Find a bijective holomorphic map \(f\) from the quadrant
\[ Q = \{ x + iy \in \mathbb{C} : x > 0, \ y > 0 \} \]
onto the unit disk \(D = \{ z \in \mathbb{C} : |z| < 1 \}\) with \(f(1+i) = 0\).
Solution
One such map is \[ \boxed{f(z)=\frac{z^2-2i}{z^2+2i}.} \]
1. Squaring maps \(Q\) biholomorphically onto the upper half-plane \(H\).
Every \(z\in Q\) has a unique polar expression \(re^{i\theta}\) with \(r>0\) and \(0<\theta<\pi/2\). Its square has argument \(2\theta\in(0,\pi)\), so lies in \(H=\{v:\operatorname{Im}v>0\}\). Conversely, for \(v=\rho e^{i\alpha}\in H\) with \(0<\alpha<\pi\), the point \(z=\sqrt\rho e^{i\alpha/2}\) is in \(Q\) and squares to \(v\). It is the only such point: the other square root is \(-z\), which is outside \(Q\). Thus squaring is a holomorphic bijection. Its derivative \(2z\) is nonzero on \(Q\), so the inverse is holomorphic by the local inverse function theorem [@SS03].
2. A fractional transformation supplies the target disk and prescribed zero.
Set \[ T(v)=\frac{v-2i}{v+2i},\qquad S(w)=2i\frac{1+w}{1-w}. \] For \(v\in H\), the denominator is nonzero and \[ 1-|T(v)|^2=\frac{8\operatorname{Im}v}{|v+2i|^2}>0. \] For \(w\in D\), \(1-w\ne0\) and \[ \operatorname{Im}S(w)=2\frac{1-|w|^2}{|1-w|^2}>0. \] Substitution gives \(T\circ S=\operatorname{id}_D\) and \(S\circ T=\operatorname{id}_H\). Hence \(T:H\to D\) is a holomorphic bijection. Composing it with step 1 gives the stated \(f\), whose denominator cannot vanish on \(Q\). Finally \((1+i)^2=2i\), so \(f(1+i)=T(2i)=0\).