Ytukyg

Let $sqrt{}:mathbb{C}setminus(-infty,0]rightarrowmathbb{C},sqrt{z}=exp(frac{1}{2}Log(z))$ be the main branch of the sqaure root and Log(z) the main branch of the logarithm. For which values $zinmathbb{C}$ applies $sqrt{z^2}=z$?

Can anyone help?

Hint: Show first that if $winBbb Csetminus (-infty,0]$ and $zinBbb C$ then

$sqrt w=z$ if and only if $z^2=w$ and $z$ has positive real part.

Since it's been suggested that there's some problem with this, here's a proof:

First note that by definition $sqrt w=sqrt re^{it/2}$ if $w=re^{it}$ with $r>0$ and $-pi<t<pi$. (The constraint on $t$ follows from the definition of the principal value (or "main branch" in the OP) logarithm.)

Suppose first that $w=z^2$ and $z$ has positive real part. Then $z=re^{it}$ where $r>0$ and $-pi/2<t<pi/2$. Now $w=r^2e^{2it}$, and since $-pi<2t<pi$ this shows that $sqrt{w}=re^{it}=z$.

Conversely, if $z=sqrt w$ then certainly $w=z^2$; also $z=sqrt re^{it/2}$, where $r>0$ and $-pi<t<pi$; hence $-pi/2<t/2<pi/2$, so $z$ has postive real part.

Cor. $sqrt{z^2}=z$ if and only if $Re z>0$.

Proof. Note first that if $Re z=0$ then $sqrt{z^2}$ is undefined. Assume $Re zne0$. Let $s=sqrt{z^2}$.

Now $s^2=z^2$, so $s=pm z$. Above we've shown that $Re s>0$, so if $s=z$ then $Re z>0$.
Conversely, if $Re z>0$ then $sne -z$ since $Re s>0$, hence $s=z$.