Is the following power series expression valid?

Question

So I know that for $z\in \mathbb{C}$ with $|z|<1$ we have that: $$-\ln(1-z) = \sum_{n=1}^{\infty}\frac{z^{n}}{n}.$$ If we let $z = xe^{it}$ where $t\in (0,\pi)$ with $|x|<1$ then $z^{n} = x^{n} \cos(nt) + ix^{n}\sin(nt).$ Furthermore if I express $$1-z = 1-x e^{it} = re^{i\theta}$$ I get that $$-\ln(1-z) = -\ln(r) -i\theta$$ where $r = \sqrt{1-2x\cos(t) + x^2} $ and $\theta = \arctan(-x\sin(t)/ (1-x\cos(t))) $ then can I say that, $$\sum_{n\geq 1}\frac{\cos(nt)}{n}x^{n} = -\frac{1}{2} \ln(1-2x\cos(t) + x^2)$$ and $$\sum_{n\geq 1}\frac{\sin(nt)}{n}x^{n} = \arctan\left(\frac{x\sin(t)}{1-x\cos(t)}\right)?$$

Answer 1

Yes, you derivation is correct.