Prove equation in complex numbers Prove that for all $z\in \mathbb{C}\setminus \mathbb{R}_{\leq 0}$ following equation is satisfied:
$$
\sqrt z=\sqrt{|z|}\frac{|z|+z}{||z|+z|}
$$
Here's what I've got so far:
$$
\sqrt z \cdot ||z|+z|=\sqrt{|z|}\cdot(|z|+z)
$$
Now I square both sides:
$$
z \cdot (|z|+z)^2=|z|\cdot(|z|+z)^2
$$
Therefore:
$$
z=|z|
$$
Which obviously isn't true for all of $z$ from the given set.
How to prove this correctly? Where did I made the mistake?
 A: Since $z\in\Bbb C\setminus\Bbb R_{\le0}$, in particular $z\neq0$ so your equation equals
$$
\sqrt{\frac{z}{|z|}}=\frac{|z|+z}{||z|+z|}\;.
$$
Write now $z$ in polar form
$$
z=re^{i\theta}
$$
and substitute it in the previous equation will lead to
\begin{align*}
e^{i\left(\frac{\theta}2+k\pi\right)}
=\frac{r(1+e^{i\theta})}{|r(1+e^{i\theta})|}
=\frac{1+e^{i\theta}}{|1+e^{i\theta}|}\;,
\end{align*}
for $k=0,1$ according to the choice of the branch of the square root.
Can you go on from here?
A: We write $z$ as $z=R\cdot e^{ix}, R\neq 0$ so $\vert z\vert = R$ and $\overline{z} = \overline{R\cdot e^{ix}}=R\cdot e^{-ix}$ and notice that $\vert z\vert =\sqrt{z\cdot \overline{z}}$ .Starting from the $RHS$:
$$\sqrt{|z|}\frac{|z|+z}{||z|+z|}=\sqrt{R}\frac{R+R\cdot e^{ix}}{R\cdot|1+e^{ix}|}=\sqrt{R}\frac{1+\cdot e^{ix}}{\sqrt{\left(1+e^{ix}\right)\left(1+e^{-ix}\right)}}=\sqrt{R\cdot \frac{1+e^{ix}}{1+e^{-ix}}}=\sqrt z$$
A: If $|z|+z\not=0$ (i.e., if $z\in\mathbb{C}\setminus\mathbb{R}_{\le0}$), then
$$\left(\sqrt{|z|}{|z|+z\over||z|+z|}\right)^2=|z|{(|z|+z)(|z|+z)\over(|z|+z)(|z|+\overline{z})}={|z|^2+z|z|\over|z|+\overline{z}}={z\overline{z}+z|z|\over|z|+\overline{z}}=z$$
Observing also that $|z|\gt|\Re(z)|$ (the real part of $z$) if $z\not\in\mathbb{R}$, so that $\Re(|z|+z)\gt0$ if $z\in\mathbb{C}\setminus\mathbb{R}_{\le0}$, it follows that
$$\sqrt{|z|}{|z|+z\over||z|+z|}=\sqrt z$$
with the function-defining convention that $\Re(\sqrt z)\gt0$ for $z\in\mathbb{C}\setminus\mathbb{R}_{\le0}$. (Note, this accords with $\sqrt{e^{i\theta}}=e^{i\theta/2}$ using the convention $-\pi\lt\theta\le\pi$, not the convention $0\le\theta\lt2\pi$.)
A: I will show the geometric intuition behind this equality. To show the equality of two complex numbers, one can equally show the equality of their (1) modules and (2) arguments. Let $z\in \mathbb{C}\setminus \mathbb{R}_{\leq 0}$ such that $z=re^{i\phi}$.
(1) Equality of modules:
$$|\sqrt{z}|=\left| \sqrt{|z|}\frac{|z|+z}{||z|+z|}\right| $$
$$\sqrt{|z|}= \sqrt{|z|}\frac{\left||z|+z\right|}{||z|+z|}$$
$$\checkmark$$
(2) Due to (1) w.l.o.g let $|z|=1$. Let's check the equality of the arguments:
$$\text{Arg} \sqrt{z}=\text{Arg} \sqrt{e^{i\phi}}=\text{Arg}e^{i\frac{\phi}{2} }=\frac{\phi}{2}$$
$$\text{Arg}\sqrt{|z|}\frac{|z|+z}{||z|+z|}=\text{Arg}(1+e^{i\phi})$$
equality $\text{Arg}(1+e^{i\phi})=\frac{\phi}{2}$shows the drawing:

