Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 1 / 12G (Complex Analysis P OR Complex Methods)

Termwise differentiation inside the disc givesConsequentlySince $f(1)=0$, the constant is one, so $e^{f(z)}=z$. Thus $f$ is an analytic branch of the logarithm on $D(1,1)$ with the required value.

Given $a\in D$, write $a=|a|e^{i{\theta }_a}$ with $0<{\theta }_a<2\pi$. On $|z/a-1|<1$, defineThen $e^{{\ell }_a(z)}=z$ and $\mathrm{Im}{\ell }_a(a)={\theta }_a$. After shrinking the neighbourhood of $a$, continuity keeps its imaginary part in $(0,2\pi )$.

Near each $a\in D$, the local branch from part (a), chosen to have value $\log |a|+i\theta (a)$ at $a$, agrees withIndeed, two logarithms of the same nonzero number differ by $2\pi i$ times an integer, and the integer is locally constant; it is zero at $a$. Hence $L$ is locally analytic and therefore analytic on $D$. This is the analytic logarithm on the positive-axis slit plane.

For $-1<\alpha <1$, an analytic branch is

The residue theorem states that if a meromorphic function has finitely many poles inside a positively oriented simple closed contour and none on it, then its contour integral is $2\pi i$ times the sum of the enclosed residues.

Apply it toon a keyhole contour around the positive real axis. The outer and inner circles vanish as their radii tend to infinity and zero because $\alpha <1$ and $\alpha >-1$, respectively. On the upper bank the numerator tends to $x^{\alpha }$, while on the lower bank it tends to $e^{2\pi i\alpha }{x}^{\alpha }$ and the direction is reversed. Therefore the limiting contour integral is

The only enclosed pole is the double pole at $z=-1$. Since $L(-1)=i\pi$,The residue theorem now givesFor $\alpha \ne 0$, division andyieldAt $\alpha =0$, the integral is one, agreeing with the continuous limit. This is the positive-axis keyhole beta integral.