For n 1 and n in N, if z_1, z_2, ldots, z_n | vedclass

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(C) Given the equation $(z+1)^n = z^n$. Since $z=0$ is not a root, we can write $(\frac{z+1}{z})^n = 1$. Let $\frac{z+1}{z} = \omega_k = e^{i \frac{2k

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$\frac{n-1}{2}(2 - \pi)$

Vedclass · Answer

(C) Given the equation $(z+1)^n = z^n$. Since $z=0$ is not a root, we can write $(\frac{z+1}{z})^n = 1$. Let $\frac{z+1}{z} = \omega_k = e^{i \frac{2k\pi}{n}}$ for $k = 1, 2, \ldots, n-1$. Then $z+1 = z \omega_k \Rightarrow z(1 - \omega_k) = -1 \Rightarrow z = \frac{1}{\omega_k - 1}$. Substituting $\omega_k = \cos(\frac{2k\pi}{n}) + i \sin(\frac{2k\pi}{n})$, we get $z = \frac{1}{\cos(\frac{2k\pi}{n}) - 1 + i \sin(\frac{2k\pi}{n})} = \frac{1}{-2 \sin^2(\frac{k\pi}{n}) + 2i \sin(\frac{k\pi}{n}) \cos(\frac{k\pi}{n})} = \frac{1}{2i \sin(\frac{k\pi}{n}) [\cos(\frac{k\pi}{n}) + i \sin(\frac{k\pi}{n})]} = \frac{1}{2i \sin(\frac{k\pi}{n}) e^{i \frac{k\pi}{n}}} = \frac{1}{2} (-i \csc(\frac{k\pi}{n})) e^{-i \frac{k\pi}{n}} = \frac{1}{2} (-i \csc(\frac{k\pi}{n})) (\cos(\frac{k\pi}{n}) - i \sin(\frac{k\pi}{n})) = \frac{1}{2} (-i \cot(\frac{k\pi}{n}) - 1)$. Thus, $\operatorname{Re}(z_k) = -\frac{1}{2}$ and $\operatorname{Im}(z_k) = -\frac{1}{2} \cot(\frac{k\pi}{n})$. The sum is over $n-1$ roots. Substituting these into the expression: $\sum_{k=1}^{n-1} \frac{\cot^{-1}(|\cot(\frac{k\pi}{n})|) - 1}{-1} = \sum_{k=1}^{n-1} (1 - \frac{k\pi}{n}) = (n-1) - \frac{\pi}{n} \frac{(n-1)n}{2} = (n-1)(1 - \frac{\pi}{2}) = \frac{n-1}{2}(2 - \pi)$.

For $n > 1$ and $n \in N$, if $z_1, z_2, \ldots, z_n$ are the roots of the equation $(z+1)^n = z^n$, then $\sum_{i=1}^{n-1} \frac{\cot^{-1}(2|\operatorname{Im} z_i|) - 1}{2 \operatorname{Re} z_i} = $

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