Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 7

For a holomorphic function with $z=r{e}^{i\theta }$, the chain rule givesWriting $f=u+iv$ and equating real parts and imaginary parts,For $r>0$, division by $r$ gives the Cauchy-Riemann equations in polar coordinates:The restriction $r>0$ reflects the coordinate singularity at the origin; the Cartesian equations have no such restriction.

Insert the specified real part into the polar Cauchy-Riemann equations. The first gives $v_r=n{r}^{n-1}\sin (n\theta )$, and integration with respect to $r$ givesThe second equation then requires $v_{\theta }=n{r}^n\cos (n\theta )$, so $C^{\prime }(\theta )=0$. The remaining imaginary additive constant is fixed by $f(0)=0$. ThusThe resulting polynomial is holomorphic at the origin too, consistently extending the calculation made in polar coordinates with $r>0$.

Using the exponential form of the complex cosine, $\cos z=0$ is equivalent to $e^{2iz}=-1$. Thus the zeros are exactlyAt each point ${\cos }^{\prime }(z_n)=-\sin z_n\ne 0$, so the cosine has a simple zero. Cubing it givesThe order of a zero of a holomorphic function is therefore three at every zero. There are no finite singularities because this is an entire function.

To classify infinity, examine ${\cos }^3(1/w)$ at $w=0$. The identity ${\cos }^{3}z=(3\cos z+\cos 3z)/4$ shows that its Laurent series contains the nonzero negative-power coefficientsThere are infinitely many terms in the principal part, the criterion for an essential singularity. Therefore $\boxed{\infty \text{ is an essential singularity, not a pole.}}$ This is also the general characterization of a nonpolynomial entire function at infinity.

If $z_0\ne 0$, factor out $z_0$ and use the geometric series:ThusThe radius of convergence is the distance to the nearest singularity, the simple pole at $z=-z_0$. The case $z_0=0$ must be treated separately: the function is $1/z$, has no Taylor series about zero and is already its own Laurent series on $0<|z|<\infty$.

For $|z|<1$, multiply the absolutely convergent Taylor series of the sine by the geometric series:The product is odd, so all even coefficients vanish. Collecting the coefficient of $z^{2m+1}$ givesFor example, the first terms are $z+5z^3/6+101z^5/120+\cdots$. The denominator vanishes at $z=\pm 1$, and the numerator is nonzero there, so these are genuine simple poles. Therefore $\boxed{R=1}$ is the radius of convergence.