# Showing a function is harmonic in the unit disc but vanishes on the boundary.

this is my first post, so please forgive me for any slip-ups. This is an exercise from Stein and Shakarchi "Complex Anaylsis" chapter 8 on conformal mappings. The question reads as follows:

Prove that a function $$u$$ defined by $$u(x,y) = \Re\left(\frac{i+z}{i-z}\right) \quad \text{ and } \quad u(0,1) = 0$$ is harmonic in the unit disc and vanishes on its boundary. Note that $$u$$ is not bounded in $$\mathbb{D}$$, the unit disc.

To show that a function is harmonic I've been trying to show $$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}= 0$$ which would mean I have to calculate those partial derivatives after plugging in $$z=x+iy$$. Although this is a gross computation I have managed to show it is true, however I would be curious to know of any cleverer ways of doing this. I've never been good at finding slick calculus tricks.

I've found that usually the trick to finding properties of a function on the unit circle is to analyze the behaviour of $$u(e^{i\theta}$$). I also considered that harmonic functions are holomorphic, but this hasnt really gotten me anywhere.

My Question: I'd like to know what the trick is to showing this function vanishes on $$\partial\mathbb{D}$$. Does it have anything to do with conformal mappings? Since $$u$$ isn't bounded on $$\mathbb{D}$$ does that make it a conformal map to $$\mathbb{C}$$ since it's holomorphic? Any hints are apppreciated.

• Direct computation: multiply $(-i-\bar z)$ to both denominator and numerator, simplify using the fact $z\bar z=1$ on boundary. – Yuval Nov 10 '19 at 1:25

To show that the function vanishes on the boundary, multiply both numerator and denominator by the complex conjugate of the numerator and use that $$z\bar z = 1$$ on the boundary (and that the real part of $$i$$ is $$0$$).
For a slick geometric reason why the function vanishes on the boundary, you might want to look into the following. The function $$f(z) = \frac{i+z}{i-z}$$ is a so-called Möbius transformation. It turns out that Möbius transformations map lines and circles to lines and circles (possibly lines to circles and vice versa), and this particular Möbius transformation maps the boundary of the disk to the imaginary axis. You could check this by plugging in $$3$$ points in the unit disk, e.g., $$1,-1$$, and $$i$$.