2
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Ah i see. Thank you very much! $\endgroup$ – Timothy Mar 2 '18 at 10:41
  • $\begingroup$ That's a fine approach, but you don't "have to use" that result. $\endgroup$ – zhw. Mar 5 '18 at 18:04
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.