residue theorem, laurent espansion, singularities and order of poles

Question

for this question i have found that for part

(i) at $z=0$ we have an essential singularity

however, I'm not sure how to solve for the residual?

(ii) at $z=0$ we have a pole of order $2$, and i have found the residual to be $1$

(iii) at $z=0$ we have an essential singularity

* however, again I'm not sure how to solve for the residual? *

Answer 1

For (i), $Res(f,0)=$coefficient of $z^{-1}=1$ , using Laurent series expansion of $\sin(1/z)$.

(iii) $Res(h,0)=\lim_{z\to 0}zh(z)=\lim_{z\to 0}\frac{z^2(e^z-e^{-z})}{1-\cos z}$ , as at $z=0$ $h$ has simple pole.

(ii) $Res(g,0)=\lim_{z\to 0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}(z-0)^mg(z)$ , for pole of order $m(>1)$ at $z=0$.

Answer 2

$$\sin \left(\frac{1}{z}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\left(\frac{1}{z}\right)^{2n+1}$$

The residue is the coefficient on the $z^{-1}$ term, which here is 1.

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Using partial fraction expansion, we see that

$$\frac{z+2}{z^2(z+1)}=\frac{-1}{z}+\frac{2}{z^2}+\frac{1}{z+1}$$

The residue is the coefficient on the $z^{-1}$ term, which here is -1.

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$$\begin{align} \frac{z(e^z-e^{-z})}{1-\cos z}&=\frac{z\sinh z}{\sin^2 (z/2)}\\\\ &=\frac{\sum_{n=0}^{\infty} \frac{z^{2n+2}}{(2n+1)!}}{\left(\sum_{n=0}^{\infty} \frac{(z/2)^{2n+1}}{(2n+1)!}\right)^2}\\\\ &=\frac{z^2\sum_{n=0}^{\infty} \frac{z^{2n}}{(2n+1)!}}{(z/2)^2\left(\sum_{n=0}^{\infty} \frac{(z/2)^{2n}}{(2n+1)!}\right)^2}\\\\ &=\frac{4\sum_{n=0}^{\infty} \frac{z^{2n}}{(2n+1)!}}{\left(\sum_{n=0}^{\infty} \frac{(z/2)^{2n}}{(2n+1)!}\right)^2} \end{align}$$

The leading term in the series of the denominator is $1$. Thus, the function of interest has no singularity at $z=0$ and therefore, the residue is $0$. In fact, the limit as $z \to 0$ of the function of interest is $4$.