Compute the modulus and principal argument of given complex numbers

For each of the following complex numbers, compute the modulus and principal argument.

$2i$ .
$-3i$ .
$-1$ .
$1$ .
$-3 + \sqrt{3}i$ .
$\frac{1+i}{\sqrt{2}}$ .
$(-1+i)^3$ .
$(-1-i)^3$ .
$\frac{1}{1+i}$ .
$\frac{1}{(1+i)^2}$ .

First, the modulus is given by
$|2i| = \sqrt{2^2} = 2.$

The principal argument is the unique $\theta \in (-\pi, pi]$ such that

$x = r \cos \theta, \qquad y = r \sin \theta.$

In this case we have $x = 0$ , $y = 2$ and $r = \sqrt{2}$ so the principal argument satisfies

$2 \cos \theta = 0, \ \ 2 \sin \theta = 2 \qquad \implies \qquad \theta = \frac{\pi}{2}.$
First, the modulus is given by
$|-3i| = \sqrt{(-3)^2} = 3.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$3 \cos \theta = 0, \ \ 3 \sin \theta = -3 \qquad \implies \qquad \theta = -\frac{\pi}{2}.$
The modulus is given by
$|-1| = \sqrt{(-1)^2} = 1.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\cos \theta = -1, \ \ \sin \theta = 0 \qquad \implies \qquad \theta = \pi.$
The modulus is given by
$|1| = \sqrt{1^2} = 1.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\cos \theta = 1, \ \ \sin \theta = 0 \qquad \implies \qquad \theta = 0.$
The modulus is given by
$|-3+\sqrt{3}i| = \sqrt{(-3)^2 + \sqrt{3}^2} = \sqrt{12} = 2 \sqrt{3}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\begin{align*} && 2 \sqrt{3} \cos \theta &= -3 & \text{and} && 2 \sqrt{3} \sin \theta &= \sqrt{3} \\ \implies && \cos \theta &= -\frac{\sqrt{3}}{2} & \text{and} && \sin \theta &= \frac{1}{2} \\ \implies && \theta &= \frac{5 \pi}{6}. \end{align*}$
The modulus is given by
$\left| \frac{1 + i}{\sqrt{2}} \right| = \sqrt{\left( \frac{1}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2} = \sqrt{1} = 1.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\cos \theta = \frac{1}{\sqrt{2}}, \ \ \sin \theta = \frac{1}{\sqrt{2}} \qquad \implies \qquad \theta = \frac{\pi}{4}.$
First, we write $(-1+i)^3$ in the form $a+bi$ ,
$(-1+i)^3 = 2 + 2i.$

Therefore, the modulus is

$|(-1+i)^3| = |2+2i| = \sqrt{2^2 + 2^2} = 2\sqrt{2}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\begin{align*} && 2 \sqrt{2} \cos \theta &= 2 & \text{and} && 2 \sqrt{2} \sin \theta &= 2 \\ \implies && \cos \theta &= \frac{1}{\sqrt{2}} & \text{and} && \sin \theta &= \frac{1}{\sqrt{2}} \\ \implies && \theta &= \frac{\pi}{4}. \end{align*}$
First, we rewrite $(-1-i)^3$ in the form $a+bi$ ,
$(-1-i)^3 = (-1-i)(-1-i)(-1-i) = (2i)(-1-i) = 2 - 2i.$

Therefore, the modulus is

$|(-1-i)^3| = |2 - 2i| = \sqrt{2^2 + (-2)^2} = 2 \sqrt{2}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\begin{align*} && 2 \sqrt{2} \cos \theta &= 2 & \text{and} && 2 \sqrt{2} \sin \theta &= -2 \\ \implies && \cos \theta &= \frac{1}{\sqrt{2}} & \text{and} && \sin \theta &= -\frac{1}{\sqrt{2}} \\ \implies && \theta &= -\frac{\pi}{4}. \end{align*}$
First, we rewrite $\frac{1}{1+i}$ in the form $a+bi$ ,
$\frac{1}{1+i} = \frac{1-i}{(1+i)(1-i)} = \frac{1-i}{2} = \frac{1}{2} - \frac{1}{2}i.$

Therefore, the modulus is

$\left| \frac{1}{1+i} \right| = \left|\frac{1}{2} - \frac{1}{2}i \right| = \sqrt{\left( \frac{1}{2} \right)^2 + \left(\frac{-1}{2}\right)^2} = \frac{1}{\sqrt{2}}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\begin{align*} && \frac{1}{\sqrt{2}} \cos \theta &= \frac{1}{2} & \text{and} && \frac{1}{\sqrt{2}} \sin \theta &= -\frac{1}{2} \\ \implies && \cos \theta &= \frac{1}{\sqrt{2}} & \text{and} && \sin \theta &= -\frac{1}{\sqrt{2}} \\ \implies && \theta &= -\frac{\pi}{4}. \end{align*}$
First, we rewrite $\frac{1}{(1+i)^2}$ in the form $a+bi$ ,
$\frac{1}{(1+i)^2} = \frac{1}{2i} = -\frac{1}{2}i .$

The modulus is

$\left| \frac{1}{(1+i)^2} \right| = \left| \frac{-1}{2}i\right| = \sqrt{\left( \frac{-1}{2} \right)^2} = \frac{1}{2}.$

The principal argument $\theta \in (-\pi, \pi]$ satisfies

$\frac{1}{2} \cos \theta = 0, \ \ \frac{1}{2} \sin \theta = -\frac{1}{2} \qquad \implies \qquad \theta = -\frac{\pi}{2}.$