Complex Roots Reciprocal Sum
The equation
\[x^{10}+(13x-1)^{10}=0\,\]has 10 complex roots $ r_1, $ $ \overline{r}_1, $ $ r_2, $ $ \overline{r}_2, $ $ r_3, $ $ \overline{r}_3, $ $ r_4, $ $ \overline{r}_4, $ $ r_5, $ $ \overline{r}_5, $ where the bar denotes complex conjugation. Find the value of
\[\frac 1{r_1\overline{r}_1}+\frac 1{r_2\overline{r}_2}+\frac 1{r_3\overline{r}_3}+\frac 1{r_4\overline{r}_4}+\frac 1{r_5\overline{r}_5}.\]
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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,
\[\frac{1}{r_1 \overline{r}_1} + \dots + \frac{1}{r_5 \overline{r}_5} = 13 \cdot 130 - 5 \cdot 169 + 5 = \boxed{850}.\]
Freeflows