Complex Algebra 2: Exponential Form

We can obtain the exponential form of a complex number using polar coordinates. The exponential form can be very useful from time to time. A complex number $z=x+iy$, in terms of polar coordinates
$$x=r\cos\theta,\ y=r\sin\theta,\ r=\sqrt{x^2+y^2},$$
can be written as
\begin{equation}
\begin{aligned}
z&=r(\cos\theta+i\sin\theta)\\
&=re^{i\theta}
\end{aligned}
\end{equation}
$\theta=\tan^{-1}\frac{y}{x}$ is called the argument. The set of all arguments is denoted by $\arg z$. The principal value of $\arg z$, denoted by $\mathrm{Arg}\ z$, is the unique value $\Theta$ such that $-\pi<\Theta\leq\pi$. Then
\begin{equation}
\arg z=\mathrm{Arg}\ z+2n\pi,\ n=0,\pm1,\pm 2,\cdots
\end{equation}
When $z$ is a negative real number, $\mathrm{Arg}\ z$ has the value $\pi$, not $-\pi$.

Example. The principal argument of $z=-1-i$ is $\Theta=-\frac{3\pi}{4}$ and so
$$\arg z=-\frac{3\pi}{4}+2n\pi$$
where $n=0,\pm 1,\pm 2,\cdots$.

Remark. $\arg z$ may not necessarily be represented by the principal argument. For example, $\arg (-1-i)$ may be written as
$$\arg (-1-i)=\frac{5\pi}{4}+2n\pi, \ n=0,\pm 1,\pm 2,\cdots$$
although $\mathrm{Arg}\ (-1-i)\ne\frac{5\pi}{4}$.

In exponential form, $-1-i$ may be written in exponential form as
$$-1-i=\sqrt{2}\exp\left[i\left(-\frac{3\pi}{4}\right)\right]$$
This is one of infinitely many possibilities for the exponential form of $-1-i$.
$$-1-i=\sqrt{2}\exp\left[i\left(-\frac{3\pi}{4}+2n\pi\right)\right]\ (n=0,\pm 1,\pm2,\cdots)$$

The equation of a circle centered at $z_0$ with radius $R$ is given by
\begin{equation}
\label{eq:circle}
|z-z_0|=R
\end{equation}
In exponential form, the above equation of a circle can be written as
\begin{equation}
\label{eq:circle2}
z=z_0+Re^{i\theta}
\end{equation}
where $0\leq\theta\leq 2\pi$.

Let $z_1=r_1e^{i\theta}$ and $z_2=r_2e^{i\theta_2}$. Then
\begin{align}
\label{eq:product2}
z_1z_2&=r_1r_2e^{i(\theta_1+\theta_2)}\\
\label{eq:quotient}
\frac{z_1}{z_2}&=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)}
\end{align}
The product and the quotient in exponential form imply the identities.
\begin{align}
\label{eq:arg}
\arg(z_1z_2)&=\arg z_1+\arg z_2\\
\label{eq:arg2}
\arg\left(\frac{z_1}{z_2}\right)&=\arg z_1-\arg z_2
\end{align}
These identities are not necessarily true when $\arg$ is replaced by $\mathrm{Arg}$ as seen in the following example.

Example. Let $z_1=-1$ and $z_2=i$. Then $\mathrm{Arg}\ z_1=\pi$ and $\mathrm{Arg}\ z_2=\frac{\pi}{2}$.
$$\mathrm{Arg}(z_1z_2)=\mathrm{Arg}(-i)=-\frac{\pi}{2}$$
and
$$\mathrm{Arg}\ z_1+\mathrm{Arg}\ z_2=\pi+\frac{\pi}{2}=\frac{3\pi}{2}$$
Thus,
$$\mathrm{Arg}(z_1z_2)\ne \mathrm{Arg}\ z_1+\mathrm{Arg}\ z_2$$

Example. Find the principal argument $\mathrm{Arg}\ z$ when $z=\frac{-2}{1+\sqrt{3}i}$.

Solution.
\begin{align*}
\arg z&=\arg\left(\frac{-2}{1+\sqrt{3}i}\right)\\
&=\arg(-2)-\arg(1+\sqrt{3}i)
\end{align*}
$\mathrm{Arg}(-2)=\pi$ and $\mathrm{Arg}(1+\sqrt{3}i)=\frac{\pi}{3}$. One value of $\arg z$ is $\pi-\frac{\pi}{3}=\frac{2\pi}{3}$. Since $-\pi<\frac{2\pi}{3}\leq\pi$, $\mathrm{Arg}\left(\frac{-2}{1+\sqrt{3}i}\right)=\frac{2\pi}{3}$.

Solution 2. $\frac{-2}{1+\sqrt{3}i}=\frac{-1+\sqrt{3}i}{2}$. So, $\mathrm{Arg}\left(\frac{-2}{1+\sqrt{3}i}\right)=\frac{3\pi}{2}$.

De Moivre's formula
\begin{equation}
\label{eq:demoivre}
(\cos\theta+i\sin\theta)^n = \cos n\theta+i\sin n\theta,\ n=0,\pm 1,\pm2,\cdots
\end{equation}
can be easily proved using the Euler's identity.

Example. Write $(\sqrt{3}+i)^7$ in the form $a+ib$.

Solution. $\sqrt{3}+i=2e^{i\frac{\pi}{6}}$.
\begin{align*}
(\sqrt{3}+i)^7&=(2e^{i\frac{\pi}{6}})^7\\
&=2^7e^{i\frac{7\pi}{6}}\\
&=(2^6e^{i\pi})(2e^{i\frac{\pi}{6}})\\
&=-64(\sqrt{3}+i).
\end{align*}
The exponential form of a complex number can be used to find the $n$-th root of a complex number. Let us consider the complex equation
\begin{equation}
\label{eq:nthroot}
z^n=z_0
\end{equation}
Plugging $z=re^{i\theta}$ and $z_0=r_0e^{i\theta_0}$ into the complex equation results in
$$r^ne^{in\theta}=r_0e^{i\theta_0}$$
Comparing the LHS and the RHS, we obtain
$$r^n=r_0\ \mbox{and}\ n\theta=\theta_0+2k\pi,\ k=0,1,2,\cdots,n-1$$
that is
\begin{equation}
\label{eq:nthroot2}
r=\root n\of{r_0}\ \mbox{and}\ \theta=\frac{\theta_0}{n}+\frac{2k\pi}{n},\ k=0,1,2,\cdots,n-1
\end{equation}
Hence, the $n$-th root of $z_0$ is given by
\begin{equation}
\label{eq:nthroot3}
z=\root n\of{r_0}\exp\left[\frac{i(\theta_0+2k\pi)}{n}\right],\ k=0,1,2,\cdots,n-1
\end{equation}

Example. Find the $n$-th root of unity, i.e. solve the equation
\begin{equation}
\label{eq:nthroot4}
z^n=1
\end{equation}

Solution. It follows from the above $n$-root of $z_0$ formula with $z_0=1$ that
the $n$-th root of unity is given by
\begin{equation}
\label{eq:nthroot5}
\omega_n^k=\exp\left(i\frac{2k\pi}{n}\right),\ k=0,1,2,\cdots,n-1
\end{equation}
The set of the $n$-th root of unity
$$E_n=\{1,\omega_n^1,\omega_n^2,\cdots,\omega_n^{n-1}\}$$
is a cyclic group of  order $n$, which is isomorphic to $\mathbb{Z}_n$. The following figures show beautiful geometric shapes formed by $E_3, E_4, E_5, E_6$.

$E_3$

 

$E_4$

$E_5$

$E_6$