Determine the nature of singularities and calculate the residue of $f(z)=\frac{e^z-\mathrm{sin}z-1}{z^5+z^3}$

Question

$$f(z)=\frac{e^z-\mathrm{sin}z-1}{z^5+z^3},\;\;\;\;\;\;\; \mathrm{Res}[f(z),0]$$

I am having trouble determining the nature of singularities. This is what I managed to do:

$$f(z)=\frac{e^z-\mathrm{sin}z-1}{z^5+z^3}=\frac{e^z-\mathrm{sin}z-1}{z^3(z^2+1)}=\frac{e^z-\mathrm{sin}z-1}{z^3(z+i)(z-i)}$$

so the singularities are:

$\bullet\;\;\; z=0 \;\;\;\;\;\;\;\;\; $ pole of order 3

$\bullet\;\;\; z=\pm i \;\;\;\;\;\;\; $ poles of order 1

But there is the numerator $e^z-\mathrm{sin}z-1$ that I don't know what to do with.

Any help is appreciated, thank you.

Answer 1

$f(z)=\dfrac{e^z-1}{z}\dfrac{1}{z^2(z+i)(z-i)}-\dfrac{\sin z}{z}\dfrac{1}{z^2(z+i)(z-i)}$

Answer 2

Expand both numerator and denominator into a Taylor series around the singularities and then perform long division to find the corresponding Laurent series. For example:

$$\exp(z)-\sin(z)-1=\frac{1}{2}z^2+\frac{1}{3}z^3+\frac{1}{24}z^4+O(z^6)$$

and the denominator is simply:

$$z^3+z^5$$

A long division gives:

$$f(z)=\frac{1}{2z}+\frac{1}{3}-\frac{11}{24}z-\frac{1}{3}z^2+O(z^3)$$

Conclude that $z=0$ is a pole of order 1 and $f(z)$ has residue $\frac{1}{2}$ there.

Continue similarly with the other singularities ($\pm i$).