Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 / 3

At $z=x+iy$, compute the complex derivative first along real increments and then along imaginary increments. Holomorphy givesEquating real and imaginary parts proves the Cauchy-Riemann equations:The partial derivatives exist because these directional difference quotients are limits of the holomorphic derivative. This argument proves the necessary equations; it does not claim that their pointwise existence alone is sufficient for holomorphy.

Use the principal complex logarithm:The excluded nonpositive real ray allows a continuous choice of argument, and $L(1)=0$. For $z=x+iy$, its real part is $u=\frac{1}{2}\log (x^2+y^2)$. Differentiating the chosen local argument givesThese continuous partial derivatives satisfy the Cauchy-Riemann equations, proving holomorphy on the slit domain. Its derivative is

A holomorphic primitive on the slit domain is $P(z)=zL(z)-z$, since the product rule and $L^{\prime }=1/z$ give $P^{\prime }=L$. The line segment stays in that domain. The fundamental theorem for line integrals therefore givesHenceThe value $L(i)=\pi i/2$ is fixed by the chosen branch.

In the upper half-plane, the argument lies strictly between zero and $\pi$, while $\log |z|$ can be any real number. Conversely, if $w=s+it$ with $0<t<\pi$, then $z=e^w$ is in the upper half-plane and $L(z)=w$.Thus the principal complex logarithm is a conformal bijection from the upper half-plane onto that horizontal strip, with inverse the exponential function restricted to the strip.

Write $\alpha =a+ib$ with $b>0$. ThenIt preserves the upper half-plane, has inverse $(z-a)/b$, and sends $i$ to $\alpha$. Its nonzero derivative $b$ verifies that this bijection is conformal.

The map $h(z)=e^{\pi z/2}$ is a conformal bijection from the width-two strip to the upper half-plane, and $h(i)=i$. Its inverse there is $h^{-1}(w)=2L(w)/\pi$.

Write $e^{\pi \beta /2}=A+iB$, where $B>0$. Conjugate the preceding affine half-plane map by $h$:Every argument of $L$ lies in the upper half-plane. The formula is therefore a conformal bijection of the strip onto itself, andThe final equality uses $0<\mathrm{Im}(\pi \beta /2)<\pi$, so no multiple of $2\pi i$ is lost. This is a strip automorphism with a prescribed image of its midpoint.