Is there an inverse for $\sqrt{z}$ in the complex plane? 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
 A: $\sqrt z$ 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 $z^2$.
A: 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 $(\pm(1+i))^2=2i$.
A: Have you ever heard something about "principal value"?
This will show you also how to deal with any complex functions.
A: 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 $(-\frac {\pi}{2}, \frac {pi}{2}]$
