Thread: f and e^f have no common pole?

1. f and e^f have no common pole?

Prove that an isolated sigularity of $f(z)$ cannot be a pole of $e^{f(z)}$.
I know the key is to prove that $f$and $e^f$have no common pole, but how??

2. Hello! Here is an idea.

It is easy to see that this is true if $z_0$ is an essential or removable singularity of $f(z)$.

In the case of an $n$-th order pole, then I claim that $z_0$ is an essential singularity of $g(z)=e^{f(z)}$. Clearly it is not a regular point, so it suffices to show that $|g(z)|$ does not tend to infinity as $z \rightarrow z_0$. To do this, it suffices to show that every neighbourhood of $z_0$ contains points $\zeta$ such that $\mbox{Re } f(\zeta), for some fixed $M$.

3. Originally Posted by Bruno J.
Hello! Here is an idea.

It is easy to see that this is true if $z_0$ is an essential or removable singularity of $f(z)$.

In the case of an $n$-th order pole, then I claim that $z_0$ is an essential singularity of $g(z)=e^{f(z)}$. Clearly it is not a regular point, so it suffices to show that $|g(z)|$ does not tend to infinity as $z \rightarrow z_0$. To do this, it suffices to show that every neighbourhood of $z_0$ contains points $\zeta$ such that $\mbox{Re } f(\zeta), for some fixed $M$.
Let $f(z)$ have a pole at $z_0$. I claim that $g(z)=e^{f(z)}$ has an essential singularity at $z_0$.
The problem is solved if we can show that there exists a sequence $\{z_n\} \rightarrow z_0$ such that $|e^{f(z_n)}|$ is bounded as $n \rightarrow \infty$; this is because if $z_0$ were a pole of $g(z)$ then we would have $g(z_n)\rightarrow \infty$ as $n \rightarrow \infty$. This is equivalent to finding a sequence such that $\mbox{Re } f(z_n)$ remains bounded, since $|e^{x+iy}|=|e^x|$. Since we know that $|f(z_n)|$ necessarily blows up to infinity as as $n \rightarrow \infty$, it is the imaginary part which must become very big.
Consider $F(z)=\frac{1}{f(z)}$. This is clearly holomorphic at $z_0$ and $F(z_0)=0$. Let $N_1\supset N_2 \supset \dots$ be a sequence of bounded open neighbourhoods of $z_0$ whose intersection is $z_0$. Then $\{F(N_n)\}$ is a sequence of neighbourhoods of $0$; in particular, each of the $F(N_n)$ is bounded, and thus $S_n=\{1/F(z) : z \in N_n\}$ is a sequence of neighbourhoods of infinity. So there are points of the imaginary axis in each of the $S_n$; pick $z_n \in N_n$ such that $1/F(z) \in i\mathbb{R}$. Then the sequence $z_n$ converges to $z$ and $\mbox{Re } 1/F(z_n)=\mbox{Re } f(z_n) =0$. Thus $|e^{f(z_n)}|=1$ for $n \geq 1$ and we are done.