Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2007 / B / 6

Conformality implies $f^{\prime }(0)\ne 0$, so $a_1\ne 0$. Apply the stated Green's theorem with $F=z^n\bar{z}$:Here $\bar{\zeta }={\zeta }^{-1}$ on the unit circle. The lowest power in $f^n{f}^{\prime }$ is ${\zeta }^n$, while the most negative power in the conjugate polynomial is ${\zeta }^{-N}$. For $n\ge N$ their product has no ${\zeta }^{-1}$ term. The residue theorem gives $M_n=0$ for all $n\ge N$.

When $N=2$, the residue for $n=0$ is $|a_1{|}^2+2|a_2{|}^2$. For $n=1$, expandOnly $a_1^2\zeta {\bar{a}}_2{\zeta }^{-2}$ contributes. ThereforeThe first is the area; the second records its complex first moment, and need not be real.

The potential plane is the strip $0<\psi <1$. The wall maps to its lower boundary, with upstream $\varphi \to -\infty$, the corner at $w=0$, and downstream $\varphi \to +\infty$. The free surface maps to its upper boundary.

Write $q=w^{\prime }=u-iv$. Far upstream $q=1$; far downstream the flow points vertically down and $q=i$. On the two wall rays, $q$ runs along the positive real and positive imaginary axes. The reentrant fluid corner has angle $3\pi /2$, so its potential behaves locally as $(z-B{)}^{2/3}$ and its speed diverges like $|z-B{|}^{-1/3}$. Thus the corner corresponds to $q=\infty$, not $q=0$. The free surface has $|q|=1$, giving the hodograph domainIt is the exterior of a quarter unit disc in the first quadrant.

Now $s=q^2$ maps this domain to the part of the upper half-plane outside the unit disc. The maptakes it to the first quadrant: its real and imaginary parts are positive because $|s|>1$ and $\Im s>0$. Finally $\eta =t^2$ maps that quadrant to the upper half-plane. The corresponding strip map is $\eta =e^{\pi w}$. Both give upstream $\eta =0$, corner $\eta =1$, and downstream $\eta =\infty$, so the three boundary normalizations fix the remaining real Möbius freedom. ConsequentlyOn the free surface $q=e^{i\theta }$, $t=i\tan \theta$, so $\eta$ is negative and $\Im w=1$ as required. On the wall $\eta >0$, giving $\Im w=0$.