Map from the upper half plane to a circle $\vert w-w_0\vert<R$ such that $T'(u_0+v_0i)>0$ and $T(i)=u_0+iv_0$

Question

Transform the upper half plane $\mathop{\mathrm{Im}} z>0$ into the circle $\vert w-w_0\vert<R$ so that the point $i$ correspond to the center of the circle and the derivative in this point is positive.

We know that the transformation $T$ from the upper half plane to the unit disk is given by $T(z)=e^{i\theta_0}\dfrac{z-z_0}{z-\overline z_0}$, where $T(z_0)=0$ and $\theta_0$ is the argument of $|k|=1$. Thus from the upper half plane to $\vert w-w_0\vert<R$ would be $T(z)=Re^{i\theta_0}\dfrac{z-z_0}{z-\overline z_0},$ where $T(z_0)=0$ and $\theta_0$ is the argument of $|k|=R$.

Let $w_0=u_0+iv_0$. As is required, $T(i)=u_0+iv_0$. Evaluating $T$ at $i$ and multiplying the reciprocal of the result, say $\lambda$, with $Re^{i\theta_0}\frac{z-z_0}{z-\overline z_0}(u_0+v_0i)$ we get the condition $T(i)=u_0+iv_0$. Hence $$T(z)=\frac{z-z_0}{z-\overline z_0}\frac{i-\overline z_0}{i- z_0}(u_0+v_0i).$$

For the last condition, it is confusing what is required, could be $T'(u_0+v_0)>0$ or $T'(i)>0$. I chose $T'(u_0+v_0)>0$. After calculations, I found that $z_0>\overline z_0$ is required to have $T'(u_0+iv_0)>0$. But I do not know how to plug this into the transformation $$T(z)=\frac{z-z_0}{z-\overline z_0}\frac{i-\overline z_0}{i- z_0}(u_0+v_0i).$$

Is this solution correct so far? How to plug the derivative condition in $T$?

I need help please help. Thank you.

Answer 1

You have a lot of parameters floating around and things can get confusing. To simplify, let's take $z_0=i$ and $\theta_0=0$ and look at

$$f(z) = \frac{z-i}{z+i}.$$

Then $f$ is a holomorphic bijection from the open upper half plane $U$ to the open unit disc $\mathbb D,$ with $f(i)=0.$ We calculate that $f'(i)=-i/2.$ It follows that if $g(z)=if(z),$ then $g:U\to \mathbb D$ is a holomorphic bijection with $g(i)=0$ and $g'(i)=1/2.$ This is looking good.

Now the standard map from $\mathbb D $ to $\{|w-w_0|<R\}$ is $z\to w_0 + Rz.$ It follows that $h(z) = w_0+Rg(z)$ is a holomorphic bijection from $U$ to $\{|w-w_0|<R\},$ with $h(i)=w_0$ and $h'(i) = R/2>0.$ Thus we can take $T=h$ to solve the problem. Unravelling the chain of maps, we see

$$T(z)=w_0 +Ri\frac{z-i}{z+i}$$

has the desired properties.

Answer 2

$\def\Im{\mathop{\mathrm{Im}}}\def\i{\mathrm{i}}\def\e{\mathrm{e}}$The problem to be solved is:

Find a mapping $T$ that maps $\{z \in \mathbb{C} \mid \Im z > 0\}$ to $\{w \in \mathbb{C} \mid |w - w_0| < R\}$ with $T(\i) = w_0$, $T'(\i) > 0$.

First, define $g(z) = Rz + w_0$. Note that if $T$ satisfies the given conditions, then $f = T \circ g^{-1}$ maps $\{z \in \mathbb{C} \mid \Im z > 0\}$ to $\{z \in \mathbb{C} \mid |z| < 1\}$ with $f(\i) = 0$, $f'(\i) > 0$, and vice versa. Now, since $f$ is a Möbius tranformation that maps $\i$ to $0$, then for some $θ \in (-π, π]$ and $z_0 \in \mathbb{C} \setminus \{\i\}$,$$ f(z) = \e^{\i θ} \cdot \frac{z - \i}{z - z_0}. \quad \forall z \in \mathbb{C} $$ Note that $f$ maps $\mathbb{R}$ to $\{z \in \mathbb{C} \mid |z| = 1\}$, then $|z - \i| = |z - z_0|$ for all $z \in \mathbb{R}$. (Geometrically this implies that $z_0 = -\i$.) Taking $z = 0$ yields $|z_0| = 1$, then for any $z \in \mathbb{R}$,$$ z^2 + 1 = |z - \i|^2 = |z - z_0|^2 = z^2 + (z_0 + \overline{z_0}) z + 1 \Longrightarrow z_0 + \overline{z_0} = 0. $$ Since $|z_0| = 1$ and $z_0 ≠ \i$, then $z_0 = -\i$. Finally, since $f'(z) = \dfrac{2\i \e^{\i θ}}{(z + \i)^2}$, then$$ f'(\i) = \frac{\e^{\i θ}}{2\i} > 0 \Longrightarrow \e^{\i θ} = \i. $$ Therefore,$$ T(z) = g(f(z)) = \i R \cdot \frac{z - \i}{z + \i} + w_0. $$