Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2011 / Analysis / 6 / Solution

A conformal map preserves angles between intersecting regular curves, including their orientation. If $f$ is holomorphic with $f^{\prime }(a)\ne 0$, the image tangent vectors are $f^{\prime }(a)$ times the original tangent vectors. Multiplication by this nonzero complex number gives a common positive scaling and rotation, so their oriented angle is unchanged. This proves conformality throughout the stated domain.

At a zero derivative this can fail: $f(z)=z^2$ at zero sends rays of arguments $0$ and $\pi /4$ to rays of arguments $0$ and $\pi /2$. The angle is doubled, so this holomorphic function is not conformal there.

For the disk map $C(z)=i(1+z)/(1-z)$, computeIts inverse is $z=(w-i)/(w+i)$; for $\mathrm{Im}w>0$, the inequality $|w-i|<|w+i|$ puts this inverse in the unit disk. Both rational maps are holomorphic on their respective domains. ThusFor the final region, writing $z=x+iy$ gives $|e^z|=e^x\in (0,1)$ and $\arg (e^z)=y\in (0,\pi )$. Hence its exponential image is exactly the upper half of the unit disk,The exponential is injective on this strip, since two equal exponential values differ by $2\pi i$ times an integer while the imaginary parts here differ by less than $\pi$. Its derivative never vanishes and its inverse on $V$ is the logarithm with argument in $(0,\pi )$.

Now $L(w)=(1+w)/(1-w)$ maps $V$ conformally onto the first quadrant. Indeedand the inverse $(s-1)/(s+1)$ has modulus below one and positive imaginary part for every first-quadrant $s$. Squaring maps that quadrant bijectively onto $\mathbb{H}$, with nonzero derivative. Therefore a required composition isEvery component is a conformal bijection between the stated domains, so the composition is one as well.