Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2010 / Analysis / 4

Write $f=u+iv$ and $z=x+iy$. The complex derivative computed along a real increment is $f^{\prime }=u_x+i{v}_x$, while along an imaginary increment it is $f^{\prime }=v_y-i{u}_y$. Equality givesThese are the Cauchy–Riemann equations at every point where $f$ is holomorphic. The directional calculations also establish the existence of those partial derivatives. More explicitly, $f(z+h)-f(z)=f^{\prime }(z)h+o(|h|)$ gives both real-direction formulas directly. If a converse is desired, continuous first partial derivatives together with these equations make the real differential multiplication by $u_x+i{v}_x$, hence give complex differentiability; the necessity above is what is requested.

The allowed theorem makes $f^{\prime }$ holomorphic and hence continuous; applying it repeatedly gives continuous derivatives of every order. Since $u_x=\Re f^{\prime }$, $v_x=\Im f^{\prime }$, $u_y=-\Im f^{\prime }$ and $v_y=\Re f^{\prime }$, the real and imaginary parts have continuous second partial derivatives. Differentiate the Cauchy–Riemann equations and commute mixed derivatives:Thus $\boxed{\Delta u=\Delta v=0}$, with the required $C^2$ regularity, proving harmonicity.

Write $f=u+iv$ and evaluate all derivatives of $\psi$ at $(u,v)$. The real chain rule givesThe Cauchy–Riemann equations give equal squared lengths $u_x^2+u_y^2=v_x^2+v_y^2=|f^{\prime }{|}^2$ and zero cross term $u_x{v}_x+u_y{v}_y=0$. Part (ii) removes the last two terms. ThereforeThe composition is $C^2$, so it is harmonic. No assumption that $f^{\prime }$ is nonzero is required.

The complex exponential function $E(z)=e^z$ maps the strip bijectively and holomorphically to the upper half-plane: its modulus is $e^{\Re z}>0$ and its argument is $\Im z\in (0,\pi )$. Its inverse is the holomorphic logarithm $L(w)=\log |w|+i\mathrm{Arg}w$ with $\mathrm{Arg}w\in (0,\pi )$.

Part (iii) applies to both maps. The required inverse correspondences areThey compose to the identity because $L\circ E$ and $E\circ L$ do. This proves injectivity and surjectivity, not merely a map between the two sets.