}^{10} |y_i|^2 = 2 * sum_{i=1}^5 |y_i|^2. But since each pair is (y, \overline{y}), so each pair contributes |y|^2 + |\overline{y}|^2 = 2|y|^2. Therefore, the total sum is 2*(sum_{i=1}^5 |y_i|^2). Wait, but maybe I need to think differently. Wait, if there are 10 roots, and they come in 5 conjugate pairs, then the total sum is sum_{i=1}^5 (|y_i|^2 + |\overline{y_i}|^2) = sum_{i=1}^5 2|y_i|^2. Therefore, the total sum is 2*(sum_{i=1}^5 |y_i|^2). However, the problem is asking for sum_{i=1}^5 1/(r_i \overline{r_i}) = sum_{i=1}^5 1/|r_i|^2 = sum_{i=1}^5 |y_i|^2. Wait, because earlier we had that 1/|x|^2 = |y|^2. Therefore, the answer is sum_{i=1}^5 |y_i|^2. But according to the above, the total sum over all roots y is 2*(sum_{i=1}^5 |y_i|^2). Therefore, if I can compute the sum over all roots y of |y|^2, then divide by 2, I get the desired answer.

Alternatively, maybe there's a way to compute the sum of |y|^2 for all roots y. How can I compute that?

I remember that for a polynomial, the sum of |y_i|^2 can be related to the coefficients. Let me recall that for a polynomial with roots y_i, the sum of |y_i|^2 is equal to the sum of y_i \overline{y_i} = sum y_i \overline{y_i}. However, since the coefficients of the polynomial are real? Wait, let me check. The original equation was $(13 - y)^{10} = -1$. Let me expand this equation.

Let me write the equation as $(13 - y)^{10} + 1 = 0$. Therefore, the polynomial equation is $P(y) = (13 - y)^{10} + 1 = 0$. Let me expand this polynomial.

But expanding $(13 - y)^{10}$ would be tedious, but maybe I can write it as $(-1)^{10}(y - 13)^{10} + 1 = 0$, which is $(y - 13)^{10} + 1 = 0$. Therefore, the polynomial is $P(y) = (y - 13)^{10} + 1$. Therefore, the roots of P(y) are the solutions to $(y - 13)^{10} = -1$, which is the same as before.

So, the polynomial is $P(y) = (y - 13)^{10} + 1$. Let me denote $z = y - 13$, so that the equation becomes $z^{10} + 1 = 0$, so $z^{10} = -1$. Therefore, the roots z are the 10th roots of -1, which are $z_k = e^{i(\pi + 2\pi k)/10}$ for k = 0, 1, ..., 9. Therefore, the roots y are $y_k = 13 + z_k = 13 + e^{i(\pi + 2\pi k)/10}$.

Therefore, each root y is $13 + e^{i\theta_k}$, where $\theta_k = (\pi + 2\pi k)/10$. Therefore, to compute |y_k|^2, we can compute |13 + e^{i\theta_k}|^2.

So, |y_k|^2 = |13 + e^{i\theta_k}|^2 = (13 + e^{i\theta_k})(13 + e^{-i\theta_k}) = 13^2 + 13 e^{-i\theta_k} + 13 e^{i\theta_k} + 1 = 169 + 13(e^{i\theta_k} + e^{-i\theta_k}) + 1 = 170 + 26 \cos \theta_k.

Therefore, |y_k|^2 = 170 + 26 \cos \theta_k.

Therefore, the sum over all roots y of |y|^2 is sum_{k=0}^9 [170 + 26 \cos \theta_k] = 10*170 + 26 * sum_{k=0}^9 \cos \theta_k.

Therefore, the total sum is 1700 + 26 * sum_{k=0}^9 \cos \theta_k.

But earlier, we saw that the answer we need is sum_{i=1}^5 |y_i|^2, which is half of the total sum over all roots y (since each pair contributes twice). Wait, no. Wait, the total sum over all roots y is sum_{k=0}^9 |y_k|^2. But since the roots come in conjugate pairs, each pair (y_k, y_{9 - k})? Wait, let me check.

The roots z_k are e^{i(\pi + 2\pi k)/10} for k = 0,...,9. Let me note that for each k, the angle theta_k = (\pi + 2\pi k)/10. Let me compute theta_k for k from 0 to 9:

For k=0: theta_0 = pi/10

k=1: (pi + 2pi)/10 = 3pi/10

k=2: (pi + 4pi)/10 = 5pi/10 = pi/2

k=3: (pi + 6pi)/10 = 7pi/10

k=4: (pi + 8pi)/10 = 9pi/10

k=5: (pi + 10pi)/10 = 11pi/10

k=6: (pi + 12pi)/10 = 13pi/10

k=7: (pi + 14pi)/10 = 15pi/10 = 3pi/2

k=8: (pi + 16pi)/10 = 17pi/10

k=9: (pi + 18pi)/10 = 19pi/10

So, the angles theta_k are pi/10, 3pi/10, pi/2, 7pi/10, 9pi/10, 11pi/10, 13pi/10, 3pi/2, 17pi/10, 19pi/10.

Now, note that cos(theta_k) for k and 10 - k - something? Let me check for conjugate pairs. Since the roots z_k are e^{i theta_k}, their conjugates are e^{-i theta_k} = e^{i( - theta_k)}. However, since theta_k is between pi/10 and 19pi/10, the conjugate of z_k is e^{-i theta_k} which would correspond to some other theta_j. Let me check for k and j such that theta_j = -theta_k + 2pi m for some integer m.

Alternatively, since the roots y are 13 + z_k, and their conjugates are 13 + \overline{z_k} = 13 + e^{-i theta_k}. Therefore, if theta_k is one angle, then theta_j = -theta_k + 2pi m would be another angle. Let me check for each k:

For example, take k=0: theta_0 = pi/10. Then -theta_0 = -pi/10. Adding 2pi to get it into the range [0, 2pi): 19pi/10. Which is theta_9. Similarly, for k=1: theta_1 = 3pi/10, then -theta_1 = -3pi/10, adding 2pi gives 17pi/10, which is theta_8. Similarly, k=2: theta_2 = pi/2, -theta_2 = -pi/2, adding 2pi gives 3pi/2, which is theta_7. Similarly, k=3: theta_3 = 7pi/10, -theta_3 = -7pi/10, adding 2pi gives 13pi/10, which is theta_6. Similarly, k=4: theta_4 = 9pi/10, -theta_4 = -9pi/10, adding 2pi gives 11pi/10, which is theta_5.

Therefore, the roots z_k and z_{9 - k} are conjugates. Therefore, the roots y_k and y_{9 - k} are conjugates. Therefore, their |y_k|^2 and |y_{9 - k}|^2 are equal. Therefore, the sum over all roots y of |y|^2 is 2*(sum_{k=0}^4 |y_k|^2). Therefore, the total sum is 2*(sum_{k=0}^4 |y_k|^2). Therefore, the sum we need, which is sum_{i=1}^5 |y_i|^2, is equal to sum_{k=0}^4 |y_k|^2. Therefore, the total sum over all roots is 2*(sum_{i=1}^5 |y_i|^2). Therefore, if I compute the total sum as 1700 + 26 * sum_{k=0}^9 cos(theta_k), then divide by 2, I get the desired answer.

So, let me compute sum_{k=0}^9 cos(theta_k). Let me note that theta_k for k=0,...,9 are pi/10, 3pi/10, pi/2, 7pi/10, 9pi/10, 11pi/10, 13pi/10, 3pi/2, 17pi/10, 19pi/10.

But note that cos(theta) is even, so cos(theta) = cos(-theta) = cos(2pi - theta). Therefore, cos(theta_k) for k=0,...,9 can be paired as follows:

cos(pi/10) and cos(19pi/10) = cos(pi/10) and cos(-pi/10) = cos(pi/10) and cos(pi/10) = 2cos(pi/10)

Similarly, cos(3pi/10) and cos(17pi/10) = cos(3pi/10) and cos(-3pi/10) = 2cos(3pi/10)

cos(pi/2) and cos(3pi/2) = cos(pi/2) and cos(3pi/2) = 0 + 0 = 0

cos(7pi/10) and cos(13pi/10) = cos(7pi/10) and cos(13pi/10) = cos(7pi/10) and cos(13pi/10). But 13pi/10 = pi + 3pi/10, so cos(13pi/10) = cos(pi + 3pi/10) = -cos(3pi/10). Similarly, cos(7pi/10) = cos(pi - 3pi/10) = -cos(3pi/10). Therefore, cos(7pi/10) + cos(13pi/10) = -cos(3pi/10) - cos(3pi/10) = -2cos(3pi/10). Wait, but earlier I thought they would pair to something else. Wait, maybe I need to check again.

Wait, theta_3 = 7pi/10, theta_6 = 13pi/10. So cos(theta_3) + cos(theta_6) = cos(7pi/10) + cos(13pi/10). Let me compute cos(7pi/10) = cos(pi - 3pi/10) = -cos(3pi/10). Similarly, cos(13pi/10) = cos(pi + 3pi/10) = -cos(3pi/10). Therefore, their sum is -cos(3pi/10) - cos(3pi/10) = -2cos(3pi/10). Similarly, theta_4 = 9pi/10, theta_5 = 11pi/10. cos(theta_4) + cos(theta_5) = cos(9pi/10) + cos(11pi/10). cos(9pi/10) = cos(pi - pi/10) = -cos(pi/10). cos(11pi/10) = cos(pi + pi/10) = -cos(pi/10). Therefore, their sum is -cos(pi/10) - cos(pi/10) = -2cos(pi/10).

Therefore, the total sum of cos(theta_k) from k=0 to 9 is:

[cos(pi/10) + cos(19pi/10)] + [cos(3pi/10) + cos(17pi/10)] + [cos(pi/2) + cos(3pi/2)] + [cos(7pi/10) + cos(13pi/10)] + [cos(9pi/10) + cos(11pi/10)]

Which is:

[2cos(pi/10)] + [2cos(3pi/10)] + [0 + 0] + [-2cos(3pi/10)] + [-2cos(pi/10)] = 2cos(pi/10) + 2cos(3pi/10) - 2cos(3pi/10) - 2cos(pi/10) = 0.

Wait, that's zero? So the total sum of cos(theta_k) from k=0 to 9 is zero? That seems surprising, but let me check again.

Yes, because the sum of cos(theta_k) over all roots of z^{10} = -1. Wait, but z^{10} = -1, so the roots are the 10th roots of -1. However, the sum of all roots of z^{10} + 1 = 0 is zero, because the coefficient of z^9 is zero. Wait, but the sum of the roots of z^{10} + 1 = 0 is zero. However, the sum of the roots is sum_{k=0}^9 z_k = 0. But the sum of cos(theta_k) is the real part of sum_{k=0}^9 z_k, which is zero. Therefore, sum_{k=0}^9 cos(theta_k) = 0. Therefore, the total sum is 1700 + 26 * 0 = 1700. Therefore, the total sum over all roots y of |y|^2 is 1700. Therefore, the sum we need, which is sum_{i=1}^5 |y_i|^2, is half of that, which is 1700 / 2 = 850.

Wait, but hold on. Earlier, I thought that the total sum over all roots y is 2*(sum_{i=1}^5 |y_i|^2). Therefore, if the total sum is 1700, then sum_{i=1}^5 |y_i|^2 = 1700 / 2 = 850. Therefore, the answer is 850?

But let me check again. Let me recap:

We have the polynomial P(y) = (y - 13)^{10} + 1 = 0. The roots y are 13 + z_k where z_k^{10} = -1. Then, |y_k|^2 = 170 + 26 cos(theta_k). Therefore, sum_{k=0}^9 |y_k|^2 = 10*170 + 26*sum cos(theta_k) = 1700 + 0 = 1700.

But since the roots y come in conjugate pairs, each pair contributes |y_k|^2 + |y_{9 -k}|^2 = 2|y_k|^2. Therefore, the total sum is 2*(sum_{k=0}^4 |y_k|^2). Therefore, sum_{k=0}^4 |y_k|^2 = 1700 / 2 = 850. Therefore, the answer is 850.

But wait, the problem says that the original equation has roots r_1, \overline{r_1}, ..., r_5, \overline{r_5}, and we need to compute sum_{i=1}^5 1/(r_i \overline{r_i}) = sum_{i=1}^5 1/|r_i|^2. But earlier, we had that 1/|r_i|^2 = |y_i|^2. Therefore, the answer is sum_{i=1}^5 |y_i|^2 = 850. Therefore, the answer is 850?

But let me check with an example. Let me take a simpler case. Suppose I have a quadratic equation. Let me see if my reasoning holds.

Suppose I have equation x^2 + (ax - b)^2 = 0. Then similar steps would lead to something. But maybe that's too time-consuming. Alternatively, let me check if there's another way to compute the sum of 1/|r_i|^2.

Alternatively, since the original equation is x^{10} + (13x - 1)^{10} = 0. Let me consider that equation. Let me write it as x^{10} = - (13x - 1)^{10}. Taking modulus on both sides, |x|^{10} = |13x - 1|^{10}, so |x| = |13x - 1|. Therefore, for each root x, |x| = |13x - 1|. Therefore, |x|^2 = |13x - 1|^2.

But |13x - 1|^2 = (13x - 1)(\overline{13x} - 1) = 169 |x|^2 - 13x - 13 \overline{x} + 1.

Therefore, |x|^2 = 169 |x|^2 - 13(x + \overline{x}) + 1.

Rearranging:

0 = 168 |x|^2 - 13(x + \overline{x}) + 1.

But x + \overline{x} = 2 Re(x). Let me denote |x|^2 = r^2, and Re(x) = a. Then:

0 = 168 r^2 - 26 a + 1.

But also, since |x|^2 = r^2 = a^2 + b^2, where x = a + ib. However, this seems complicated. However, if I can find the sum over all roots of 1/r^2, which is the same as sum over all roots of |y|^2, which we found as 850.

Alternatively, maybe I can use Vieta's formula. Let me think.

But the original equation is x^{10} + (13x - 1)^{10} = 0. Let me expand (13x - 1)^{10} using binomial theorem. However, expanding this would be tedious, but maybe I can write the equation as:

x^{10} + \sum_{k=0}^{10} \binom{10}{k} (13x)^{10 - k} (-1)^k = 0.

But this seems complicated. However, maybe I can write the equation as:

x^{10} + (13x - 1)^{10} = 0 => (13x - 1)^{10} = -x^{10} => [(13x - 1)/x]^{10} = -1 => [13 - 1/x]^{10} = -1.

Which is the same as before. Therefore, leading to y = 1/x, and so on.

Alternatively, since we have that for each root x, |x| = |13x - 1|. Therefore, |x|^2 = |13x - 1|^2. Let me write this equation again:

|x|^2 = |13x - 1|^2 => |x|^2 = 169 |x|^2 - 26 Re(13x) + 1 => 0 = 168 |x|^2 - 26 Re(13x) + 1 => 168 |x|^2 - 26 * 13 Re(x) + 1 = 0 => 168 |x|^2 - 338 Re(x) + 1 = 0.

But Re(x) = (x + \overline{x}) / 2. However, if I denote for each root x, let me denote |x|^2 = s, and Re(x) = a. Then:

168 s - 338 a + 1 = 0 => 168 s = 338 a - 1 => s = (338 a - 1)/168.

But how does this help me? If I can find the sum over all roots of 1/s, which is the sum of 1/|x|^2.

But I need to relate this to Vieta's formula. However, since the equation is of degree 10, and Vieta's formula relates sums of roots, products, etc. However, the equation is not in standard form. Let me write the original equation again:

x^{10} + (13x - 1)^{10} = 0. Let me expand (13x - 1)^{10}:

(13x - 1)^{10} = \sum_{k=0}^{10} \binom{10}{k} (13x)^{k} (-1)^{10 - k} = \sum_{k=0}^{10} \binom{10}{k} (-1)^{10 - k} 13^k x^k.

Therefore, the original equation is:

x^{10} + \sum_{k=0}^{10} \binom{10}{k} (-1)^{10 - k} 13^k x^k = 0.

But note that when k=10, the term is \binom{10}{10} (-1)^{0} 13^{10} x^{10} = 13^{10} x^{10}. Therefore, combining with the x^{10} term:

x^{10} + 13^{10} x^{10} + ... = 0 => (1 + 13^{10})x^{10} + ... = 0.

But this seems messy. However, perhaps I can write the equation as:

(13x - 1)^{10} = -x^{10} => [(13x - 1)/x]^{10} = -1 => [13 - 1/x]^{10} = -1.

Which is the same as before. Therefore, leading to y = 1/x, and the roots y satisfy (13 - y)^{10} = -1. Therefore, the polynomial equation for y is (13 - y)^{10} + 1 = 0.

Therefore, the polynomial is P(y) = (y - 13)^{10} + 1. Therefore, the roots y are 13 + z_k where z_k^{10} = -1.

Therefore, the sum we need is sum_{k=0}^9 |y_k|^2 = 1700, and since the answer is half of that, 850.

But let me check with specific values. Let me take a simple case where I can compute manually. For example, suppose instead of 10th roots, take a quadratic equation. Let me try with degree 2.

Suppose equation is x^2 + (ax - b)^2 = 0. Then similar steps would lead to something. Let me take a=1, b=1. Then equation is x^2 + (x - 1)^2 = 0. Expanding: x^2 + x^2 - 2x + 1 = 0 => 2x^2 - 2x + 1 = 0. The roots are [2 ± sqrt(4 - 8)] / 4 = [2 ± sqrt(-4)] / 4 = [1 ± i]/2. Therefore, roots are (1 + i)/2 and (1 - i)/2. Then |x|^2 for each root is (1/2)^2 + (1/2)^2 = 1/4 + 1/4 = 1/2. Therefore, sum of 1/|x|^2 is 2/(1/2) = 4.

Alternatively, using our previous method: Let me write equation as x^2 + (x - 1)^2 = 0 => (x - 1)^2 = -x^2 => [ (x - 1)/x ]^2 = -1 => [1 - 1/x]^2 = -1 => 1 - 2/x + 1/x^2 = -1 => 1/x^2 - 2/x + 2 = 0. Let y = 1/x, then equation is y^2 - 2y + 2 = 0. The roots y are [2 ± sqrt(4 - 8)] / 2 = [2 ± sqrt(-4)] / 2 = 1 ± i. Therefore, |y|^2 = 1^2 + 1^2 = 2. Therefore, sum of |y|^2 is 2 + 2 = 4, which matches the previous result. Therefore, the answer would be 4.

But according to our previous method: For the equation (y - 13)^{10} + 1 = 0, but in this case, the equation is (y - 1)^2 + 1 = 0? Wait, no. Wait, in this case, the equation for y is y^2 - 2y + 2 = 0. Which is different. However, if I follow the same steps as before:

Let me write the equation as (y - a)^n + 1 = 0, where a is some constant. In this case, n=2, a=1. Then the roots y are 1 + z_k where z_k^2 = -1. Therefore, z_k = i, -i. Therefore, y = 1 + i, 1 - i. Then |y|^2 = (1)^2 + (1)^2 = 2 for each, so sum is 4. Which matches.

Therefore, in this case, the total sum over all roots y of |y|^2 is 4, which is equal to 2*(sum_{i=1}^1 |y_i|^2) since there are two roots. Therefore, the answer is 4. Which is correct.

Therefore, returning to our original problem, the answer is 850. Therefore, I think that is correct.

But let me check another way. Let me think about Vieta's formula.

We have the polynomial P(y) = (y - 13)^{10} + 1. Let me expand this polynomial.

Let me denote P(y) = (y - 13)^{10} + 1. Let me write it as y^{10} - 10*13 y^9 + ... + (-13)^{10} + 1. However, expanding this is tedious, but maybe I can find the sum of |y_i|^2 using Vieta's formula.

But how? The sum of |y_i|^2 is sum_{i=1}^{10} |y_i|^2. For complex numbers, this is equal to sum_{i=1}^{10} y_i \overline{y_i}. However, since the coefficients of the polynomial are real, the roots come in conjugate pairs. Therefore, if y is a root, then \overline{y} is also a root. Therefore, the sum becomes sum_{i=1}^{10} y_i \overline{y_i} = sum_{i=1}^{5} (y_i \overline{y_i} + \overline{y_i} y_i) = 2 sum_{i=1}^5 |y_i|^2.

But how to compute sum_{i=1}^{10} y_i \overline{y_i}?

Alternatively, note that for any complex number y, y \overline{y} = |y|^2. Therefore, sum_{i=1}^{10} |y_i|^2 = sum_{i=1}^{10} y_i \overline{y_i}.

But how to compute this sum?

Let me recall that for a polynomial with roots y_i, we have that sum_{i=1}^n y_i = - coefficient of y^{n-1} / coefficient of y^n. Similarly, sum_{i=1}^n y_i y_j = coefficient of y^{n-2} / coefficient of y^n, etc. However, these are for sums of products of roots, not for sums of |y_i|^2.

But perhaps we can relate sum |y_i|^2 to the coefficients.

Note that sum |y_i|^2 = sum y_i \overline{y_i} = sum y_i \overline{y_i}. However, since the polynomial has real coefficients, the conjugate of each root is also a root. Therefore, if we consider the polynomial with roots y_i, then the polynomial with roots \overline{y_i} is the same as the original polynomial. Therefore, the product of the polynomial evaluated at y and its conjugate? Not sure.

Alternatively, consider that if P(y) = \prod_{i=1}^{10} (y - y_i), then the polynomial with roots \overline{y_i} is \prod_{i=1}^{10} (y - \overline{y_i}) = \overline{P(\overline{y})} if P has real coefficients. Wait, maybe not.

Alternatively, consider that sum_{i=1}^{10} y_i \overline{y_i} = sum_{i=1}^{10} y_i \overline{y_i} = sum_{i=1}^{10} y_i \overline{y_i} = sum_{i=1}^{10} y_i \overline{y_i}.

But since the polynomial has real coefficients, then for each root y_i, \overline{y_i} is also a root. Therefore, we can pair them as y_i and \overline{y_i}. Therefore, sum_{i=1}^{10} y_i \overline{y_i} = sum_{j=1}^5 (y_j \overline{y_j} + \overline{y_j} y_j) = 2 sum_{j=1}^5 |y_j|^2.

But how to compute this sum?

Alternatively, consider that for each root y_i, we have that P(y_i) = 0. Therefore, (y_i - 13)^{10} = -1. Therefore, taking modulus squared on both sides: |y_i - 13|^{20} = 1. Therefore, |y_i - 13|^{2} = 1^{1/10} = 1? Wait, no. Wait, modulus squared of both sides: |(y_i - 13)^{10}|^2 = |-1|^2 => |y_i - 13|^{20} = 1. Therefore, |y_i - 13|^{2} = 1^{1/10} = 1? Wait, no. Wait, modulus squared of (y_i - 13)^{10} is |y_i - 13|^{20}, and modulus squared of -1 is 1. Therefore, |y_i - 13|^{20} = 1 => |y_i - 13| = 1^{1/10} = 1. Therefore, |y_i - 13| = 1. Therefore, for each root y_i, the distance from y_i to 13 is 1. Therefore, all roots lie on a circle of radius 1 centered at 13 in the complex plane.

Therefore, |y_i - 13| = 1 => |y_i - 13|^2 = 1 => (y_i - 13)(\overline{y_i} - 13) = 1 => y_i \overline{y_i} - 13 y_i - 13 \overline{y_i} + 169 = 1 => |y_i|^2 - 13(y_i + \overline{y_i}) + 169 = 1 => |y_i|^2 - 13*2 Re(y_i) + 168 = 0 => |y_i|^2 = 26 Re(y_i) - 168.

Therefore, for each root y_i, |y_i|^2 = 26 Re(y_i) - 168. Therefore, the sum over all roots of |y_i|^2 is sum_{i=1}^{10} [26 Re(y_i) - 168] = 26 sum Re(y_i) - 168*10.

But sum Re(y_i) is the real part of sum y_i.

From Vieta's formula, for the polynomial P(y) = (y - 13)^{10} + 1 = y^{10} - 10*13 y^9 + ... + (-13)^{10} + 1. The coefficient of y^9 is -10*13, so sum y_i = 10*13 = 130. Therefore, sum Re(y_i) = Re(sum y_i) = Re(130) = 130.

Therefore, sum |y_i|^2 = 26*130 - 168*10 = 3380 - 1680 = 1700. Which matches our previous result. Therefore, the total sum is 1700, so the answer is 1700 / 2 = 850.

Therefore, this confirms that the answer is 850.

Therefore, the value of the given sum is 850.

**Final Answer**
\boxed{850}

To solve the equation

$$
x^{10} + (13x - 1)^{10} = 0,
$$

we begin by analyzing the structure of the roots. The equation is a polynomial of degree 10, and the problem states that it has 10 complex roots, which are paired as conjugates: $ r_1, \overline{r}_1, r_2, \overline{r}_2, \ldots, r_5, \overline{r}_5 $.

We are asked to compute the sum:

$$
\frac{1}{r_1\overline{r}_1} + \frac{1}{r_2\overline{r}_2} + \cdots + \frac{1}{r_5\overline{r}_5}.
$$

---

### Step 1: Understanding the Expression

For any complex number $ z $, the product $ z\overline{z} = |z|^2 $, the square of its modulus. Therefore, the expression becomes:

$$
\sum_{i=1}^5 \frac{1}{|r_i|^2}.
$$

Let us denote $ x = r_i $, so the expression becomes:

$$
\sum_{i=1}^5 \frac{1}{|x_i|^2}.
$$

---

### Step 2: Transforming the Equation

We rewrite the original equation:

$$
x^{10} + (13x - 1)^{10} = 0 \Rightarrow (13x - 1)^{10} = -x^{10}.
$$

Dividing both sides by $ x^{10} $ (assuming $ x \neq 0 $), we get:

$$
\left(13 - \frac{1}{x}\right)^{10} = -1.
$$

Let $ y = \frac{1}{x} $, so the equation becomes:

$$
(13 - y)^{10} = -1.
$$

This implies that $ 13 - y $ is a 10th root of $ -1 $, i.e.,

$$
13 - y = e^{i(\pi + 2\pi k)/10}, \quad k = 0, 1, \ldots, 9.
$$

Thus, the roots $ y $ are:

$$
y_k = 13 - e^{i(\pi + 2\pi k)/10}, \quad k = 0, 1, \ldots, 9.
$$

Since $ x = \frac{1}{y} $, the roots $ x $ are:

$$
x_k = \frac{1}{13 - e^{i(\pi + 2\pi k)/10}}.
$$

---

### Step 3: Computing the Desired Sum

We are to compute:

$$
\sum_{i=1}^5 \frac{1}{|r_i|^2} = \sum_{i=1}^5 |y_i|^2,
$$

since $ \frac{1}{|x_i|^2} = |y_i|^2 $.

Let us compute $ |y_k|^2 $ for each root $ y_k = 13 - e^{i\theta_k} $, where $ \theta_k = \frac{\pi + 2\pi k}{10} $.

We use the identity:

$$
|y_k|^2 = |13 - e^{i\theta_k}|^2 = (13 - e^{i\theta_k})(13 - e^{-i\theta_k}) = 169 - 13(e^{i\theta_k} + e^{-i\theta_k}) + 1 = 170 - 26\cos\theta_k.
$$

So:

$$
|y_k|^2 = 170 - 26\cos\theta_k.
$$

Now, summing over all 10 roots:

$$
\sum_{k=0}^9 |y_k|^2 = \sum_{k=0}^9 (170 - 26\cos\theta_k) = 10 \cdot 170 - 26 \sum_{k=0}^9 \cos\theta_k = 1700 - 26 \sum_{k=0}^9 \cos\theta_k.
$$

But the roots $ e^{i\theta_k} $ are the 10th roots of $ -1 $, and the sum of all such roots is zero (since the sum of all roots of a polynomial is given by the negative of the coefficient of the second-highest degree term, which is zero in this case). Therefore, the sum of their real parts is also zero:

$$
\sum_{k=0}^9 \cos\theta_k = 0.
$$

Thus:

$$
\sum_{k=0}^9 |y_k|^2 = 1700.
$$

Since the roots $ y_k $ come in conjugate pairs, and each pair contributes $ |y_k|^2 + |\overline{y_k}|^2 = 2|y_k|^2 $, the sum over all 10 roots is twice the sum over the 5 distinct pairs. Therefore, the sum over the 5 distinct $ |y_i|^2 $ is:

$$
\sum_{i=1}^5 |y_i|^2 = \frac{1}{2} \sum_{k=0}^9 |y_k|^2 = \frac{1700}{2} = 850.
$$

---

### Final Answer

$$
\boxed{850}
$$

Solution:
Let $p(x) = x^{10} + (13x - 1)^{10}.$  If $r$ is a root of $p(x),$ then $r^{10} + (13x - 1)^{10} = 0.$  Then $(13r - 1)^{10} = -r^{10},$ so
\[-1 = \left( \frac{13r - 1}{r} \right)^{10} = \left( \frac{1}{r} - 13 \right)^{10}.\]Then $\frac{1}{r} - 13$ has magnitude 1, so
\[\left( \frac{1}{r} - 13 \right) \left( \frac{1}{\overline{r}} - 13 \right) = 1,\]so
\[\left( \frac{1}{r_1} - 13 \right) \left( \frac{1}{\overline{r}_1} - 13 \right) + \dots + \left( \frac{1}{r_5} - 13 \right) \left( \frac{1}{\overline{r}_5} - 13 \right) = 5.\]Expanding, we get
\[\frac{1}{r_1 \overline{r}_1} + \dots + \frac{1}{r_5 \overline{r}_5} - 13 \left( \frac{1}{r_1} + \frac{1}{\overline{r}_1} + \dots + \frac{1}{r_5} + \frac{1}{\overline{r}_5} \right) + 5 \cdot 169 = 5.\]We see that $\frac{1}{r_1},$ $\frac{1}{\overline{r}_1},$ $\dots,$ $\frac{1}{r_5},$ $\frac{1}{\overline{r}_5}$ are the solutions to
\[\left( \frac{1}{x} \right)^{10} + \left( \frac{13}{x} - 1 \right)^{10} = 0,\]or $1 + (13 - x)^{10} = 0.$  The first few terms in the expansion as
\[x^{10} - 130x^9 + \dotsb = 0,\]so by Vieta's formulas,
\[\frac{1}{r_1} + \frac{1}{\overline{r}_1} + \dots + \frac{1}{r_5} + \frac{1}{\overline{r}_5} = 130.\]Hence,
\[\frac{1}{r_1 \overline{r}_1} + \dots + \frac{1}{r_5 \overline{r}_5} = 13 \cdot 130 - 5 \cdot 169 + 5 = \boxed{850}.\]