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

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

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

2. Hello! Here is an idea.

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

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

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

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

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