Show that there is no harmonic function $v$ in $\mathbb{C}^*$ such that $u+iv$ is holomorphic in $\mathbb{C}^*$

Question

The question:

Let $u$ be a harmonic function in $\mathbb{C}^*$ such that $\lim_{z\rightarrow 0} u(z) = +\infty$. Show that there is no harmonic function $v$ in $\mathbb{C}^*$ such that $u+iv$ is holomorphic in $\mathbb{C}^*$

How do i start with such a question? I think contradiction will have to be used, say $f=u+iv$ with $f$ holomorphic in $\mathbb{C}^*$ I tried considering $\dfrac{e^{f}}{u}$ but it doesnt seem to lead anywhere.. Any ideas/hints?

Answer 1

You can use the following

Theorem: If $f$ is holomorphic in $\mathbb{C}^*$ and if $f$ has in $0$ a non-removable singularity, then $e^f$ has an essential singularity in $0$.

To your question: If $f=u+iv$ is holomorphic in $\mathbb{C}^*$ and if $\lim_{z\rightarrow 0} u(z) = +\infty$, then

$|e^{f(z)}|= e^{u(z)} \to + \infty$ as $z \to 0$. This shows that $e^f$ has a pole at $0$. The above theorem shows that this is impossible.

Answer 2

Lemma: Suppose $f$ is holomorpic in $\{0<|z|<R\}$ and $f$ has a pole at $0.$ Then for all $0<r\le R,$ there exists $z_r\in \{0<|z|<r\}$ such that $\text { Re }f(z_r)<0.$

Proof: Let $N\in \mathbb N$ be the order of the pole. Then there exists $c\ne 0$ such that

$$f(z)=\frac{c}{z^N} + O\left (\frac{1}{|z|^{N-1}}\right) \text { as }z \to 0.$$

Write $c=|c|e^{it}$ and consider $z= re^{i(t+\pi)/N}.$ Then

$$f(re^{i(t+\pi)/N})=-\frac{|c|}{r^N} + O\left (\frac{1}{r^{N-1}}\right).$$

As $r\to 0^+$ the real part of the expression on the right is negative, proving the lemma.

The lemma solves your problem: If there were such a $v$ as in the statement of your problem, then $f=u+iv$ would have a pole at $0,$ proving that $u$ takes on negative values in every deleted neighborhood of $0,$ contradiction.