Consider the Mobius transformation $$f(z) = \frac{1+z}{1-z}$$
Use this map to find a function $f(x,y)$ which is harmonic in $x^2+y^2<1$ and on the boundary $x^2+y^2=1$ takes values $+1$ when $y>0$ and $−1$ when $y<0$.
My work: this is part (b) of a question that I am working on. For part (a), I described the conformal mapping, showing that it maps the unit disk to the right half-plane.
I know that $f$ harmonic means that for every $(x,y)$ in the interior of the disk, we need that $f_{xx}(x,y) + f_{yy}(x,y) = 0$. It also needs to take values +1 for $(x,y)$ on the upper semi-circle, and -1 on the lower semi-circle.
I tried writing $f(z)$ as $$f(x,y) =\frac{(1+x)+iy}{(1-x)-iy}$$
How can I proceed?
I tried setting $f(x,y) = 1$ and got the equation $2x = -2iy$. So I made a function $f(x,y)= 2x + i2y$, which happens to be harmonic. But I don't know how to make it so that it takes the value $+1$ on the boundary $x^2+y^2=1$, $y>0$ (and similarly, $-1$ on the boundary $x^2+y^2=1$, $y<0$).
Thanks,
How does $\log(f)$ behave, particularly on the unit circle?