I want to know, why $Re\Big\{ \frac{zf''(z)}{f'(z)}\Big\}=|z|\frac{\partial}{\partial|z|} Re(log(f'(z)))$ I don't know how to prove this, $Re\Big\{ \frac{zf''(z)}{f'(z)}\Big\}=|z|\frac{\partial}{\partial|z|} Re(log(f'(z)))$.
I know it is related whit cauchy riemann equations and partial derivation.
 A: Let $g=u+iv$ be an analytic function with real part $u$ and imaginary part $v$.
We note the operator identity 
$$\left(|z|\frac{\partial }{\partial |z|}\right) \lbrace\cdot\rbrace =\left(x\frac{\partial }{\partial x}+y\frac{\partial }{\partial y}\right) \lbrace\cdot\rbrace \tag 1$$
First, we apply $(1)$ to the real part of $g$ and find that
$$\left( |z|\frac{\partial }{\partial |z|}\right) \lbrace \text{Re}\left(g\right)\rbrace =\left(x\frac{\partial }{\partial x}+y\frac{\partial }{\partial y}\right) \lbrace u \rbrace \tag 2$$
Next, we note that 
$$\begin{align}
\text{Re}\left(z\frac{dg}{dz}\right)&=\text{Re}\left((x+iy)\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)\right) \tag 3\\\\
&=\text{Re}\left((x+iy)\left(\frac{\partial u}{\partial x}-i\frac{\partial u}{\partial y}\right)\right) \tag 4\\\\
&=\left(x\frac{\partial }{\partial x}+y\frac{\partial }{\partial y}\right) \lbrace u \rbrace \tag 5
\end{align}$$
where we used the Cauchy-Riemann equations in going from $(3)$ to $(4)$.
Comparing $(2)$ and $(5)$ reveals the coveted identity
$$\bbox[5px,border:2px solid #C0A000]{\left( |z|\frac{\partial }{\partial |z|}\right) \lbrace \text{Re}\left(g\right)\rbrace =\text{Re}\left(z\frac{dg}{dz}\right)}$$

NOTE:
For the problem of interest, $g=\log f'$ with $\frac{d}{dz}\log f'=\frac{f''}{f'}$.
A: $\log(z)=\log(|z|)+i\arg(z)$. Also $\frac{d}{dz}|z|=\frac{z}{|z|}.$ Can you finish it now?
