# Need a hint with this question

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?

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