In a complex variable class, a professor asked us the following.
Describe the function $$ f(z)=\frac{1}{(z-i)^n} $$ geometrically.
I tried writing $z=x+iy$ to see what the function does to lines and circles. Initially I got
\begin{equation} f(z)=\frac{(\bar{z}+i)^n}{\lvert z-i\rvert^{2n}}=\frac{1}{(x^2+(y-1)^2)^n}\sum_{k=1}^n\binom{n}{k}x^{n-k}(1-y)^ki^k \end{equation}
but it got me nowhere. I then tried restricting the domain to the real line and obtained
\begin{align} f(x+0i)=\frac{1}{(x^2+1)^n}\sum_{k=1}^n\binom{n}{k}x^{n-k}i^k&=\frac{1}{(x^2+1)^n}\sum_{\quad \,\,k=1\\k\equiv 0,2\operatorname{mod}4}^n\binom{n}{k}x^{n-k}(-1)^k\\ &\quad+\frac{i}{(x^2+1)^n}\sum_{\quad \,\,k=1\\k\equiv 1,3\operatorname{mod}4}^n\binom{n}{k}x^{n-k}(-1)^k \end{align}
And now I think the problem might not have a solution. I'm starting to think there's no clear way to describe this geometrically. Any ideas are gladly appreciated.
As you were told, it is geometric interpretation that you have been asked. It is best to take an example that you should follow by drawing in the $OJI$ plane($O=0, J=1$ and $I=i$): Let $n=6$ and $z=A=\frac12+\frac12i$. $z-i=\frac12+\frac12i-i=\frac12-\frac12i=A-I=\vec{IA}=\vec{OA'}=A'-O=A'$, with $A'=\frac12-\frac12i=\frac{\sqrt2}{2}e^{-i\frac{\pi}{4}}$
Then $\frac{1}{z-i}=\frac{1}{\frac{\sqrt2}{2}e^{-i\frac{\pi}{4}}}=\frac{1}{\frac{\sqrt2}{2}}e^{i\frac{\pi}{4}}=\sqrt2e^{i\frac{\pi}{4}}$(observe the effect on the argument and the modulus.) Then I built the example for this, everything works fine, but it remains anecdotal for the explanation: $\frac{1}{z-i}=1+i$, which is a point that we denote $A'':=1+i=\sqrt2e^{i\frac{\pi}{4}}$
Then $\frac{1}{(z-i)^n}=(\frac{1}{(z-i)})^n=A''^6=(\sqrt2e^{i\frac{\pi}{4}})^6=(\sqrt2)^6(e^{i\frac{\pi}{4}})^6=8e^{3i\frac{\pi}{2}}=-8i$(observe the effect on the argument and the modulus.)
This type of explanation, even a mathematician like Adrien Douady did not shy away from giving them in his famous film "the dynamics of the rabbit" (sorry, it's in French)