Compute the image of the triangle with corners \(z = 0, z = 1/2, z = 1/2 + 1/2 i\) under inversion across the unit circle. Where does the interior go? Are the angles between the sides preserved in the image?
2.
Show that a linear fractional transform can have at most two points so that \(T(z) = z\text{.}\) Write an example of a LFT with \(T(\infty) = \infty\text{.}\)
3.
Show that the map
\begin{equation*}
z \mapsto z + \frac{1}{z}
\end{equation*}
takes the unit circle to the segment \([-2, 2] \subset \R\text{.}\) Where is this map conformal?
4.
Compute the real and imaginary parts of the composition of a uniform flow on the upper half plane with the map \(z \mapsto z + 1/z\text{.}\) Use your favorite visualizer to confirm that the level curves for these functions are orthogonal. You might also consider adding an image of the unit circle to your plot. (I used desmos for this).
5.
Prove that the map given by \(u(x, y) = \frac{1}{\pi} \theta\) is harmonic on the upper half plane, where \(\theta = \arg (x + iy)\text{.}\) Show that this function has boundary values of \(u(x, 0) = 1\) for \(x \lt 0\) and \(u(x,0) = 0\) for \(x > 0\text{.}\)
6.
Use the solution above to find a harmonic solution \(u\) to the Dirichlet problem on the disk where \(u = 0\) on the upper half of the unit circle, and \(u = 1\) on the lower half of the unit circle. Hint: consider the conformal map
\begin{equation*}
T = i \frac{1+ z}{1-z}.
\end{equation*}
7.
Find the image of the upper half of the unit disk under the map
\begin{equation*}
z \mapsto \operatorname{Log} z.
\end{equation*}
(That is, it isn't enough to know what the answer is, you need to show it!) Is this map conformal? What is the inverse map?
8.
Determine the image of the map \(z \mapsto \sin z\) on the half-strip with boundary \(y = 0, x = 0, x = \pi/2\) with \(0 \leq x \leq \pi/2\) and \(y > 0\) (your image should be a quarter-plane: make sure to track where the boundary pieces map). Use this conformal map to find a harmonic function giving the temperature on the first quadrant, with boundary conditions \(T = 1\) on the positive imaginary axis, \(T\) is insulated on the interval \(0 \leq x \leq 1\text{,}\) and \(T = 0\) on the rest of the real axis. (Bonus, use a visualizer to draw the flow lines by finding the level curves of the harmonic conjugate of \(T\text{.}\))
