2022TrialShort answer

Q16 (a) (iii)

3 marks

Question

Using parts (i) and (ii), find the roots of
$\left(\frac{2+z}{2-z}\right)^5=1.$

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Solution

Since $w^5=1$ ,

\frac{2+z}{2-z}=w

\begin{aligned} 2+z&=w(2-z)\\ 2+z&=2w-zw\\ zw+z&=2w-2\\ z(w+1)&=2(w-1)\\ z&=\frac{2(w-1)}{w+1}. \end{aligned}

Since $w^5=1$ and from part (ii),

w=\cos\theta+i\sin\theta,\qquad \theta=\frac{2k\pi}{5},\quad k=0,\pm1,\pm2.

And from part (i),

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

For $\theta=\frac{2k\pi}{5}$ , where $k=0,\pm1,\pm2$ ,

z=2i\tan\left(-\frac{2\pi}{5}\right),\quad 2i\tan\left(-\frac{\pi}{5}\right),\quad 2i\tan(0),\quad 2i\tan\left(\frac{\pi}{5}\right),\quad 2i\tan\left(\frac{2\pi}{5}\right).

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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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