Show that $z=0$ is a pole for $$f(z) = \frac {\sin z}{e^{\tan z} - e^{\sin z}}$$ and find the order of the pole.
$\lim_{z\to 0} \sin z = 0$, and $\lim_{z\to0} e^{\tan z} - e^{\sin z} = 0$ , so we can apply l'hopital's rule $$\frac {\cos ^3 z}{e^{\tan z} - e^{\sin z} \cos^3z}$$ which diverges to infinity.
So $\lim_{z\to0}f(z) = \infty$, which means it has a pole at $z=0$.
Now i stacked in second part, in finding order.
I know that it is enough to find zero order of $\frac 1{f(z)}$ function, but it doesn't helped me much.
Any help or advice is appreciated.
Numerator $$ \sin z=z\,f(z), \quad \text{with $f(0)\ne 0$.} $$ Denominator $$ g(z)=e^{\tan z}-e^{\sin z} $$ with $g(0)=0$, $$ g'(z)=e^{\tan z}\sec^2 z-e^{\sin z}\cos z, \quad \text{and $g'(0)=0$,} $$ $$ g''(z)=e^{\tan z}\sec^4 z+2e^{\tan z}\sec^3z\sin z-e^{\sin z}\cos^2 z+2e^{\sin z}\sin z $$ and $g''(0)=0$.
Finally $$ g'''(z)=8e^{\tan z}\sec^6 z+7e^{\tan z}\sec^5z\sin z-e^{\sin z}\cos^4 z+2e^{\sin z}\sin z\cos z+2e^{\sin z}\sin z\cos^2 z+2e^{\sin z}\cos z $$ and hence $g'''(0)=9\ne 0.$ Hence $g(z)=z^3h(z)$, with $h(0)\ne 0$.
Thus $$ \frac{\sin z}{e^{\tan z}-e^{\sin z}}=\frac{zf(z)}{z^3h(z)}=\frac{w(z)}{z^2}, $$ where $w(0)\ne 0$.
Hence second order pole.