Prove the identity $\frac{d}{dt} |z(t)|= |z(t)|Re \frac{z'(t)}{z(t)}$ Let $t \mapsto z(t)$ be a complex-valued function of the real variable $t$. Assume further that $z'(t) \not= 0$ for every $t$. Prove the identity $\frac{d}{dt} |z(t)|= |z(t)|Re \frac{z'(t)}{z(t)}$. By the way what kind of identity is this I can not find anything about this problem.
Let $z(t)=x(t)+iy(t) \neq 0 \ \forall t\ $, then 
$$
\frac{d}{dt}|z(t)| = \frac{d}{dt} \sqrt{x^2(t)+y^2(t)} = \frac{1}{2} \frac{2[x(t)x'(t)+y(t)y'(t)]}{\sqrt{x^2(t)+y^2(t)}} = \frac{x(t)x'(t)+y(t)y'(t)}{|z(t)|} \ \ \ \ \ \ (1)
$$
On the other hand, 
$$
\frac{z'(t)}{z(t)} = \frac{x'(t)+iy'(t)}{x(t)+iy(t)}=\frac{x(t)x'(t)+y(t)y'(t)+ i [x(t)y´(t)-y(t)x'(t)]} {x^2(t)+y^2(t)}
$$
then 
$$
Re\left(\frac{z'(t)}{z(t)}\right)=\frac{x(t)x'(t)+y(t)y'(t)} {x^2(t)+y^2(t)}=\frac{x(t)x'(t)+y(t)y'(t)} {|z(t)|^2}
$$
which gives indeed
$$
|z(t)| \cdot Re\left(\frac{z'(t)}{z(t)}\right) = \frac{x(t)x'(t)+y(t)y'(t)} {|z(t)|} \overset{(1)}{=}\frac{d}{dt}|z(t)|
$$
you have $$|z|^2 = z \bar z \to 2|z|\frac{d|z|}{dt}=\frac{d}{dt}|z|^2 = z \frac{d\bar z}{dt} + \bar z\frac{dz}{dt} =2 Re\left(\bar z\frac{dz}{dt}\right) $$ dividing both side by $2 z\bar z$ gives you the desired result.