I have a regular, square, cartesian grid. Let us call the bottom and the left hand boundaries of the grid B and L. Then the angle between B and L is obviously 90 degrees, and also between the vertical and horizontal internal gridlines. Now, I want to map it to a distorted grid such that the following is true:
- B and L are mapped to straight lines.
- The angle between B and L is (90+$\theta$)degrees for a known, constant $\theta$.
- The angles between all the internal grid lines remain 90degrees.
I don't really care how the other two boundaries look after the mapping, but hopefully the shape still resembles a box. I have been told that I can accomplish this using a conformal mapping. Can anyone give me pointers on how I would go about constructing such a mapping for this particular case? Or tell me how to do it? ;)
Thank you.
Let $z_0$ be the intersection of $B$ and $L$. The mapping cannot be conformal in a neighborhood of $z_0$ since the angle at the intersection is not preserved.
Consider the function $$ f(z)=(z-z_0)^{1+\theta/90^\circ}=e^{(1+\theta/90^\circ)\log(z-z_0)} $$ where the branch cut for $\log(z-z_0)$ is from $z_0$ to $z_0-\infty$ (that is, in the direction of the negative real axis). It should be verified that $f(z)$ will map the two lines to lines at the specified angle, and since it is conformal inside the box, all of the internal grid lines will remain perpendicular.
Here is how the unit square in the first quadrant gets mapped when $\theta=45^\circ$: