Complex differential equations involving $\bar{f}$ and $f'$ I am trying to prove that given a holomorphic function $f$, $u(x, y)=|f(x+iy)|$, $F=u^2$, we have
$$\frac{\partial u}{\partial x}=\frac{\text{Re} (\bar{f}f')}{|f|}$$
$$\frac{\partial u}{\partial y}=-\frac{\text{Im}(\bar{f}f')}{|f|}$$
and 
$$\frac{\partial^2F}{\partial x^2}+\frac{\partial^2F}{\partial y^2}=4|f'(z)|^2$$
I tried to use the equation
$$\text{Re} (\bar{f}f')=\text{Re}(\overline{f})\text{Re}(f')-\text{Im}(\overline{f})\text{Im}(f')$$
But I don't have an idea how to get $\text{Re}(f')$ and $\text{Im}(f')$ into something I recognize.
Any hints?
 A: Let $z=x+iy$ and $\bar z=x-iy$.  Then,from the chain rule, we have
$$\begin{align}
\frac{\partial G(z,\bar z)}{\partial x}&=\frac{\partial G(z,\bar z)}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial G(z,\bar z)}{\partial \bar z}\frac{\partial \bar z}{\partial x}\\\\
&=\frac{\partial G(z,\bar z)}{\partial z}+\frac{\partial G(z,\bar z)}{\partial \bar z}\\\\
\end{align}$$
and 
$$\begin{align}
\frac{\partial G(z,\bar z)}{\partial y}&=\frac{\partial G(z,\bar z)}{\partial z}\frac{\partial z}{\partial y}+\frac{\partial G(z,\bar z)}{\partial \bar z}\frac{\partial \bar z}{\partial y}\\\\
&=i\frac{\partial G(z,\bar z)}{\partial z}+i\frac{\partial G(z,\bar z)}{\partial \bar z}\\\\
\end{align}$$
If $G$ is analytic, then $\frac{\partial G(z,\bar z)}{\partial \bar z}=0$.
Now suppose that $u(x,y)=|f(z)|$.  Then, we have 
$$\begin{align}
\frac{\partial u(x,y)}{\partial x}&=\frac{\partial \sqrt{f(z)\overline{ f(z)}}}{\partial x}\\\\
&=\frac{1}{2|f(z)|}\frac{\partial f(z)\overline{ f(z)}}{\partial x}\\\\
&=\frac{1}{2|f(z)|}\left(f(z)\frac{\partial \overline{ f(z)}}{\partial x}+\overline{ f(z)}\frac{\partial  f(z)}{\partial x}\right)\\\\
&=\frac{1}{2|f(z)|}2\text{Re}\left(\overline{ f(z)}\frac{\partial f(z)}{\partial x}\right)\\\\
&=\frac{1}{|f(z)|}\text{Re}\left(\overline{ f(z)}\left(\frac{\partial f(z)}{\partial z}+\frac{\partial f(z)}{\partial \bar z}\right)\right)\\\\
&=\frac{1}{|f(z)|}\text{Re}\left(\overline{ f(z)}\left(\frac{\partial f(z)}{\partial z}+0\right)\right)\\\\
&=\frac{1}{|f(z)|}\text{Re}\left(\overline{ f(z)}f'(z)\right)
\end{align}$$
We also have
$$\begin{align}
\frac{\partial u(x,y)}{\partial y}&=\frac{\partial \sqrt{f(z)\overline{ f(z)}}}{\partial y}\\\\
&=\frac{1}{2|f(z)|}\frac{\partial f(z)\overline{ f(z)}}{\partial y}\\\\
&=\frac{1}{2|f(z)|}\left(f(z)\frac{\partial \overline{ f(z)}}{\partial y}+\overline{ f(z)}\frac{\partial  f(z)}{\partial y}\right)\\\\
&=\frac{1}{2|f(z)|}2\text{Re}\left(\overline{ f(z)}\frac{\partial f(z)}{\partial y}\right)\\\\
&=\frac{1}{|f(z)|}\text{Re}\left(i\overline{ f(z)}\left(\frac{\partial f(z)}{\partial z}+\frac{\partial f(z)}{\partial \bar z}\right)\right)\\\\
&=\frac{1}{|f(z)|}\text{Re}\left(i\overline{ f(z)}\left(\frac{\partial f(z)}{\partial z}+0\right)\right)\\\\
&=\frac{1}{|f(z)|}\text{Re}\left(i\overline{ f(z)}f'(z)\right)\\\\
&=-\frac{1}{|f(z)|}\text{Im}\left(\overline{ f(z)}f'(z)\right)
\end{align}$$
A: For the first equation only. Let $f=a+i b $ where $a,b $ are real. For brevity let $\frac {\partial a}{\partial x}=a_1 $ and $\frac {\partial b}{\partial x}=b_1. $ When $0\ne u=|f|=\sqrt {a^2+b^2} $ we have $$\frac {\partial u}{\partial x}=\frac {2a a_1+2b b_1}{2 \sqrt {a^2+b^2}}=\frac {a a_1+b b_1}{|f|}.$$ On the other hand  $$\text {Re}(\bar f f')=Re ((a-i b)(a_1+i b_1))=a a_1+b b_1.$$
