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I need to express $ \cos(\varphi) $ with $z = e^{i \varphi}$ in order to use the Cauchy integral formula on the following integral:

$ \int^{2 \pi}_0 \frac{1}{3+2\cos(\varphi)} \,d \varphi $

I got:

$ \int_{|z|=1} \frac{e^{-i \varphi}}{3+2\cos(\varphi)} e^{i \varphi}\, d \varphi = -i \int_{|z|=1} \frac{z^{-1}}{3+2\cos(\varphi)} e^{i \varphi} dz $

But I don't know what to do with the cos.

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$\cos \phi =\frac {z+\overline z} 2=\frac {z+\frac 1z} 2$ when $|z|=1$.

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If you define$$f(z)=\frac1z\cdot\frac1{3+z+1/z},$$then, if $\varphi\in[0,2\pi]$,$$f(e^{i\varphi})=e^{-i\varphi}\frac1{3+\cos\varphi}$$and therefore\begin{align}\int_0^{2\pi}\frac1{3+2\cos\varphi}\,\mathrm d\varphi&=\frac1i\int_{|z|=1}f(z)\,\mathrm dz\\&=2\pi\operatorname{res}\left(0,f(z)\right)\\&=\frac{2\pi}{\sqrt5}.\end{align}

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