# Polar form and graphical representation of complex numbers

Being that ${\displaystyle \mathbb {C} \backsimeq \mathbb {R} ^{2}\ ,}$ we can think at representing complex numbers on a plane similar to the Cartesian one. The complex plane is known as Argand-Gauss plane.

This plane as the real part of the chosen complex number as first coordinate, and the imaginary part as the second one.

From this graphical representation it is easily derived the polar form of a complex number. In particular, if ${\displaystyle z=a+ib\,\in \mathbb {C} \ ,}$ we set:

${\displaystyle {\begin{cases}a=Re(z)=r\cos \theta \\b=Im(z)=r\sin \theta \end{cases}}}$
where
${\displaystyle {\begin{cases}r^{2}=a^{2}+b^{2}\equiv \mid z\mid ^{2}\ \\\theta =\arctan \left({\frac {b}{a}}\right)+2n\pi ,\,n\in \mathbb {Z} \end{cases}}}$

Actually, in order to well-define a complex number in polar coordinates we have to choose an interval of definition for the angle ${\displaystyle \theta }$, for instance let ${\displaystyle \theta }$ vary in ${\displaystyle [-\pi ;\pi ].}$ The angle ${\displaystyle \theta }$ is called the argument of the complex number ${\displaystyle z}$.

Theorem (Euler's Formula)

We see that every complex number in trigonometric form can be rewritten as:

${\displaystyle z=r(\cos \theta +i\sin \theta )=re^{i\theta }\ .}$

We provide two alternative proofs of the Euler's Formula.

Proof

We consider the series ${\displaystyle \sum _{n=0}^{\infty }{\frac {(i\theta )^{n}}{n!}}}$.

This is absolutely convergent since the series of the modules:

${\displaystyle \sum _{n=0}^{\infty }\mid {\frac {(i\theta )^{n}}{n!}}\mid =\sum _{n=0}^{\infty }\mid {\frac {\theta ^{n}}{n!}}\mid }$

converges in ${\displaystyle \mathbb {R} .}$ Moreover, it is known that the sum of the series is ${\displaystyle e^{i\theta },}$ so:

${\displaystyle e^{i\theta }=\sum _{n=0}^{\infty }{\frac {(i\theta )^{n}}{n!}}\ .}$

Now, we can rewrite the series by separating the terms in an even and odd position as follows:

${\displaystyle e^{i\theta }=\sum _{n=0}^{\infty }{\frac {(i\theta )^{2n}}{(2n)!}}+\sum _{n=0}^{\infty }{\frac {(i\theta )^{(2n+1)}}{(2n+1)!}}=}$

${\displaystyle =\sum _{n=0}^{\infty }(-1)^{n}{\frac {\theta ^{2n}}{(2n)!}}+i\sum _{n=0}^{\infty }(-1)^{n}{\frac {\theta ^{2n+1}}{(2n+1)!}}\ .}$

The two series are both power convergent series (it has been proven in the course of Multivariate Calculus). In fact ${\displaystyle \sin x}$ and ${\displaystyle \cos x}$ are two examples of functions of class ${\displaystyle C^{\infty }}$ that satisfy the right conditions on the growth of the ${\displaystyle n}$-th order derivative, so that we can write them as sum of power series. In particular, we have that:

${\displaystyle \cos(\theta )=\sum _{n=0}^{\infty }(-1)^{n}{\frac {\theta ^{2n}}{(2n)!}}\ ,}$

${\displaystyle \sin(\theta )=\sum _{n=0}^{\infty }(-1)^{n}{\frac {\theta ^{2n+1}}{(2n+1)!}}\ .}$

And so we achieve that:

${\displaystyle e^{i\theta }=\cos(\theta )+i\sin(\theta )\ .}$

Proof

Given the exponential function ${\displaystyle e^{i\theta }}$, we expext that ${\displaystyle \exists \,A,B:\mathbb {R} \rightarrow \mathbb {C} \,\colon \,e^{i\theta }=A(\theta )+iB(\theta ).}$

We want to prove that ${\displaystyle A(\theta )=\cos \theta \ }$ and ${\displaystyle B(\theta )=\sin \theta .}$

We take the derivative of ${\displaystyle e^{i\theta }}$ with respect to ${\displaystyle \theta }$:

${\displaystyle ie^{i\theta }=A^{\prime }(\theta )+iB^{\prime }(\theta )\ .}$

By taking a second derivative, we have that:

${\displaystyle -e^{i\theta }=A^{\prime \prime }(\theta )+iB^{\prime \prime }(\theta ).}$

We comare the two equations we have with ${\displaystyle e^{i\theta }=A(\theta )+iB(\theta )}$ and we obtain the following system of differential equations:

${\displaystyle {\begin{cases}A^{\prime \prime }(\theta )=-A(\theta )\\B^{\prime \prime }(\theta )=-B(\theta )\\A(0)=1&A^{\prime }(0)=0\\B(0)=0&B^{\prime }(0)=1\end{cases}}}$

Where we have imposed the initial conditions for ${\displaystyle A,\,A^{\prime },\,B,\,B^{\prime }}$.

By solving these equations we have that:

${\displaystyle A(\theta )=\cos \theta \quad {\mbox{e}}\quad B(\theta )=\sin \theta \ .}$

#### Example 1

For the imaginary unit ${\displaystyle z=i\ ,}$ we have that:

${\displaystyle z=i=(0,1)\Rightarrow z=e^{i{\frac {\pi }{2}}}\ .}$