I'm not familiar with complex analysis. While playing with Mathematica (a mathematics software), I found that it keeps spitting out unexpected results, and the reason was that it considers differently from me of the inverse of certain polynomial functions such as $f(x)=x^3$.
Mathematica evaluates the following line
N[InverseFunction[Function[x,x^3]][-1]]
as 0.5 + 0.866025 I. The above means to evaluate numerically the value of $f^{-1}(-1)$ where $f(x)=x^3$.
I was thinking naively that an inverse function is plotted as symmetric to $y=x$ from its original form, so that the result above ought to be $-1$. What did I miss?
It depends which root of the equation the software considers by default. The point is that for real $y$, the equation $x^3-y=0$ always has $3$ complex roots. If $y>0$, there is no controversy about taking the positive real root, which we normally call $\sqrt[3]{y}$. On the other hand, if $y<0$, then there are no positive real roots; instead the roots are $$ -\sqrt[3]{-y}, \quad -\omega \sqrt[3]{-y}, \quad -\omega^2 \sqrt[3]{-y}, $$ where $\omega$ is a complex number which satisfies $\omega^3=1$.
Mathematica's convention is to take the root with the smallest complex argument; this happens to be $-\omega^2 \sqrt[3]{-y}$ when $y$ is negative, for a particular choice of $\omega$. Other calculators will use $-\sqrt[3]{-y}$, but they normally only have to deal with real numbers: the point of Mathematica's convention is that it is consistent and continuous in most of the complex plane for any power of $x$, with the result that the odd integers do not behave quite as you expect.
On the other hand, if you plot the same thing multiplied by $i\sqrt{3}-1$, you will find the negative part reappears where you expect, so it is where it ought to be with Mathematica's definition.