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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 .

If singularity is pole then it has an answer here:pole becomes essential singularity when lifting by exponential

If singularity is essential then it has an answer here:If $z=a$ is not the removable singularity of $f$, show that $e^{f(z)}$ has essential singularity at $z=a$.

But I am not able to find an answer when it's removable singularity .

Can anyone please answer it ? I shall be really thankful as I am not good in dealing with singularities.

Thanks

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If $e^f$ has a pole at $a$, then $|e^f|$ goes to infinity at $a$. Thus the real part of $f$ must go to infinity at $a$. This shows that, at $a$, $f$ cannot have a removable singularity, nor an essential singularity (by, eg, Casorati-Weierstrass – in the neighborhood of an essential singularity, a dense subset of $\mathbb{C}$ is reached infinitely many times). Thus $f$ has a pole.

Write $f(z)=\frac{g(z)}{(z-a)^k}$ with $g$ holomorphic at $a$ and $g(a) \neq 0$. Let $\tau$ be such that $R=g(a)e^{-i\tau} > 0$. Then, with $z=a+\epsilon e^{i(\pi+\tau)/k}$, $f(z)$ is $-\epsilon^{-k}(R+o(1))$ so its real part goes to $-\infty$, a contradiction.

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