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I've been self-studying some complex variables and just started on residues. I'm looking at a problem where I've been asked to calculate the residue of:

$$f(z) = \frac{z}{1-\cos(z)}$$

at $z=0$. I'm not really sure how to find the Laurent series of this function, and I can't really apply Cauchy's integral formula either. So I would appreciate if anyone can get me started.

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The first question you should ask is what type of singularity $f(z)$ has at $z = 0$. Since $z$ has a zero of order $1$ and $1 - cos(z)$ has a zero of order $2$, $f(z)$ has a simple pole. This means $zf(z)$ has a removable singularity, and relating the power series expansion at $0$ of $f(z)$ and $zf(z)$, you know that the residue of $f(z)$ at $z = 0$ is the value of $zf(z)$ at $z = 0$. Now you need to find this value.

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  • $\begingroup$ Thanks, it seems like I was doing this problem before I had read the relevant section (i just read the definition of a residue and started this problem). This is what happens when you're reading a different text while doing problems from another source. $\endgroup$ – Ahsan Jul 8 '12 at 20:15
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Write the (first few terms of the) Taylor series for $\cos z$, and subtract this from 1. You should find that the result can be written in the form $1-\cos z=az^2+bz^4+\cdots$ for some constants $a$ and $b$, and where the dots represent higher powers of $z$. Then think about using the binomial theorem to pin down the coefficient of $z^{-1}$ in $$ f(z) = \frac{z}{1-\cos z}=\frac{z}{az^2(1+\frac{b}{a}z^2+\cdots)}.$$

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  • $\begingroup$ Very helpful! Turns out there was a more direct way of doing it. I forgot about my binomial expansions. $\endgroup$ – Ahsan Jul 8 '12 at 20:27
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Using the power series $\cos(z)=1-\frac12z^2+O(z^4)$ yields $$ \begin{align} f(z) &=\frac{z}{1-(1-\frac12z^2+O(z^4))}\\ &=\frac{1}{\frac12z+O(z^3)}\\ &=\frac{1}{z}\frac{1}{\frac12+O(z^2)}\\ &=\frac{1}{z}(2+O(z^2)) \end{align} $$ Thus the residue is $2$


Another thought is to write $$ \begin{align} \frac{z}{1-\cos(z)} &=\frac{z^2}{\sin^2(z)}\frac{1+\cos(z)}{z}\\ &=\frac{1}{1+O(z^2)}\cdot\frac{2+O(z^2)}{z} \end{align} $$ again giving a residue of $2$.

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  • $\begingroup$ You beat me to it! (+1) $\endgroup$ – user26872 Jul 8 '12 at 20:26
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$\displaystyle\text{Res}=\lim_{z\to0}(z-0)f(z)=\lim\frac{z^2}{1-\cos z}$. Now use l'Hopital's rule twice.

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