Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 7

For a holomorphic function, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. ConsequentlyA gradient is normal to its level set, so these identities give orthogonal level curves wherever the gradients are nonzero. At a critical point where $f^{\prime }=0$, both gradients vanish and a regular tangent to a level set need not exist; the geometric orthogonality statement is restricted to regular intersections.

Differentiate the Cauchy-Riemann equations and use equality of mixed partial derivatives:Thus the real part is a harmonic function. The same calculation gives $\Delta v=0$. Every holomorphic function has the required smoothness.

Inverting $u+iv=\mathrm{arcosh}z$ gives $z=\cosh (u+iv)$, hence the elliptic coordinatesEliminating $v$ at fixed $u>0$ gives the ellipseIts focal distance is $\sqrt{{\cosh }^{2}u-{\sinh }^{2}u}=1$. At fixed $v$, eliminating $u$ instead gives the hyperbolaIts focal distance is also one, so both families have foci $(\pm 1,0)$. A fixed $v$ traces an appropriate half of the hyperbola; complementary coordinate values complete it. Degenerate values give the focal segment and portions of the coordinate axes. Away from these degeneracies the curves are orthogonal by part (i), or by ${\mathbf{r}}_u\cdot {\mathbf{r}}_v=0$ in the displayed parametrization. A branch of the inverse is used locally; its real part with $u\ge 0$ is single-valued outside the focal segment.

On the specified ellipse, the semiaxes are $\sqrt{2}$ and $1$, so its elliptic coordinate is $u_0=\mathrm{arcosh}\sqrt{2}=\log (\sqrt{2}+1)$. Thus in the exterior,is a harmonic function with a homogeneous Dirichlet boundary condition on that ellipse. Choose the square root with $\sqrt{z^2-1}\sim z$ at infinity. Although a global complex logarithm requires a cut, its real part $\log |z+\sqrt{z^2-1}|$ is well defined throughout the exterior and gives the same $\varphi$.

Expanding at large $|z|$ yields $z+\sqrt{z^2-1}=2z[1-1/(4z^2)+O(|z{|}^{-4})]$. ThereforeIn particular the potential grows logarithmically in the distance, in every direction at infinity. It is the exterior solution; no harmonic continuation through the focal segment in the ellipse's interior is claimed.

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$, for which $\Delta =4{\partial }_z{\partial }_{\bar{z}}$. Since $H$ and $G$ are holomorphic,The Laplacian has real coefficients, so taking real and imaginary parts provesThis is the complex analytic representation of biharmonic functions; $F$ itself is generally not holomorphic because of its $\bar{z}$ factor.