2015 H2 Maths Paper 1, Question 9

Let $w = a + ib$, where $a$ and $b$ are non-zero real numbers, so $w^{\text{*}} = a - ib$.

Since $\frac{w^{2}}{w^{\text{*}}}$ is purely imaginary, its real part is zero:

Since $a^{2} + b^{2} \neq 0$ (as $a,b$ are non-zero real numbers):

Since $a \neq 0$:

Note: substituting $b^{2} = \frac{a^{2}}{3}$ into the imaginary part gives $3a^{2}b - b^{3} = b\left( 3a^{2} - b^{2} \right) = \frac{8a^{2}}{3}b$, which is non-zero for $a,b \neq 0$, so $\frac{w^{2}}{w^{\text{*}}}$ is genuinely purely imaginary (not zero) for these values.

Since $- 32i = 32\left( \cos\left( - \frac{1}{2}\pi \right) + i\sin\left( - \frac{1}{2}\pi \right) \right)$, the equation $z^{5} = - 32i$ gives

By De Moivre’s theorem, the five roots are

so each $z$ has modulus $32^{\frac{1}{5}} = 2$ and argument $- \frac{\pi}{10} + \frac{2k\pi}{5}$.

Taking $k = 0, \pm 1, \pm 2$ gives the five arguments, all already in $( - \pi,\pi\rbrack$:

Since $\frac{1}{2}\pi < \arg z_{1} < \pi$, we take $\arg z_{1} = \frac{7\pi}{10}$, so $z_{1} = 2\left( \cos\frac{7\pi}{10} + i\sin\frac{7\pi}{10} \right)$.

Since $- \pi < \arg z_{2} < - \frac{1}{2}\pi$, we take $\arg z_{2} = - \frac{9\pi}{10}$, so $z_{2} = 2\left( \cos\frac{9\pi}{10} - i\sin\frac{9\pi}{10} \right)$, using $\cos( - \theta) = \cos\theta$ and $\sin( - \theta) = - \sin\theta$.

Using $\cos A - \cos B = - 2\sin\frac{A + B}{2}\sin\frac{A - B}{2}$ and $\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2}$, with $A = \frac{7\pi}{10}$, $B = \frac{9\pi}{10}$, so $\frac{A + B}{2} = \frac{4\pi}{5}$ and $\frac{A - B}{2} = - \frac{\pi}{10}$:

Since $\sin\frac{\pi}{10} = \cos\left( \frac{1}{2}\pi - \frac{\pi}{10} \right) = \cos\frac{2\pi}{5}$ and $\cos\frac{\pi}{10} = \sin\left( \frac{1}{2}\pi - \frac{\pi}{10} \right) = \sin\frac{2\pi}{5}$, and $\sin\frac{4\pi}{5} = \sin\left( \pi - \frac{4\pi}{5} \right) = \sin\frac{\pi}{5} > 0$:

This is already in the form $r\left( \cos\alpha + i\sin\alpha \right)$ with $r = 4\sin\frac{\pi}{5} > 0$, so we read off the modulus and argument directly.