Polar Representation of Complex Number Review

# Polar Representation of Complex Number Review

We will now review some of the recent material regarding the polar representation of complex numbers.

- Recall from
**The Polar Representation of Complex Numbers**page that if $z = a + bi \in \mathbb{C}$ and $z \neq 0$ then the**Argument**of $z$ denoted $\arg (z)$ is the angle $\theta$ that is made between the positive real axis and the position vector $(a, b)$ in the complex plane.

- Moreover, the notation "$\mathrm{Arg}(z)$" denotes the
**Principal Branch of the Argument Function**where we restrict $0 \leq \mathrm{Arg} (z) < 2\pi$.

- With this, for $z \neq 0$, we were able to define the
**Polar Representation**of $z$ where $r = \mid z \mid$ and $\theta = \arg (z)$ to be:

\begin{align} \quad z = r (\cos \theta + i \sin \theta) \end{align}

- On the
**Polar Representation with Multiplication of Complex Numbers**page we were able to find a nice geometric interpretation of complex number multiplication with regards to the polar representation of complex numbers. If $z, w \in \mathbb{C}$, $z, w \neq 0$, we proved that then:

\begin{align} \quad \mid z \cdot w \mid = \mid z \mid \mid w \mid \end{align}

(3)
\begin{align} \quad \arg (z \cdot w) = \arg(z) + \arg(w) \end{align}

- Therefore, if two complex numbers $z$ and $w$ are multiplied together then their product $z \cdot w$ is the complex number whose absolute value / modulus is the product of the moduli of $z$ and $w$, and whose argument is the sum of the arguments of $z$ and $w$ as illustrated below:

- We were also able to prove that if $z, w \in \mathbb{C}$ and $z, w \neq 0$ then:

\begin{align} \quad \biggr \lvert \frac{z}{w} \biggr \rvert = \frac{\mid z \mid}{\mid w \mid} \end{align}

- On the
**De Moivre's Formula for the Polar Representation of Powers of Complex Numbers**page we proved a very important formula known as De Moivre's formula which says that if $z \in \mathbb{C}$ ($z \neq 0$), $r = \mid z \mid$, and $\theta = \arg (z)$ then for all $n \in \mathbb{N}$ we have that:

\begin{align} \quad z^n = r^n (\cos (n \theta) + i \sin (n \theta)) \end{align}

- On the
**nth Roots of Complex Numbers**page we used the formula above to obtain a formula for the $n^{\mathrm{th}}$ roots of a complex number. If $z = r(\cos \theta + i \sin \theta) \in \mathbb{C}$ and $n \in \mathbb{N}$ then the $n$, $n^{\mathrm{th}}$ roots of $z$ are given by the following formula for each $k \in \{ 0, 1, ..., n - 1 \}$:

\begin{align} \quad r^{1/n} \left ( \cos \left ( \frac{\theta}{n} + \frac{k}{n} 2 \pi \right ) + i \sin \left ( \frac{\theta}{n} + \frac{k}{n} 2 \pi \right ) \right ) \end{align}