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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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That power series is missing some powers, isn't it? It should be $\sum z^n/n$, I think. –  Dylan Moreland May 13 '12 at 15:10
    
You're correct, I've fixed it –  Maciek May 13 '12 at 15:16
2  
You know $\ln\,z=\ln\,|z|+i\arg\,z$, yes? –  J. M. May 13 '12 at 15:21

2 Answers 2

up vote 3 down vote accepted

Hint:

Given that $\log{z} = \log|z| + i\arg{z}$, we have:

\begin{align} \Re\left(-\log\left(1-\frac{e^{i\theta}}{2}\right)\right) &= -\log\left|1-\frac{e^{i\theta}}{2}\right| \\ &= -\log\left|1 - \frac{\cos{\theta} + i \sin{\theta}}{2}\right| \\ &= -\log\left|\left(1 - \frac{\cos\theta}{2}\right) - i\frac{\sin\theta}{2}\right| \end{align}

Now simplify.

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Hint:

$$\frac{2}{\sqrt{5-4\cos\theta}}=\sqrt{\frac{4}{(2^2+1)-2e^{i\theta}-2e^{-i\theta}}}=\sqrt{\frac{2\cdot2}{(2-e^{i\theta})(2-e^{-i\theta})}}=\sqrt{\frac{1}{1-z}}\sqrt{\frac{1}{1-\bar{z}}}.$$

Keep in mind the logarithm rules and how trig functions decompose into complex exponentials.

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