Is there an inverse for $\sqrt{z}$ in the complex plane?

Question

Usually some people think about the square root as a 'multivalent' function (don't know how to call it in ensligh). That is, $\sqrt{4}$ is either $2$ or $-2$. This is compatible with the notion of complex roots: given a root $\sqrt{z}$ where $z$ is complex, there are always $2$ possible solutions

My book is asking me if it's possible to consider a set where $f$ is invertible. I didn't understand what this means. For example, given $\sqrt{x}$ as a multivalued function with $x$ real(is this the name?), what should be an inverse set for it? The $\{x;'x>0\}$? How it should be in the complex plane?

I'm confused because my book didn't define the inverse of multivalued functions

Answer 1

$\sqrt z$ can be considered multivalued precisely when it represents all solutions of

$$(\sqrt z)^2=z.$$

Then clearly, the function $z\to z^2$ maps all square roots of $z$ to $z$ itself and the inverses of all branches of $\sqrt z$ are $z^2$.

Answer 2

The inverse of the multivalued square root is simply $y=x^2$.

See that $\sqrt4=\pm2$, and $(\pm2)^2=4$.

This holds further into complex numbers. See that $\sqrt{2i}=\pm(1+i)$ and that $(\pm(1+i))^2=2i$.

Answer 3

Have you ever heard something about "principal value"?

This will show you also how to deal with any complex functions.

Answer 4

The way out with real numbers is to define the square root function to be the positive square root. i.e. $\sqrt {x^2} = |x|$ And then for any $x\ge0$ there is only one square root.

So what to do with complex numbers? Pick one of the two roots, to be the primary root. If $(a+bi)^2 = z$ then $(-a - bi)^2 = z,$ which one do you want to be primary. Traditionally, it is the one with a positive real part, or if it has a 0 real part a positive imaginary part.

If you are in polar / exponential form, the one whose argument is in $(-\frac {\pi}{2}, \frac {pi}{2}]$