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Is the residue of $\cos(\frac{z}{1-z})$ at z=1 : sin(1)?

i.e $\frac{1}{2\pi i}\int_c\cos(\frac{z}{1-z})dz=Res(\frac{1}{z^2}f(\frac{1}{z}),0)$=$\frac{1}{2\pi i}\int_c\frac{\cos(\frac{1}{z-1})}{z^2}dz$

Just wanted to clarify the method.

Edit

Using the following theorem: enter image description here

Jhon Doe
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  • Isolated essential singularity. Try Expanding the function about 1. Or Substitute $z-1=w$ and expand about zero. See https://math.stackexchange.com/questions/168355/residue-of-fz-fracz1-cosz-at-z-0 – Sujit Bhattacharyya Sep 16 '18 at 14:08

1 Answers1

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Note that\begin{align}\cos\left(\frac z{1-z}\right)&=\cos\left(-1-\frac1{1-z}\right)\\&=\cos\left(1-\frac1{z-1}\right)\\&=\cos(1)\cos\left(\frac1{z-1}\right)+\sin(1)\sin\left(\frac1{z-1}\right)\\&=\cos(1)\left(1-\frac1{2!}\frac1{(z-1)^2}+\cdots\right)+\sin(1)\left(\frac1{z-1}-\frac1{3!}\frac1{(z-1)^3}+\cdots\right)\end{align}and therefore the residue is $\sin(1)$, since that's the coefficient of $\frac1{z-1}$.

José Carlos Santos
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