= Solution
{c}
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 $f'(z)\ne0$ 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,
$$
w=e^{iz}=e^{-y}(\cos x+i\sin x).
$$
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 $x=\arg w$, $y=-\log|w|$. Hence
$$
\boxed{e^{iS}=D_+=\{w:|w|<1,\ \operatorname{Im}w>0\}.}
$$
The <complex exponential> is <injective> on $S$, and its <derivative> $ie^{iz}$ never vanishes.
On $D_+$, use the <Joukowski map> with a sign change,
$$
J(w)=-\left(w+\frac1w\right).
$$
Since
$$
\operatorname{Im}J(w)=\operatorname{Im}w\left(\frac1{|w|^2}-1\right)>0,
$$
it maps into the <upper half-plane>. To prove <injectivity>, $J(w_1)=J(w_2)$ gives
$$
(w_1-w_2)\left(1-\frac1{w_1w_2}\right)=0.
$$
Both moduli are below one, so $w_1w_2\ne1$ 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 $\operatorname{Im}w>0$, so it belongs to $D_+$. Finally $J'(w)=w^{-2}-1$ 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 map>
$$
\boxed{F(z)=J(e^{iz})=-e^{iz}-e^{-iz}=-2\cos z:\ S\longrightarrow H.}
$$
Its <derivative> $2\sin z$ is nonzero on $S$, since the <zeros> of the complex <sine> are the real integer multiples of $\pi$.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2012-half-strip-map.png]
{title=The half-strip maps to the upper half-disc by the complex exponential and then to the upper half-plane by a Joukowski map. Matching colours follow curves through both maps.}
Back to article page
