Help with complex logarithms

Question

For real $x$ what does $-\ln(1-e^{2\pi i x})$ equal so that it agrees with the series expansion, how would I find the real and imaginary parts. $$-\ln(1-e^{2\pi ix})=\sum_{n=1}^\infty\frac{e^{2\pi i n x}}{n}$$

Answer 1

By Euler's formula, $$ 1-e^{2\pi ix}=1-\cos(2\pi x)-i\sin(2\pi x) $$ Using $\log(z)=\log(|z|)+i\arg(z)$, we get $$ \begin{align} &-\log(1-e^{2\pi ix})\\ =&-\frac12\log\left((1-\cos(2\pi x))^2+\sin^2(2\pi x)\right)-i\tan^{-1}\left(\frac{-\sin(2\pi x)}{1-\cos(2\pi x)}\right)\\ =&-\frac12\log\left(2-2\cos(2\pi x)\right)+i\tan^{-1}(\cot(\pi x))\\ =&-\frac12\log\left(4\sin^2(\pi x)\right)+i\tan^{-1}(\cot(\pi x))\\ =&-\log\left(2|\sin(\pi x)|\right)+i\tan^{-1}(\cot(\pi x))\\ \end{align} $$ for all $x$. However, for $x\in\left(-\tfrac12,+\tfrac12\right)$, $$ \begin{align} \tan^{-1}(\cot(\pi x))&=\frac\pi2\mathrm{sgn}(x)-\pi x \end{align} $$ and $-\log(1-e^{2\pi ix})$ has period $1$.

The series converges, for $x\not\in\mathbb{Z}$, by the Dirichlet Test.