# A question on using Residues Theorem

I am having a doubt on an example on Conway's "Funtions of one Complex Variables". At example 2.9, page 117. At some point, it is stated that, since $$a+\cos(\theta)=\dfrac{z^2+2az+1}{2z}$$ it must imply that $$\int_0^{\pi}\dfrac{d\theta}{a+cos(\theta )}=\dfrac{1}{2}\int_0^{2\pi}\dfrac{d\theta}{a+cos(\theta)}=-i\int_{\gamma}\dfrac{dz}{z^2+2az+1}$$ where $\gamma (t)=e^{it}$, $t\in \left[0,2\pi \right]$.

I could get all that, except for what happens between the last two integrals; I think that there should be a "$2z$" before the $dz$, but when I insert it there the calculations doesn't end up right. Can someone explain me what I'm missing?

Thanks.

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$$\int\frac{d\theta}{2a+2\cos\theta}=\int\frac{d\theta}{2a+e^{i\theta}+e^{-i\theta}}=\int\frac{dz}{iz}\cdot\frac{1}{2a+z+\frac{1}{z}}.$$ Then $$\int\frac{dz}{iz}\cdot\frac{1}{2a+z+\frac{1}{z}}=-i\int\frac{dz}{z^2+2az+1}.$$

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Great! Thank you. – Marra Feb 13 '13 at 3:08

$$a+\cos(\theta)=\dfrac{z^2+2az+1}{2z}\Longrightarrow -\sin\theta\,d\theta=\left(\frac{1}{2}-\frac{1}{2z^2}\right)dz=\frac{z^2-1}{2z^2}dz\Longrightarrow$$

$$d\theta=-\frac{z^2-1}{2z^2\sin\theta}dz=-\frac{z^2-1}{2z^2\sqrt{1-\cos^2\theta}}dz$$

But

$$1-\cos^2\theta=1-\left(\frac{z^2+2az+1}{2z}-a\right)^2=1-\left(\frac{z^2+1}{2z}\right)^2=$$

$$=-\frac{(z-1)^2(z+1)^2}{4z^2}\Longrightarrow d\theta=\frac{z^2-1}{2z^2}dz\frac{2z}{z^2-1}=\frac{dz}{z}$$

so

$$\frac{1}{2}\int\limits_0^{2\pi}\frac{d\theta}{a+\cos\theta}=\frac{1}{2}\int\limits_\gamma\frac{dz}{z}\frac{2z}{z^2+2az+1}=\ldots$$

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