202211211714 Solution to 1976-CE-AMATH-I-XX

Express in polar form the following complex numbers:

$z=-\sqrt{3}-\mathrm{i}$ and $w=1-\mathrm{i}$ .

Hence or otherwise express $\displaystyle{\frac{z^3}{w^4}}$ in the form of $a+\mathrm{i}b$ .

Roughwork.

$\begin{aligned} \mathrm{mod}(z) & = |z| \\ & = |-\sqrt{3}-\mathrm{i}|\\ & = \sqrt{(-\sqrt{3})^2+(-1)^2} \\ & = 2 \\ \mathrm{arg}(z) & = \arctan\bigg(\frac{-1}{-\sqrt{3}}\bigg) \\ & = \frac{4\pi}{3} \\ \mathrm{mod}(w) & = |w| \\ & = |1-\mathrm{i}| \\ & = \sqrt{(1)^2+(-1)^2} \\ & = \sqrt{2} \\ \mathrm{arg}(w) & = \arctan\bigg(\frac{-1}{1}\bigg) \\ & = \frac{3\pi}{4} \\ \end{aligned}$

$\begin{cases} z=2e^{\mathrm{i}\frac{4\pi}{3}} \\ w=\sqrt{2}e^{\mathrm{i}\frac{3\pi}{4}} \\ \end{cases}$

$\begin{aligned} \frac{z^3}{w^4} & = z^3w^{-4} \\ & = \big( 2e^{\mathrm{i}\frac{4\pi}{3}}\big)^3\big(\sqrt{2}e^{\mathrm{i}\frac{3\pi}{4}}\big)^{-4} \\ & = \dots \\ \end{aligned}$

This problem is not to be attempted.

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