# 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

• "...$\sqrt{4}$ is either $4$ or $−4$..." please can you better explain this? – the_candyman Aug 31 '16 at 19:04
• @the_candyman it's not the usual definition for a function, but a multivalued function. i know that the funciton $\sqrt{x}$ has only one result, always, that's why it's a function – user365279 Aug 31 '16 at 19:05
• They call it multivalued. – Yves Daoust Aug 31 '16 at 19:05
• "...$\sqrt{4}$ is either $4$ or $−4$..." or "...$\sqrt{4}$ is either $2$ or $−2$..."??? – the_candyman Aug 31 '16 at 19:06
• @the_candyman sorry, now I see. Updated. – user365279 Aug 31 '16 at 19:08

$\sqrt z$ 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 $z^2$.

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 $(\pm(1+i))^2=2i$.

Have you ever heard something about "principal value"?

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

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 $(-\frac {\pi}{2}, \frac {pi}{2}]$