Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 1 / 16B (Complex Methods)

Let two regular differentiable paths through $z_0$ have nonzero tangent vectors $v_1,v_2$. The chain rule gives tangent vectors $f^{\prime }(z_0)v_1$ and $f^{\prime }(z_0)v_2$ for their images. Multiplication by a nonzero complex number is a positive scaling and rotation: writing $f^{\prime }(z_0)=r{e}^{i\theta }$ scales lengths by $r>0$ and adds the same angle $\theta$ to both tangents. Equivalently,Thus the oriented angle between the paths is preserved, proving $f$ is conformal at $z_0$. The nonzero derivative also gives local invertibility; a zero derivative could multiply angles through branching instead.

Use the Möbius transformationThe lens's boundary intersections are $z=\pm 1$. Write $w=u+iv$. Substitution in the inverse gives $\mathrm{Im}z=2v/|w+1{|}^2$, andConsequently the two domain inequalities are exactly $0<v<u$, or $0<\arg w<\pi /4$. This algebra proves that $T$ maps the lens bijectively onto the quarter-sector, not merely that its boundary curves go to rays.

The map $w\mapsto w^4$ maps this sector bijectively onto the upper half-plane: the argument increases from $(0,\pi /4)$ to $(0,\pi )$, and the positive modulus map is bijective. Its inverse is the fourth-root branch with argument in $(0,\pi /4)$. Therefore the quarter-sector mapping of a circular lens isIts derivative is nonzero in the lens, since $T^{\prime }(z)=2/(1-z{)}^2$ and $T(z)\ne 0$ there. Finally the Cayley transform $\zeta \mapsto (\zeta -i)/(\zeta +i)$ maps $H$ bijectively to the unit disc, because $|\zeta -i|<|\zeta +i|$ exactly when $\mathrm{Im}\zeta >0$. A required conformal bijection is thusAll poles and branching endpoints of the construction lie outside the open domain.