Complex derivative in terms of partial derivatives

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!

Solution 1:

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

Solution 2:

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.