This particular problem was asked in my assignment which could not be discussed due to pendamic.
Question : If f is holomorphic on $\Omega$/{a} prove that $e^{f(z)}$ cannot have a pole at a .
What I tried :If f has a pole of order k at a and let $|e^{f(z)}|\to \infty$ as z$\to$ a. then I took $g(z)=\frac{f(z)} {(z-a)^k}$. Then $\frac{g(z)}{(z-a)^k} \to \infty $ as z$\to a$ .( But I don't know how to proceed in this case from here).
Let f have an isolated singularity. I am confused in this part on which property should be used to prove that pole does not exists .
If f has an essential singularity at a then also , I am confused on which result to use . Can it be proved that $e^{f(z)}$ will also have an isolated singularity at z$\to $ a ? If yes kindly give some hints .
Any guidence will be really appreciated .