Can we make $f(z) = \sqrt{ z^2 }$ holomorphic on a subset of $\mathbb{C}$?

Question

It is well-known that the modulus function $|z|$ is not holomorphic.

My question is; can we make $f(z) = \sqrt{z^2}$ [which we can also write as $\sqrt{ x^2 - y^2 + 2 i x y } = f(x+iy)$] holomorphic on some subset of $\mathbb{C}$?

In some way $f(z)$ "looks like" the modulus function along the real line. Does it possess the same issues as $|z|$ at $z=0$? What about elsewhere?

The reason I am asking: I would like to perform a contour integration of some function $g( \sqrt{z^2} )$, and safely use the residue theorem (which would require $\sqrt{z^2}$ to be holomorphic). Part of my contour runs along the real line (and the rest of the contour is elsewhere in the complex plane) - the contour never hits $0$, which I am guessing is the only place where $\sqrt{z^2}$ might not be holomorphic.

Answer 1

If a function is positive we can talk about "the" square root, that being the positive square root. But for a complex-valued function there really is no similar notion of "the" square root, we just talk about "a" square root.

The function $z^2$ certainly has a square root that's holomorphic in the entire plane, namely $z$. (In particular, $\sqrt{z^2}$ really has nothing to do with $|z|$; I don't know why you say it "looks like" that. If $x$ is real then $|x|$ is the positive square root of $x^2$, but if $z$ is complex then $|z|$ is not even a square root of $z^2$.)

Answer 2

For $x\geq0$ the term $\sqrt{x}$ denotes the unique nonnegative real number $y$ satisfying $y^2=x$. But for all other $c$, in particular: all $c\in{\mathbb C}$, the exact meaning of the term $\sqrt{c}$ has to be divined from the context. In any case we should insist on $\bigl(\sqrt{c}\bigr)^2=c\>$.

In the case at hand we are looking for a holomorphic function $z\mapsto f(z)$, defined on some domain $\Omega\subset{\mathbb C}$, and satisfying$$\bigl(f(z)\bigr)^2=z^2\ .$$ This necessitates $$\bigl(f(z)-z\bigr)\bigl(f(z)+z\bigr)\equiv0\ ,$$ so that from general principles about such functions we can conclude that either $f(z)\equiv z$ or $f(z)\equiv-z$ in $\Omega$.