Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading something, and don't understand why a certain equality comes up.

Suppose $u:\mathbb{C}\setminus\{0\}\to\mathbb{R}$ given by $u(z)=\ln(|z|^2)$ is a harmonic function. We want to see if $u$ has a harmonic conjugate $v$. If it does, then $f=u+iv$ is holomorphic.

But then I read that since the differential of $f$ is complex linear, then $$ \frac{\partial}{\partial\theta}f(re^{i\theta})=i\frac{\partial}{\partial r}f(re^{i\theta}). $$

I don't follow this. Even writing it out with the chain rule in terms of $u$ and $v$ doesn't make it clear to me. Why does this equality follow? Thanks.

share|cite|improve this question
There seems to be missing a factor of $r$ to the right of the $=$. – Henning Makholm Apr 5 '12 at 0:12
up vote 4 down vote accepted

I think this is an error in what you're reading. You can see that without going into details simply because $\partial/\partial r$ has units of inverse length whereas $\partial/\partial \theta$ doesn't. The derivatives are

$$\frac{\partial}{\partial\theta}f(r\mathrm e^{\mathrm i\theta})=\frac{\partial(r\mathrm e^{\mathrm i\theta})}{\partial\theta}f'(r\mathrm e^{\mathrm i\theta})=r\mathrm i\mathrm e^{\mathrm i\theta}f'(r\mathrm e^{\mathrm i\theta})$$


$$\frac{\partial}{\partial r}f(r\mathrm e^{\mathrm i\theta})=\frac{\partial(r\mathrm e^{\mathrm i\theta})}{\partial r}f'(r\mathrm e^{\mathrm i\theta})=\mathrm e^{\mathrm i\theta}f'(r\mathrm e^{\mathrm i\theta})\;,$$

so the correct relationship is

$$\frac{\partial}{\partial\theta}f(r\mathrm e^{\mathrm i\theta})=r\mathrm i\frac{\partial}{\partial r}f(r\mathrm e^{\mathrm i\theta})\;.$$

share|cite|improve this answer
Thank makes sense, thanks. – ankitha Apr 5 '12 at 0:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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