Finding the residue of $\frac{e^z-1-z}{\sin z -z}$ at $z=0$ by taylor expansion.

Question

I am trying to find the residue of $\frac{e^z-1-z}{\sin z -z}$ at $z=0$. What I have done so far is to taylor expand the numerator and denominator:

$$\frac{\frac{1}{2}z^2+O(z^3)}{-\frac{1}{6}z^3+O(z^5)}$$

However, I am now stuck and dont see how to divide this rational function to find the coefficient of the $z^{-1}$ term.

Any tips? Any other methods to suggest?

Answer 1

You can notice that $0$ is a zero of order $2$ of the numerator and of order $3$ of the denominator. Hence it is a simple pole and the residue is simply given by $$\text{Res}_0(f) = \lim_{z\to 0} zf(z).$$ In your case \begin{align*}\lim_{z\to 0} z\frac{e^z-1-z}{\sin z -z}&=\lim_{z\to 0}\frac{(e^z-1-z)+z(e^z-1)}{\cos(z)-1}\\ & = \lim_{z\to 0} \frac{e^z-1+ze^z+e^z-1}{-\sin(z)}\\ & = \lim_{z\to 0} \frac{3e^z+ze^z}{-\cos(z)}\\ & = -3 \end{align*}

where I have applied multiple times the complex version of De l'Hospital theorem.

Answer 2

Since the power series expansion of $\sin(z)$ is $$ \sin(z)=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots $$ the power series expansion of $\sin(z)-z$ is $$ \sin(z)-z=-\frac{z^3}{3!}+\frac{z^5}{5!}-\cdots=z^3\left(-\frac{1}{3!}+\frac{z^2}{5!}-\cdots\right) $$ Therefore, $\sin(z)-z=z^3g(z)$ where $g(z)$ is an entire function and $g(0)=-\frac{1}{3!}\not=0$.

On the other hand, the power series expansion of $e^z$ is $$ e^z=1+z+\frac{z^2}{2}+\frac{z^3}{3!}+\cdots. $$ Therefore, the power series expansion of $e^z-1-z$ is $$ e^z-1-z=\frac{z^2}{2}+\frac{z^3}{3!}+\cdots $$

Therefore, $$ \frac{e^z-1-z}{\sin(z)-z}=\frac{\frac{z^2}{2}+\frac{z^3}{3!}+\cdots}{z^3g(z)}=\frac{1}{z}\frac{1}{2g(z)}+\frac{1}{3!g(z)}+\cdots. $$ Thus, the residue at $z=0$ is $\frac{1}{2g(0)}-\frac{3!}{2}=-3$.

Alternately, the numerator can be written as $e^z-1-z=z^2h(z)$ where $h(0)=\frac{1}{2}$, then your residue becomes $\frac{h(0)}{g(0)}$.