2023TrialShort answer

Q12 (a) (ii)

2 marks

Question

Let $\alpha$ be a real number and suppose that $z$ is a complex number such that

z+\frac{1}{z}=2\cos\alpha.

You may assume that

z^n+\frac{1}{z^n}=2\cos(n\alpha)

for all positive integer $n$ . Let

\omega=z+\frac{1}{z}.

Find the ninth roots of unity.

Try the question before revealing the answer.

Solution

(ii)

z^9=1

z^9=\cos0+i\sin0

Let $z=r(\cos\theta+i\sin\theta)$ .

r^9(\cos9\theta+i\sin9\theta)=\cos0+i\sin0\quad\text{(De Moivre's)}

r=1

9\theta=0+2k\pi,\quad k\in\mathbb{Z}

\theta=\frac{2k\pi}{9}

Therefore roots are:

\begin{aligned} &1,\ \cos\frac{2\pi}{9}+i\sin\frac{2\pi}{9},\ \cos\left(\frac{2\pi}{9}\right)+i\sin\left(-\frac{2\pi}{9}\right),\\ &\cos\frac{4\pi}{9}+i\sin\frac{4\pi}{9},\ \cos\left(-\frac{4\pi}{9}\right)+i\sin\left(-\frac{4\pi}{9}\right),\\ &\cos\frac{6\pi}{9}+i\sin\frac{6\pi}{9},\ \cos\left(-\frac{6\pi}{9}\right)+i\sin\left(-\frac{6\pi}{9}\right),\\ &\cos\frac{8\pi}{9}+i\sin\frac{8\pi}{9},\ \cos\left(-\frac{8\pi}{9}\right)+i\sin\left(-\frac{8\pi}{9}\right) \end{aligned}

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Complex Numbers

Solving equations with complex numbersUse De Moivre's theorem and solve equations and polynomial problems involving complex and conjugate roots.

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