November 21, 2022 – Physicspupil

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.

Let the complex number $z_1=-1+\mathrm{i}\sqrt{3}$ .

i. Represent $z_1$ and its conjugate $\overline{z_1}$ on the Argand diagram.
ii. Calculate the modulus and the argument of $z_1$ .
iii. If $z_2=(z_1)^2$ , express $z_2$ in the form of $a+\mathrm{i}b$ .
iv. Show that both $z_1$ and $\displaystyle{\frac{1}{2}z_2}$ are roots of the equation $z^3-8=0$ .

Roughwork.

$\begin{aligned} \textrm{modulus }\mathrm{mod}(z_1) & = |z_1| \\ & = |-1+\mathrm{i}\sqrt{3}| \\ & = \sqrt{(-1)^2+(\sqrt{3})^2} \\ & = 2 \\ \textrm{argument }\mathrm{arg}(z_1) & = \arctan\bigg(\frac{\sqrt{3}}{-1}\bigg) \\ & = \frac{2\pi}{3}\\ \end{aligned}$

Thus, $z_1=2e^{\mathrm{i}\frac{2\pi}{3}}$ .

As is given,

$\begin{aligned} z_2 & = (z_1)^2 \\ & = \big( 2e^{\mathrm{i}\frac{2\pi}{3}}\big)^2 \\ & = 4e^{\mathrm{i}\frac{4\pi}{3}} \\ \mathrm{mod}(z_2) & = 4 \\ \mathrm{arg}(z_2) & = \frac{4\pi}{3} \\ \end{aligned}$

or,

$\begin{aligned} z_2 & = (z_1)^2 \\ & = (-1+\mathrm{i}\sqrt{3})^2 \\ & = (-1)^2 +2(-1)(\mathrm{i}\sqrt{3}) + (\mathrm{i}\sqrt{3})^2 \\ & = 1 -\mathrm{i}(2\sqrt{3}) -3 \\ & = -2-\mathrm{i}(2\sqrt{3}) \\ \end{aligned}$

This problem is not to be attempted.

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Day: November 21, 2022

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

202211211635 Solution to 1976-CE-AMATH-I-XX