Find a conformal map of the unit disk onto the upper half plane.
Solution
Let \[ \DD=\{z\in\CC:\abs z<1\}, \qquad \mathcal H=\{w\in\CC:\operatorname{Im}w>0\}, \] and define \[ T(z)=i\frac{1+z}{1-z}. \]
1. The map \(T\) is holomorphic on \(\DD\) and\[ T(\DD)\subseteq\mathcal H. \]
The only pole of \(T\) is at \(z=1\), which does not lie in \(\DD\), so \(T\) is holomorphic on \(\DD\).
For \(z\in\DD\), \[ \begin{aligned} \operatorname{Im}T(z) &= \operatorname{Re}\frac{1+z}{1-z}\\ &= \operatorname{Re} \frac{(1+z)(1-\bar z)}{\abs{1-z}^2}\\ &= \frac{1-\abs z^2}{\abs{1-z}^2}\\ &> 0. \end{aligned} \] Thus \(T(z)\in\mathcal H\).
2. For every \(w\in\mathcal H\), the equation \(w=T(z)\) has the uniquesolution\[ z=\frac{w-i}{w+i}, \]and this solution belongs to \(\DD\).
Solving \[ w=i\frac{1+z}{1-z} \] for \(z\) gives the displayed formula. Since \(w\in\mathcal H\), one has \(w\neq-i\), so the denominator is nonzero.
Write \[ w=u+iv, \qquad v>0. \] Then \[ \begin{aligned} \abs{w+i}^2-\abs{w-i}^2 &= \bigl(u^2+(v+1)^2\bigr) - \bigl(u^2+(v-1)^2\bigr)\\ &= 4v\\ &> 0. \end{aligned} \] Hence \[ \abs{\frac{w-i}{w+i}}<1, \] so \(z\in\DD\). The algebraic solution is unique.
3. The derivative of \(T\) is nowhere zero on \(\DD\).
Direct differentiation gives \[ T'(z)=\frac{2i}{(1-z)^2}. \] Since \(1\notin\DD\), this derivative is nonzero at every point of \(\DD\).
4. A conformal bijection of the unit disk onto the upper half-plane is\[ \boxed{ T(z)=i\frac{1+z}{1-z} }. \]
Step 1 shows that \(T\) maps \(\DD\) into \(\mathcal H\). Step 2 gives an inverse defined on all of \(\mathcal H\), so \(T\) is bijective onto \(\mathcal H\). Step 3 shows that its derivative never vanishes. Hence \(T\) is conformal.
5. Q.E.D.
Step 4 gives the requested map.