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Let $f: \mathbb{C} \rightarrow \mathbb{C}$ holomorphic. Now, if we write $f(x+iy) = u(x,y) + iv(x,y)$ with $u,v$ harmonic, is there a way to write the complex derivative $\frac{d f}{d z}$ in terms of the partial derivatives of $u,v$?

Thanks for your help!

baguette
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    According to Stein, proposition 2.3 on page 12 says that if $f$ is holomorphic at $z_0$, then we have $f'(z_0)=\frac{\partial d}{\partial z}(z_0) = 2\frac{\partial u}{\partial z}(z_0)$ where it has been defined previously that $\frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x}+\frac{1}{i}\frac{\partial}{\partial y}\right)$. – Kenny Hegeland Aug 12 '13 at 17:09
  • I'm sorry, what is $d$? – baguette Aug 12 '13 at 17:15
  • Oops, that should read $\frac{\partial f}{\partial z}(z_0)$, I'm writing up a more thorough answer currently. – Kenny Hegeland Aug 12 '13 at 17:16
  • Also, the fact that $u,v$ are harmonic isn't even necessary, for $f$ holomorphic implies that $u$ and $v$ are harmonic. – Kenny Hegeland Aug 12 '13 at 17:36

3 Answers3

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$$\frac{d f}{d z}(z_0)=\frac{\partial u}{\partial x}(z_0) + i\frac{\partial v}{\partial x}(z_0)$$.

Georges Elencwajg
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First we will define $$ \frac{\partial}{\partial z} = \frac{1}{2}\left(\frac{\partial}{\partial x}+ \frac{1}{i}\frac{\partial}{\partial y}\right) $$ and $$ \frac{\partial}{\partial \bar{z}}=\frac{1}{2}\left(\frac{\partial}{\partial x}-\frac{1}{i}\frac{\partial}{\partial y}\right) $$

I'm skipping some of the stuff in proposition 2.3 of Stein, however, the part I believe you will be interested in is the following,

If $f$ is holomorphic, then we have $$ \frac{\partial f}{\partial \bar{z}}(z_0) = 0 $$

and

$$ f'(z_0) = \frac{\partial f}{\partial z}(z_0) = 2\frac{\partial u}{\partial z}(z_0). $$

The Cauchy-Riemann equations give us $$ f'(z_0) = \frac{\partial f}{\partial x}(z_0) = \frac{1}{i}\frac{\partial f}{\partial y}(z_0)\Rightarrow f'(z_0) = \frac{1}{2}\left(\frac{\partial f}{\partial x}(z_0) + \frac{1}{i}\frac{\partial f}{\partial y}(z_0)\right)=\frac{\partial f}{\partial z}(z_0). $$

Now if we use the Cauchy-Riemann equations we see that $\frac{\partial f}{\partial z}(z_0) = 2\frac{\partial u}{\partial z}(z_0)$. It is useful (and quite simple) to derive this and I suggest you do so to get more comfortable with Cauchy-Riemann equations.

Kenny Hegeland
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If you look up The Cauchy-Riemann equations on Wikipedia you'll see how to use the chain rule to do this complex differentiation.

JohnS
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