Partial Derivative of Modulus

Question

Let $f(z) = u(x,y) + i v(x,y)$. Show that $\frac{\partial}{\partial z} |f(z)| = \frac{1}{2} |f(z)| \frac{f'(z)}{f(z)}$.

Any tip/s how can I solve this item. So far, what I was to use the identity $z \overline{z} = |z|^2$, then I differentiate both sides of the equation and got stuck.

Another question before this is to prove $\frac{\partial f}{\partial z} = \frac{1}{2} \left( \frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y}\right)$, which I already solved. Maybe it can help.

Answer 1

If $f$ is holomorphic then $$\frac{\partial \overline{f}}{\partial z} = 0$$ and so $$ \frac{\partial f}{\partial z} \cdot \overline{f} = \frac{\partial \,|f|^2}{\partial z} = 2\, |f| \cdot \frac{\partial \, |f|}{\partial z} $$

or

$$ \frac{\partial \, |f|}{\partial z} = \frac{1}{2} \frac{\partial f}{\partial z} \frac{\overline{f}}{|f|} = \frac{1}{2} \frac{\partial f}{\partial z} \frac{|f|}{f} $$