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    Construct conformal maps between disks, half-planes, and sectors

    P-BKF78-14

    Area
    Prelim
    Seen at
    BERKELEY
    Years
    1978
    Status
    draft

    Give examples of conformal maps as follows:

    1. from \(\{z:|z|<1\}\) onto \(\{z:\operatorname{Re}z<0\}\);

    2. from \(\{z:|z|<1\}\) onto itself, with \[ f(0)=0, \qquad f(1/2)=i/2; \]

    3. from \[ \left\{z:z\ne0,\ 0<\arg z<\frac{3\pi}{2}\right\} \] onto \[ \left\{z:z\ne0,\ 0<\arg z<\frac{\pi}{2}\right\}. \]

    Solution

    Write \[ \DD=\{z\in\CC:\abs z<1\}, \qquad H_-=\{w\in\CC:\Re w<0\}, \] and for \(0<\alpha<2\pi\) write \[ S_\alpha = \left\{ re^{i\theta}:r>0,\ 0<\theta<\alpha \right\}. \]

    1. A conformal map \(\DD\to H_-\) is\[ \boxed{ f_1(z)=-\frac{1+z}{1-z} }. \]

    For \(z\in\DD\), \[ \Re\left(\frac{1+z}{1-z}\right) = \frac{1-\abs z^2}{\abs{1-z}^2} >0, \] so \(f_1(z)\in H_-\). Solving \(w=f_1(z)\) for \(z\) gives \[ z=\frac{w+1}{w-1}. \] If \(\Re w<0\), then \[ 1-\abs{\frac{w+1}{w-1}}^2 = \frac{-4\Re w}{\abs{w-1}^2} >0, \] so this inverse lies in \(\DD\). Hence \(f_1\) is a bijection \(\DD\to H_-\). Finally, \[ f_1'(z)=-\frac{2}{(1-z)^2}\neq0 \] on \(\DD\), so \(f_1\) is conformal.

    2. A conformal automorphism of \(\DD\) satisfying the prescribedvalues is\[ \boxed{ f_2(z)=iz }. \]

    Multiplication by \(i\) is a rotation, hence a conformal bijection of \(\DD\) onto itself. Moreover, \[ f_2(0)=0, \qquad f_2(1/2)=i/2. \]

    3. A conformal map \(S_{3\pi/2}\to S_{\pi/2}\) is\[ \boxed{ f_3(z)=\exp\left(\frac13 L(z)\right) }, \]where\[ L(re^{i\theta})=\log r+i\theta, \qquad r>0,\quad0<\theta<\frac{3\pi}{2}. \]

    The function \(L\) is the holomorphic logarithm on \(S_{3\pi/2}\) with argument in \((0,3\pi/2)\). If \[ z=re^{i\theta}, \qquad r>0, \qquad 0<\theta<\frac{3\pi}{2}, \] then \[ f_3(z)=r^{1/3}e^{i\theta/3}, \] whose argument lies in \((0,\pi/2)\). Thus \(f_3(S_{3\pi/2})\subseteq S_{\pi/2}\).

    For \(w=\rho e^{i\phi}\in S_{\pi/2}\), we have \(0<3\phi<3\pi/2\), so \(z=w^3=\rho^3e^{3i\phi}\) lies in \(S_{3\pi/2}\). The chosen branch gives \[ L(w^3)=3\log\rho+3i\phi, \] and therefore \(f_3(w^3)=w\). Conversely, \(f_3(z)^3=z\), so \(w\mapsto w^3\) is the inverse of \(f_3\). Finally, \[ f_3'(z)=\frac{f_3(z)}{3z}\neq0 \] on \(S_{3\pi/2}\). Hence \(f_3\) is conformal and onto \(S_{\pi/2}\).

    4. Q.E.D.

    Steps 1, 2, and 3 give the requested maps in parts 1, 2, and 3, respectively.