Past exam of the mathematical methods I course of the University of Oxford / 2022 / 9

The right-hand side has modulus two and argument $\pi /3$. Its fifth roots therefore haveThey are the vertices of a regular pentagon on the circle of radius $2^{1/5}$, beginning at angle $\pi /15$. These five roots are distinct, and a degree-five polynomial has no further complex roots.

The base is $2e^{i\pi /4}$. Its complex logarithms are $\ln 2+i(\pi /4+2\pi k)$, $k\in \mathbb{Z}$. Exponentiating $i$ times each logarithm gives all values of the multivalued power:The values form a discrete geometric sequence along the Argand ray of angle $\ln 2$. Consecutive radii differ by $e^{-2\pi }$. The origin is an accumulation point but is not itself a value. If the power is instead defined using the principal logarithm, only $k=0$ is retained. The phrase “all values” requires this branch distinction.

For $z\ne -1$, multiply numerator and denominator by $\overline{z}+1$. The real part isThus its zero locus isIt is the unit circle with the point $-1$ removed. That point cannot be restored by the zero numerator obtained after clearing a denominator: the original expression is undefined there.

Use the imaginary part of the finite complex sum:The identity $1+e^{i\theta }=2e^{i\theta /2}\cos (\theta /2)$ then givesHere $N$ is a nonnegative integer. Keeping the signed cosine raised to the integer power makes the result valid even when $\cos (\theta /2)<0$; inserting an absolute value would lose the sign for odd $N$.