Computing the $\operatorname{res}_{z=0}\frac{e^z-1-z}{(1-\cos(2z))\sin z}$

Question

Compute the $$\operatorname{res}_{z=0}\frac{e^z-1-z}{(1-\cos(2z))\sin z}$$

I simplified the expression in the following way:

$\frac{e^z-1-z}{(1-\cos(2z))\sin z}=\frac{e^z-1-z}{2\sin^3 z}$

However since I cannot use l'hopital's rule, I have no idea on how to compute the limit in order to determine the singularity type of $0$.

Question:

How should I determine the singularity type at 0? Which kind of procedure shall I use to compute $\lim_{z\to 0}\frac{e^z-1-z}{(1-\cos(2z))\sin z}=\frac{e^z-1-z}{2\sin^3 z}$?

Thanks in advance!

Answer 1

Since$$e^z-1-z=\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots$$and\begin{align}\bigl(1-\cos(2z)\bigr)\sin(z)&=\left(\frac{4z^2}{2!}-\frac{16z^4}{4!}+\cdots\right)\left(z-\frac{z^3}{3!}+\cdots\right)\\&=2z^3-z^5+\cdots\end{align}you have\begin{align}\frac{e^z-1-z}{\bigl(1-\cos(2z)\bigr)\sin(z)}&=\frac{\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots}{2z^3-z^5+\cdots}\\&=\frac1z\times\frac{\frac1{2!}+\frac z{3!}+\cdots}{2-z^2+\cdots}\end{align}and therefore$$\operatorname{res}_{z=0}\left(\frac{e^z-1-z}{\bigl(1-\cos(2z)\bigr)\sin(z)}\right)=\frac{\frac1{2!}}2=\frac14.$$

Answer 2

Use the known limit

$$\lim_{x\to0}\frac{\sin(x)}x=1$$

and with $z\mapsto-z$, we get

$$L=\lim_{z\to0}\frac{e^z-1-z}{z^2}=\lim_{z\to0}\frac{e^{-z}-1+z}{z^2}$$

which, when added together, gives us

$$2L=\lim_{z\to0}\frac{e^z-2+e^{-z}}{z^2}=\left[\lim_{z\to0}\frac{e^{z/2}-1}z+\frac{e^{-z/2}-1}{-z}\right]^2$$

which can be easily computed, assuming $L$ exists. Then rewrite your function as

$$\frac12\frac{e^z-1-z}{z^2}\frac{z^3}{\sin^3(z)}\frac1z$$

and observe the behavior as $z\to0$.