A question about analytic functions on the unit disk with $\Re[h(z)]=0$ and a double pole at $1$ Let $\bar{D}(0,1)$ denote the closed unit disk around $0$ and $D$ the unit circle. 
I am interested in obtaining a complex function, say h, with the following properties:


*

*$h(z)$ is analytic for $z\in D$ and continuous on $\bar{D}(0,1)\setminus\{1\}$; 

*$\Re[h(z)]=0$, $z\in D\setminus\{1\}$ (has a zero real part on the unit circle);

*$\lim_{z\to 1}(z-1)^2 h(z)=c\neq 0$ (has a double root at $1$).


I think that this problem fits into the category "Dirichlet problem with a pole", but I am not 100% sure on how to obtain the function $h(z)$.

How about $h(z)= i\bigl(\frac{z+1}{z-1}\bigr)^2$?

(Reasoning: Consider what happens to the imaginary part of $h$ as we go around the unit circle. If it goes to $+\infty$ as we approach $1$ both from north and south, it must have a minimum somewhere on the circle. Wlog the function value at that minimum could be 0, in which case it would be a double root. For symmetry, let's suppose it is at $z=-1$. This leads to contemplating $z\to\frac{(z+1)^2}{(z-1)^2}$; what do we have to do to that to make the function value purely imaginary on the unit circle? Nothing except multiplying it by $i$ -- it is the square of the Möbius transformation that takes the unit circle to the imaginary axis.)
