I'm looking over one of my past papers and I'm having some trouble with the following question.
By considering the series expansion of: $\ln(1-z)$, where $z=\frac{e^{i\theta}}{2}$, show that $\ln(\frac2{\sqrt{5-4\cos(\theta)}})=\sum_{n=1}^{\infty}\frac{\cos n\theta}{n2^n}$
I'm not really sure what to do. I know that $ -\ln(1-z)=\sum_{n=1}^{\infty}\frac{z^n}{n} $ therefore I was trying to get $\dfrac{\sqrt{5-4\cos(\theta)}}{2}$ into the form $1-z$ for some $z$.
Also $ -\ln(1-\frac{e^{i\theta}}{2})=\sum_{n=1}^{\infty}\frac{\cos(n\theta)+i\sin(n\theta)}{n2^n} $ So I *know* that $ \ln(\frac{2}{\sqrt{5-4\cos(\theta)}})=\sum_{n=1}^{\infty}\frac{\cos n\theta}{n2^n}=\Re(-\ln(1-\frac{e^{i\theta}}{2})) $ but I didn't have much success either way I've tried it. Could anyone point me in the right direction?