Inverse of $z^{2}$.

Question

How one can find a inverse function of a complex function?

If we have complex function say $f:\mathbb{C}\rightarrow\mathbb{C}$, $z\mapsto z^{2}$ then how one can find the inverse of it?

The components of function $f$ are $u(x,y)=x^2 + y^2$ and $v(x,y)=2xy$.

The function $f$ is not one-to-one since for example image of $-1$ is the set $\{i,-i \}$.

How to proceed? Solve for $u$ and $v$?

What if we limit ourselves into $-\pi/2 < \arg\varphi < \pi/2$?

Shouldn't we then be able to have a inverse function?

You simply can't. If a function is not injective, you have no chance of finding an inverse (unless you restrict the domain). — b00n heT, Nov 28 '16 at 11:27
Maybe relevant: http://math.stackexchange.com/questions/44406/how-do-i-get-the-square-root-of-a-complex-number — Hans Lundmark, Nov 28 '16 at 11:42

score 1 · Answer 1 · answered Nov 28 '16 at 20:22

In general : let $$ D_{0,n}=\left \{z\in \mathbb{C} : \frac{-\pi}{n}<Arg(z)\leq \frac{\pi}{n} \right \}\cup\left \{0 \right \} $$ then : $$ f :D_{0,n}\rightarrow \mathbb{C} \ , f=z^n $$ is 1-1 ,onto and has inverse function : $$ g:\mathbb{C} \rightarrow D_{0,n} $$ where : $$g(w_{n})=|w| ^{\frac{1}{n}}(cos(\frac{\phi}{n})+i\ sin(\frac{\phi}{n})), \ w=\rho(cos\phi + i \ sin\phi ) , g(0)=0.$$

YukiJ · Answer 2 · 2017-01-05T11:35:55.757

0

If the image space of -1 of your function $f$ indeed is {-i,i}, then your function does not have an inverse. What you call "one-to-one" is called injective which means that the function is both injective and surjective. A function can only have an inverse, if it is bijective (i.e. $f$ being bijective is a necessary condition so that the inverse exists)

edited Jan 05 '17 at 11:35

answered Nov 28 '16 at 11:29

YukiJ

2,529

Inverse of $z^{2}$.

2 Answers2