I need to find a conformal bijection of the form $f(z)=λz+μ$ from the upper half complex plane $H$ $(im(z)>0)$ to itself, mapping $a$ to $b$, where $a$ and $b$ are both in $H$.
We know $f(a)=λa+μ=b$, so $μ=b-λa$, so $f(z)=λz+b-λa$. I was trying to find conditions on $λ$ so that $f(z)>0$ for all $z$, but have not succeeded. Any hints/suggestions on how to proceed?
Continuing upon my comment, we want $\lambda$ to be $Im(b)/Im(a)$. This is well-defined, since $a,b$ are both fixed and with positive imaginary parts. Then $\lambda a=\lambda Re(a)+ \lambda Im(a) = \lambda Re(a)+Im(b)$. Let $\mu=Re(b)-\lambda Re(a)$, and we're done.