Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 7

Use $\pi /12=\pi /3-\pi /4$ to multiply the elementary complex exponentials:Collecting real and imaginary parts gives

Factor $1-z^3=(1-z)(1+z+z^2)$. The geometric series gives, for $|z|<1$,Thus the first three nonzero terms are $1-z+z^3$. If “first three terms” means the coefficients through degree two, that Taylor polynomial is $1-z+0z^2$. The distinction matters because the quadratic coefficient vanishes. The nearest singularities are the two nontrivial cube roots of unity, so the radius of convergence is exactly one.

Put $w=z-\pi$. The identity ${\cosh }^2(1/w)=[1+\cosh (2/w)]/2$ gives a Laurent series containing a constant and only even negative powers $w^{-2},w^{-4},\dots$. There is no $w^{-1}$ coefficient, soThe singularity is an essential singularity, not a pole: infinitely many negative Laurent coefficients are nonzero. The residue is still defined as the coefficient of $w^{-1}$.

For $z=r{e}^{i\theta }\ne 0$, fix one argument $\theta$. Every value of the complex logarithm is $\log r+i(\theta +2\pi k)$, $k\in \mathbb{Z}$. By complex exponentiation, $z^z=\exp [z(\log r+i(\theta +2\pi k))]$. Taking the real part of its exponent gives all possible moduli:Equivalently these are $r^{\mathrm{Re}z}{e}^{-\mathrm{Im}z(\theta +2\pi k)}$. If $z$ is real and nonzero the modulus is the single value $r^{\mathrm{Re}z}$; otherwise the values form a geometric progression with ratio $e^{-2\pi \mathrm{Im}z}$. Changing the initially chosen argument just reindexes the same set. The logarithmic definition excludes $z=0$. This is the modulus of a complex self-power.