Answer: (a) \(\displaystyle \sum_{r=1}^{n} z^{4r}=\frac{z^{4n+2}-z^2}{z^2-z^{-2}}\).

(b) For \(z=\cos\theta+\mathrm{i}\sin\theta\), if \(\sin 2\theta\neq 0\), then \(\displaystyle \sum_{r=1}^{n}\sin(4r\theta)=\frac{\cos 2\theta-\cos(4n+2)\theta}{2\sin 2\theta}\).

(a) The series

\(\displaystyle \sum_{r=1}^{n} z^{4r}=z^4+z^8+\cdots+z^{4n}\)

is geometric with first term \(z^4\) and common ratio \(z^4\). Therefore

\(\displaystyle \sum_{r=1}^{n} z^{4r}=\frac{z^4(z^{4n}-1)}{z^4-1}.\)

Now multiply numerator and denominator by \(z^{-2}\):

\(\displaystyle \frac{z^4(z^{4n}-1)}{z^4-1}\times\frac{z^{-2}}{z^{-2}}=\frac{z^{4n+2}-z^2}{z^2-z^{-2}}.\)

Hence

\(\displaystyle \sum_{r=1}^{n} z^{4r}=\frac{z^{4n+2}-z^2}{z^2-z^{-2}},\)

as required, for \(z^2\ne z^{-2}\).

(b) Let \(z=\cos\theta+\mathrm{i}\sin\theta\). Then by de Moivre's theorem,

\(z^2=\cos 2\theta+\mathrm{i}\sin 2\theta\),

\(z^{-2}=\cos 2\theta-\mathrm{i}\sin 2\theta\),

so

\(z^2-z^{-2}=2\mathrm{i}\sin 2\theta.\)

Also, from part (a),

\(\displaystyle \sum_{r=1}^{n} z^{4r}=\frac{z^{4n+2}-z^2}{z^2-z^{-2}}.\)

Using de Moivre again,

\(z^{4n+2}=\cos((4n+2)\theta)+\mathrm{i}\sin((4n+2)\theta).\)

Therefore

\(\displaystyle \sum_{r=1}^{n} z^{4r}=\frac{\cos((4n+2)\theta)+\mathrm{i}\sin((4n+2)\theta)-\cos 2\theta-\mathrm{i}\sin 2\theta}{2\mathrm{i}\sin 2\theta}.\)

But

\(\displaystyle \sum_{r=1}^{n} z^{4r}=\sum_{r=1}^{n}\bigl(\cos 4r\theta+\mathrm{i}\sin 4r\theta\bigr),\)

so the imaginary part of the left-hand side is \(\displaystyle \sum_{r=1}^{n}\sin(4r\theta)\).

Write the numerator on the right as

\(\bigl(\cos((4n+2)\theta)-\cos 2\theta\bigr)+\mathrm{i}\bigl(\sin((4n+2)\theta)-\sin 2\theta\bigr).\)

Dividing by \(2\mathrm{i}\sin 2\theta\), its imaginary part is

\(\displaystyle \frac{\cos((4n+2)\theta)-\cos 2\theta}{-2\sin 2\theta}.\)

Hence

\(\displaystyle \sum_{r=1}^{n}\sin(4r\theta)=\frac{\cos((4n+2)\theta)-\cos 2\theta}{-2\sin 2\theta}=\frac{\cos 2\theta-\cos((4n+2)\theta)}{2\sin 2\theta}.\)

This is the required result.