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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.

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    There seems to be missing a factor of $r$ to the right of the $=$.2012-04-05

1 Answers 1

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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})

and

\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

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

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    Thank makes sense, thanks.2012-04-05