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Consider the function $\cos\left(\frac{z}{z+1}\right)$, which has an essential singularity at the point $z=-1$.

How does one compute its residue at $z=-1$.

2 Answers 2

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Let's shift $z$ by $1$ and equivalently find the residue of $\cos((z-1)/z)=\cos(1-1/z)$ at $z=0$. From the Taylor series for $\cos z$, we have

$\cos\left(1-\frac1z\right)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\left(1-\frac1z\right)^{2k}\;.$

The residue at $0$ is the sum of the coefficients of $1/z$, which is

$ \begin{eqnarray} \sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}(-2k) &=& \left[-\frac{\mathrm d}{\mathrm dz}\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}z^{2k}\right]_{z=1} \\ &=& \left[-\frac{\mathrm d}{\mathrm dz}\cos z\right]_{z=1} \\ &=& \sin 1 \\ &\approx& 0.841471\;. \end{eqnarray}$

Here's a check of the result by contour integration using Wolfram|Alpha.

Note that the specific form of the Taylor series for $\cos z$ wasn't actually used, so more generally the residue at $0$ of $f(1+1/z)$ for any entire function $f(z)$ is f'(1). Actually we can see this more directly by expanding $f(z)$ around $z=1$; then

f\left(1+\frac1z\right)=f(1)+f'(1)\frac1z+\frac1{2!}f''(1)\frac1{z^2}+\dotso\;,

from which the residue is again f'(1).

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    @Robert: I see. Thanks, I fixed it.2012-03-19
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$ A=\cos\left(\frac{z}{z+1}\right) = \cos\left(\frac{z\color{red}{+1-1}}{z+1}\right)=\cos \left( 1- \frac{1}{z+1} \right) $ $ \cos(a\pm b)=\cos(a)\cos(b) \mp \sin(a)\sin(b)$ $A=\cos(1)\cos\left(\frac{1}{z+1}\right)+\sin(1)\sin\left(\frac{1}{z+1}\right) $ $ \cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-... \hspace{10mm} \sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-... $ $ \cos\left(\frac{1}{z+1}\right)=1-\frac{1}{(z+1)^2 2!}+\frac{1}{(z+1)^4 4!}-... $ $ \sin\left(\frac{1}{z+1}\right)=\frac{1}{z+1}-\frac{1}{(z+1)^3 3!}+\frac{1}{(z+1)^5 5!}-... $ $ A=\cos(1) \left[ 1-\frac{1}{(z+1)^2 2!}+\frac{1}{(z+1)^4 4!}-...\right]+\sin(1)\left[ \frac{1}{z+1}-\frac{1}{(z+1)^3 3!}+\frac{1}{(z+1)^5 5!}-... \right]$ $ \text{ so residue is coefficient of term } \frac{1}{z+1} \rightarrow \sin(1)$