If $f$ is holomorphic at $z_0$, then $\frac{\partial f}{\partial \bar{z}}=0$ and $f'(z_0)=\frac{\partial f}{\partial z}(z_0)=2\frac{\partial u}{\partial z}(z_0)$
Also if we write $F(x,y)=f(z)$, then $F$ is differentiable in the sense of real variables and $\det J_F(x_0,y_0)=\vert f'(z_0)\vert ^2$.
Where $\frac{\partial}{\partial z}= \frac{1}{2}(\frac{\partial}{\partial} z +\frac{1}{i}\frac{\partial}{\partial y})$ and $\frac{\partial}{\partial \bar{z}}=\frac{1}{2}(\frac{\partial}{\partial x}-\frac{1}{i}\frac{\partial }{\partial y})$ are differential operators.
This is from Stein and Sharkarchi, Complex analysis. Previous to this the book showed the Cauchy Riemann equations $\frac{\partial f}{\partial x}(z_0)=\frac{1}{i}\frac{\partial f}{\partial y}(z_0)$
I see that $\frac{\partial f}{\partial \bar{z}}(z_0)=\frac{1}{2}(\frac{\partial f}{\partial x}-\frac{1}{i}\frac{\partial f}{\partial y})=0$ directly from the cauchy riemann equations.
And that $f'(z_0)==\frac{\partial f}{\partial z}(z_0)$ also from cauchy riemann equations.
Then also $\frac{\partial f}{\partial z}=2\frac{\partial u}{\partial z}$
Which I think comes from this:
$\frac{1}{2}(\frac{\partial f}{\partial x}(z_0)+\frac{1}{i}\frac{\partial f}{\partial y}(z_0))=\frac{1}{2}\biggr((u_x (x,y),v_x(x,y))+i(u_y(x,y), v_y(x,y))\biggr)$
But you would write it as $u_x+iv_x+-iu_y+v_y=(u_x+v_y)+i(v_x-u_y)$ which by Cauchy Riemann equations $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$
So I end up with $(u_x+v_y)+i(v_x-u_y)=2u_x-2iu_y$ which I thought I should have $2\frac{\partial u}{\partial x}=2u_x+2iu_y$
Im not sure if when $\frac{\partial}{\partial z}$ is written is this supposed to mean the partials written as a complex number with $z=x+iy$? Or does this mean the partial derivative as a vector of the component functions partial derivatives?
Best Answer
We have (omitting the arguments for brevity) $$ \frac{\partial}{\partial z} f = \frac{\partial}{\partial z}(u +iv) = \frac12 \left((u_x +iv_x) + \frac 1i (u_y + iv_y)\right) = \frac12 \left(u_x +v_y + i(v_x-u_y)\right) \, . $$ Using the Cauchy-Riemann equations this is equal to $$ \frac12 \left(u_x +u_x + i(-u_y-u_y)\right) = u_x + \frac 1i u_y = 2\frac{\partial}{\partial z} u \, . $$