(i) Write down the five fifth roots of unity - CIE - A-Level Further Maths - Question 9 - 2010 - Paper 1

To find the roots, rewrite the equation as:

$z^5 = -16 - \frac{16}{\sqrt{3}}i.$

Calculate the modulus:

$r = |z^5| = \sqrt{(-16)^2 + \left(-\frac{16}{\sqrt{3}}\right)^2} = \sqrt{256 + \frac{256}{3}} = \sqrt{\frac{768}{3}} = \frac{16\sqrt{3}}{3}.$

Taking the fifth root, we get:

$|z| = \sqrt[5]{\frac{16\sqrt{3}}{3}}.$

Next, find the argument:

$\theta = \tan^{-1}\left(\frac{-\frac{16}{\sqrt{3}}}{-16}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}.$

Thus, the argument for $z^5$ is:

$\arg(z^5) = \pi + \frac{\pi}{6} = \frac{7\pi}{6}.$

Therefore, the roots are given by:

$z_k = \sqrt[5]{\frac{16\sqrt{3}}{3}} e^{i\left(\frac{7\pi}{6} + \frac{2k\pi}{5}\right)}, \quad k = 0, 1, 2, 3, 4.$

On the Argand diagram, these roots are evenly spaced around a circle with radius $\sqrt[5]{\frac{16\sqrt{3}}{3}}$.

Choose $|w| = r$. The sum is a geometric series:

$\sum_{k=0}^4 r^k = \frac{1 - r^5}{1 - r}.$

Therefore,

$\sum_{k=0}^4 \left( \frac{|w|}{2} \right)^k = \sum_{k=0}^4 \left( \frac{r}{2} \right)^k = \frac{1 - (\frac{r}{2})^5}{1 - \frac{r}{2}} = \frac{1 - \frac{r^5}{32}}{1 - \frac{r}{2}}.$

Using the property that $w$ is a root of the original equation and substituting values, show that:

$\frac{3 + i/3}{2 - w}.$

From the roots determined in part (ii), compute $|2 - w_k|$ for each root $w_k$. This can be simplified and evaluated to find which root minimizes the distance. The minimum occurs when $w$ is chosen such that:

$|2 - w| = \min_k |2 - w_k|,$

and substituting the values from the earlier calculations will give you the least value.