Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2012 / 6

A conformal map preserves oriented angles between intersecting smooth curves at each point, with a nonzero local scale factor. A sufficient criterion is that $f$ be a holomorphic function on the domain and that $f^{\prime }(z)\ne 0$ everywhere there. Its first-order action then multiplies every tangent direction by the same nonzero complex number, combining a rotation and a dilation. A conformal bijection additionally requires injectivity and surjectivity onto the target.

For $z=x+iy$ in the given half-strip,Thus $0<|w|<1$ and $0<\arg w<\pi$. Conversely every such $w$ has a unique argument in $(0,\pi )$ and unique radius in $(0,1)$, giving $x=\arg w$, $y=-\log |w|$. HenceThe complex exponential is injective on $S$, and its derivative $i{e}^{iz}$ never vanishes.

On $D_{+}$, use the Joukowski map with a sign change,Sinceit maps into the upper half-plane. To prove injectivity, $J(w_1)=J(w_2)$ givesBoth moduli are below one, so $w_1{w}_2\ne 1$ and $w_1=w_2$.

For surjectivity, take any $\zeta$ in the upper half-plane and solve $w^2+\zeta w+1=0$. The two roots have product one. Neither can lie on the unit circle, since there $-(w+w^{-1})$ is real. Exactly one root therefore has modulus below one. For this root, the displayed imaginary-part identity forces $\mathrm{Im}w>0$, so it belongs to $D_{+}$. Finally $J^{\prime }(w)=w^{-2}-1$ has no zero in $D_{+}$. This proves the Joukowski map from the upper half-disc to the upper half-plane is a conformal bijection.

Composing gives the required conformal mapIts derivative $2\sin z$ is nonzero on $S$, since the zeros of the complex sine are the real integer multiples of $\pi$.