I learned a bit about Mobius transformations. I am having the following problem: If I take the following points (in a straight line):
$$1-1000i,1-999i,\dots,1+999i,1+1000i$$
And graph them after applying $f(x)=\frac{1}{z}$, I'll get the following image:
The points are very unevenly distributed so I thought the following: I could write a set of points in term of a real function $f$ in the following manner:
$$1+f(-1000)i,1+f(-999),\dots,1+f(999)i,1+f(1000)i$$
And perhaps, there is a function where the points get more evenly distributed. After some guesswork, I noticed that functions such as $f(n)=\frac{n}{m}$ with $m>1$ make it a bit better. For example: If $m=10$, we get:
My question is: Is there a function $f$ such that the points are perfectly evenly distributed?
Perhaps this is very easy but I couldn't figure it out. I tried the following: The coordinates of our points are:
$$\left(\Re\left[\frac{1}{1+f(n)i}\right],\Im\left[\frac{1}{1+f(n)i}\right]\right)$$
Working out the computations, we get:
$$P(n)=\left(\frac{1}{f(n)^2+1},-\frac{f(n)}{f(n)^2+1}\right)$$
I tried to use something like:
$$d(P(n),P(n+1))=d(P(n+1),P(n+2))$$
After making the computations, I obtained the following simplified form (using Mathematica):
$$\frac{\frac{(f(n)-f(n+1))^2}{f(n)^2+1}-\frac{(f(n+1)-f(n+2))^2}{f(n+2)^2+1}}{f(n+1)^2+1}=0$$
Which is very bad looking and doesn't seems to give me a hint of where to start looking for such an $f$. But at least explains why the choice f$(x)=\frac{x}{100}$ is better than $f(x)=x$: Plotting the graph of the LHS of the previous equation, we find:
Which is "very close" to 0 for all $n$. If we do that with $f(x)=x$, it gets further from $0$.