The square root of $z^2$ is analytic?

Question

Let us consider a disk $|z|<\epsilon$. Clearly, the square root of $z$ is not analytic on this disk. In order to show it, I compute the intergral over a path $\gamma$ containing $z=0$ in $Int(\gamma)$. But, what can I say of the principal branch of the square root of $z^2$? Is it holomorphic in the disk $|z|<\epsilon$?

Answer 1

Your function $z\to z^2$ is $1$ at a neighbourhhod of $1$, then if you are in such of the analytic continuation of $\sqrt{z^2}$ which coincides with $z$ in a neighbourhood of $1$ (using the expansion of $\sqrt{1+h}$), it is $z$ and then analytic.

As regards the slit plane for $z^2$, $\mathbb{C}\setminus ]-\infty,0]$, your function $\exp(\frac{1}{2}\log(z^2))$ is defined on $\mathbb{C}\setminus i\mathbb{R}$, so you must choose which part you want. If you stick at the neighbourhood of one (then $\Re(z)>0$) you get $z$, on the other hand (then $\Re(z)<0$), you get $-z$.

Answer 2

The proposed function is not continuous on the imaginary axis, negating any holomorphy claims on this basic level. Indeed, the square of $z=\pm\delta+ib$ wth $0<δ\ll|b|$ is $$ z^2=δ^2-b^2\pm i2bδ. $$ which lie on opposite sides of the negative real ray, the branch cut for the usual complex square root.

The sign the main branch of the square root takes values on the half-plane with positive real values, which would be $$ +δ+ib\mapsto δ+ib\\ -δ+ib\mapsto δ-ib $$ which gives a jump in the sign while crossing the imaginary axis.