From the unit disk to the right half plane and $T(0)=3$ Find a Mobius transformation $T$ from the unit disk to the right half plane with condition $T(0)=3$.
First, the transformation from the unit circle to the upper half plane is $T_1(z)=(1-i)\frac{z-i}{z-1}$.
So from the unit circle to the right half plane, $T_2(z)=-i(1-i)\frac{z-i}{z-1}$
How can I introduce the condition $T(0)=3$ ?
$T(0)=1-i\neq3$
 A: For a purely algebraic derivation, consider the general form of the Möbius transformation $\,T(z)=\dfrac{az+b}{cz+d}\,$. Both $\,a\,$ and $\,c\,$ cannot be $\,0\,$, otherwise it would be a constant transformation. The right half-plane is invariant to the inversion $\,T(z) \to \dfrac{1}{T(z)}\,$ so it can be assumed WLOG that $\,a \ne 0\,$, then after normalizing it can be assumed WLOG that $\,a=1\,$. The condition $\,T(0)=3\,$ translates to $\,b = 3d\,$, so in the end $\,T(z)=\dfrac{z+3d}{cz+d}\,$ for some $\,c, d \in \Bbb C\,$ with $\,d \ne 0\,$.
The unit circle must transform into the imaginary axis, so for $\,|z|=1\,$:
$$
\begin{align}
0 = 2 \operatorname{Re}\left(T(z)\right) &= \dfrac{z+3d}{cz+d} + \dfrac{\bar z+3 \bar d}{\bar c \bar z+ \bar d} \\
 &= \frac{(z+3d)(\bar c \bar z + \bar d)+(\bar z + 3 \bar d)(cz + d)}{|cz+d|^2} \\
 &= \frac{(c+\bar c)|z|^2+ 6 |d|^2+(\bar d + 3c\bar d) z+(3 \bar cd +d)\bar z}{|cz+d|^2} \\
 &= \frac{2 \operatorname{Re}(c)+ 6 |d|^2+(3c+1)\bar d z+(3 \bar c +1)d\bar z}{|cz+d|^2}
\end{align}
$$
It follows that $\,3c+1=0\,$ for the numerator to not depend on $\,z\,$, and $\,2 \operatorname{Re}(c)+ 6 |d|^2=0\,$ for the numerator to be $\,0\,$. The first equation gives $\,c = -\dfrac{1}{3}\,$, and the second one $\,|d|=\dfrac{1}{3}\,$. Therefore, defining $\,\omega = 3d\,$ the general solution is:
$$
T(z) \;=\; \frac{z + 3d}{-\dfrac{1}{3}z+d} \;=\; 3\,\dfrac{z + \omega}{-z + \omega} \quad\quad\style{font-family:inherit}{\text{where}}\;\; |\omega|=1
$$

[ EDIT ]   For quick verification of the form above:

$$\small
\frac{1}{3}T(z) = \dfrac{z + \omega}{-z + \omega} \color{red}{\cdot \frac{\bar \omega}{\bar \omega}} = \frac{1+\bar \omega z}{1 - \bar \omega z} \color{red}{\cdot \frac{1 - \omega \bar z}{1 - \omega \bar z}} = \frac{1 - |\omega|^2|z|^2+\bar \omega z - \omega \bar z }{|1 - \bar \omega z|^2} = \frac{1 - |z|^2+ 2i \operatorname{Im}(\bar \omega z)}{|1 - \bar \omega z|^2}
$$
Therefore $\,\small\operatorname{Re}(T(z)) = 3\,\dfrac{1 - |z|^2}{|1 - \bar \omega z|^2} \ge 0\,$ iff $\,\small|z| \le 1\,$, and of course $\,\small T(0) = 3\,$.
A: Let $\mathbb D$ be the open unit disc, $U$ the open right half plane. Suppose $f:\mathbb D\to U$ is bilholomorphic, with $f(0)= 1-i$ (just as your map $T_2$ does.) How do we then find a biholomorphic map $g:\mathbb D\to U$ with $g(0)= 3?$
There are two families of biholomorphic maps from $U$ to $U$ that are simple and will be helpful here:
i) Vertical translations. These are the maps $v_c(z)=z+ic,$ where $c$ is a real constant.
ii) Positive dilations: These are the maps $d_r(z)= rz,$ where $r$ is a positive real constant.
Claim: $g(z) = (d_3\circ v_{1}\circ f)(z)$ has the desired properties.
Proof: It should be clear that $g:\mathbb D\to U$ is biholomorphic. And we see
$$g(0) = d_3(v_1(f(0)))) = d_3(v_1(1-i)) = d_3((1-i)+i))= d_3(1)=3.$$
So $g$ does the job. For completeness, note $g(z) = 3(f(z) +i).$
A: You took only one possible transformation that maps the unit disk to the right half-plane. Generally, since the conjugation wrt the unit circle becomes the conjugation wrt the imaginary axis, $T(0) = 3$ implies $T(\infty) = -3$. Therefore, $T(z) = 3 - 6 z/(z + a)$. Then $T(a) = 0$, therefore $|a| = 1$ gives the general form of $T$.
A: It is well worth knowing that the conformal map $z \rightarrow \frac{1+z}{1-z}$ permutes the regions shown below $(1234)$ and act similarly on the lower half plane.

In particular the unit circle is mapped to the right half plane, indeed one can check that the unit circle ($z=e^{i \theta} =\cos( \theta)+i \sin( \theta)$ is mapped to the imaginary axis
\begin{eqnarray*}
\frac{1+z}{1-z}= \frac{1+\cos( \theta)+i \sin( \theta)}{1-\cos( \theta)-i \sin( \theta)} \times \frac{1-\cos( \theta)+i \sin( \theta)}{1-\cos( \theta)+i \sin( \theta)} = \cdots
\end{eqnarray*}
Now in order to fulfill the requirement that $T(0)=3$ simply multiply this function by $3$ (and note that this will still map the unit circle to the imaginary axis) & so ... 
\begin{eqnarray*}
T(z)=3 \frac{1+z}{1-z}
\end{eqnarray*}
will do the trick.
