# Why does $\frac{\partial}{\partial\theta}f(re^{i\theta})=i\frac{\partial}{\partial r}f(re^{i\theta})$?

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 $=$. –  Henning Makholm Apr 5 '12 at 0:12

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. –  ankitha Apr 5 '12 at 0:19