Singularity of $f(z)=\frac {\sin z}{e^{-z}+z-1}$

Question

What's the kind of singularity $z_0=0$ of $f(z)$: $$f(z)=\frac {\sin z}{e^{-z}+z-1}$$

I know I should write it in the form:

$$f(z)=\frac{\phi(z)}{(z-z_0)^n}$$

but I'm unable to do it.

Answer 1

Here is how you advance. Recalling the Taylor series, we have $$ \frac {\sin z}{e^{-z}+z-1} = \frac{ \sin z}{(1-z+z^2/2-\dots)+z-1}=\frac{\sin(z)}{z^2/2-z^3/3!+\dots} $$

$$ = \frac{\sin(z)}{z^2( 1/2-z/3!+\dots)} . $$

Now, you can see that the order of the pole is $1$.

Answer 2

Another approach:

$$\lim_{z\to 0}\,z\,f(z)=\frac{z\sin z}{e^{-z}+z-1}\stackrel{\text{l'Hospital}}=\lim_{z\to 0}\frac{\sin z+z\cos z}{-e^{-z}+1}\stackrel{\text{l'H}}=$$

$$=\lim_{z\to 0}\frac{2\cos z-z\sin z}{e^{-z}}=2$$

and thus this is a pole of order one and residue $\;2\;$ .

Caution: The above approach is works nicely mostly when we have a pole of low order ($\;1,2,3\;$) , otherwise it can be weary to repeat the above with $\;z^2f(z)\;,\;\;z^3f(z)\;$ and etc. But it is usually so simple that is, imo., worth a shot.