Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 7

The Cauchy-Riemann equations give $u_x=v_y$ and $u_y=-v_x$. ThereforeTaking the nonnegative square root yieldsThe two gradients are also orthogonal, since $\nabla u\cdot \nabla v=v_y{v}_x-v_x{v}_y=0$, although equality of their lengths alone is the requested result.

Differentiate the prescribed imaginary part with respect to $y$:The Cauchy-Riemann equation $u_x=v_y$ integrates toIts $y$ derivative is $u_y=-e^x[(x+1)\sin y+y\cos y]+F^{\prime }(y)$. On the other hand, $-v_x=-e^x[(x+1)\sin y+y\cos y]$. The second Cauchy-Riemann equation therefore requires $F^{\prime }(y)=0$. Combining the real and imaginary parts,Thus the most general answer on a connected domain isA real constant preserves the given imaginary part. Conversely, the difference of any two such analytic functions has zero imaginary part, and its Cauchy-Riemann derivatives force it to be constant. This proves completeness, rather than only exhibiting one possible harmonic conjugate.

For coefficients $a_n=1/\ln n$, the Cauchy-Hadamard theorem gives $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. HereThereforeOmitting the first two powers has no effect on the radius of convergence.

Let $a_n=(1+p/n{)}^{n^2}$. For every fixed real $p$, the base is positive once $n$ is sufficiently large; any initial negative or zero bases affect only finitely many coefficients. For that tail,The Cauchy-Hadamard theorem therefore yieldsFor example $p=0$ recovers the geometric-series radius one. The stronger expansion $a_n=e^{pn-p^2/2+O(1/n)}$ also shows that, on $|z|=e^{-p}$, the magnitudes of the individual terms tend to $e^{-p^2/2}>0$. Thus this series diverges at every point on its convergence circle; the radius itself does not require that extra boundary analysis.