Answer: EITHER: the three roots written down are \(1,e^{2\pi i/3},e^{-2\pi i/3}\). The other three roots are \(-1,\frac{1}{3}+\frac{2\sqrt2}{3}i,\frac{1}{3}-\frac{2\sqrt2}{3}i\).

OR: one possible choice is \(\mathbf P=\begin{pmatrix}-2&0&0\\1&3&0\\0&4&1\end{pmatrix}\) and \(\mathbf D=\begin{pmatrix}3^n&0&0\\0&7^n&0\\0&0&1\end{pmatrix}\). The final range is \(-\frac{1}{7}\lt k\lt\frac{1}{7}\).

EITHER

(i) Since \(z=e^{i\theta}\), write \(z=e^{i\theta/2}e^{i\theta/2}\). Then

\(\frac{z-1}{z+1}=\frac{e^{i\theta/2}-e^{-i\theta/2}}{e^{i\theta/2}+e^{-i\theta/2}}.\)

Using \(e^{iu}-e^{-iu}=2i\sin u\) and \(e^{iu}+e^{-iu}=2\cos u\),

\(\frac{z-1}{z+1}=\frac{2i\sin(\theta/2)}{2\cos(\theta/2)}=i\tan\frac{\theta}{2}.\)

(ii) If \(z=1\), then each numerator is zero, so the sum is zero.

For the other cube roots of unity, write \(z=e^{i\theta}\), where \(\theta=\pm\frac{2\pi}{3}\). Since \(z^3=1\), the first term is zero. Also, by part (i),

\(\frac{z^2-1}{z^2+1}=i\tan\theta,\qquad \frac{z-1}{z+1}=i\tan\frac{\theta}{2}.\)

If \(\theta=\frac{2\pi}{3}\), then \(\tan\theta=-\sqrt3\) and \(\tan\frac{\theta}{2}=\sqrt3\). If \(\theta=-\frac{2\pi}{3}\), the two values are reversed in sign. In both cases, their sum is zero. Therefore

\(\frac{z^3-1}{z^3+1}+\frac{z^2-1}{z^2+1}+\frac{z-1}{z+1}=0.\)

(iii) From part (ii), the three cube roots of unity are roots of the polynomial equation. Thus

\(z=1,\quad z=e^{2\pi i/3},\quad z=e^{-2\pi i/3}\)

are three roots.

Expanding the left-hand side gives

\(3z^6+z^5+z^4-z^2-z-3=0.\)

This factorises as

\(3z^6+z^5+z^4-z^2-z-3=(z+1)(z^3-1)(3z^2-2z+3).\)

The roots not already listed are therefore

\(z=-1\)

and the roots of \(3z^2-2z+3=0\). Hence

\(z=\frac{2\pm\sqrt{4-36}}{6}=\frac{2\pm i\sqrt{32}}{6}=\frac{1}{3}\pm\frac{2\sqrt2}{3}i.\)

So the other three roots are

\(\boxed{-1,\ \frac{1}{3}+\frac{2\sqrt2}{3}i,\ \frac{1}{3}-\frac{2\sqrt2}{3}i}.\)

OR

(i) Another eigenvector corresponding to \(\lambda\) is any non-zero scalar multiple of \(\mathbf e\), for example \(2\mathbf e\).

(ii) Since \(\mathbf A\mathbf e=\lambda\mathbf e\), repeated application gives

\(\mathbf A^n\mathbf e=\lambda^n\mathbf e.\)

So \(\mathbf e\) is an eigenvector of \(\mathbf A^n\), with eigenvalue \(\lambda^n\).

(iii) The matrix \(\mathbf A\) is triangular, so its eigenvalues are \(3,7,1\).

For \(\lambda=3\), an eigenvector is \((-2,1,0)^T\).

For \(\lambda=7\), an eigenvector is \((0,3,4)^T\).

For \(\lambda=1\), an eigenvector is \((0,0,1)^T\).

Therefore one suitable matrix is

\(\mathbf P=\begin{pmatrix}-2&0&0\\1&3&0\\0&4&1\end{pmatrix}.\)

For \(\mathbf A^n\), the diagonal matrix is

\(\mathbf D=\begin{pmatrix}3^n&0&0\\0&7^n&0\\0&0&1\end{pmatrix}.\)

Thus

\(\mathbf A^n=\mathbf P\mathbf D\mathbf P^{-1}.\)

(iv) The partial sum is

\(\sum_{n=1}^{N}k^n(\mathbf A^n-k\mathbf A^{n+1})=k\mathbf A-k^{N+1}\mathbf A^{N+1}.\)

For the infinite sum to equal \(k\mathbf A\), we need

\(k^{N+1}\mathbf A^{N+1}\to \mathbf 0\)

as \(N\to\infty\). Since the largest eigenvalue of \(\mathbf A\) in modulus is \(7\), this requires

\(|7k|\lt1.\)

Hence

\(\boxed{-\frac{1}{7}\lt k\lt\frac{1}{7}}.\)