Answer: (a) \(z+z^2+z^3+\ldots+z^n=\dfrac{z-z^{n+1}}{1-z}\).

(b) Since \(z^n=1\) and \(z\neq 1\), it follows that \(1+z+z^2+\ldots+z^{n-1}=0\).

(c) \(\displaystyle \sum_{m=1}^{\infty}3^{-m}\cos m\theta=\frac{3\cos\theta-1}{10-6\cos\theta}\).

(a) This is a geometric series with first term \(z\), common ratio \(z\), and \(n\) terms.

So

\(z+z^2+z^3+\ldots+z^n=\dfrac{z(1-z^n)}{1-z}=\dfrac{z-z^{n+1}}{1-z}\), for \(z\neq 1\).

(b) If \(z\) is an \(n\)th root of unity, then \(z^n=1\).

Using the result from part (a),

\(z+z^2+\ldots+z^n=\dfrac{z-z^{n+1}}{1-z}.\)

But \(z^{n+1}=z\cdot z^n=z\cdot 1=z\), so

\(z+z^2+\ldots+z^n=0.\)

Hence

\(z(1+z+z^2+\ldots+z^{n-1})=0.\)

Since \(z\neq 0\), it follows that

\(1+z+z^2+\ldots+z^{n-1}=0.\)

(c) We are given

\(z=\dfrac{1}{3}(\cos\theta+\mathrm{i}\sin\theta).\)

By de Moivre's theorem,

\(z^m=\left(\dfrac{1}{3}\right)^m(\cos m\theta+\mathrm{i}\sin m\theta).\)

So

\(\sum_{m=1}^{\infty}3^{-m}\cos m\theta=\operatorname{Re}\left(\sum_{m=1}^{\infty}z^m\right).\)

Since \(|z|=\dfrac13\lt 1\),

\(\sum_{m=1}^{\infty}z^m=\dfrac{z}{1-z}.\)

Substituting for \(z\),

\(\sum_{m=1}^{\infty}z^m=\frac{\frac13(\cos\theta+\mathrm{i}\sin\theta)}{1-\frac13(\cos\theta+\mathrm{i}\sin\theta)}=\frac{\cos\theta+\mathrm{i}\sin\theta}{3-\cos\theta-\mathrm{i}\sin\theta}.\)

Rationalise the denominator:

\(\sum_{m=1}^{\infty}z^m=\frac{(\cos\theta+\mathrm{i}\sin\theta)(3-\cos\theta+\mathrm{i}\sin\theta)}{(3-\cos\theta)^2+\sin^2\theta}.\)

Expand the numerator:

\((\cos\theta+\mathrm{i}\sin\theta)(3-\cos\theta+\mathrm{i}\sin\theta)\)

\(=3\cos\theta-\cos^2\theta-\sin^2\theta+\mathrm{i}\sin\theta\cos\theta+\mathrm{i}\sin\theta(3-\cos\theta).\)

Thus the real part of the numerator is

\(3\cos\theta-(\cos^2\theta+\sin^2\theta)=3\cos\theta-1.\)

Also, the denominator is

\((3-\cos\theta)^2+\sin^2\theta=9-6\cos\theta+\cos^2\theta+\sin^2\theta=10-6\cos\theta.\)

Therefore

\(\operatorname{Re}\left(\sum_{m=1}^{\infty}z^m\right)=\frac{3\cos\theta-1}{10-6\cos\theta}.\)

Hence

\(\sum_{m=1}^{\infty}3^{-m}\cos m\theta=\frac{3\cos\theta-1}{10-6\cos\theta}.\)