# Polar representation and complex exponential

Polar coordinates use a fixed point called the pole, and a fixed direction, the initial line to describe the position of other points. In the Cartesian plane, the pole is the origin and the initial line is the positive $$x$$-axis.

The polar coordinate of a point $$A$$ is written $$(r, \theta)$$. It consists of the radius $$r\ge 0$$, which is distance from the pole $$O$$, and the angle $$\theta$$ that the ray $$OA$$ makes with the initial line.

The polar coordinates $$(r, \theta)$$ is equivalent to the Cartesian coordinates $$$(r\cos \theta, r\sin \theta).$$$

The $$r$$-coordinate is unique, but the $$\theta$$-coordinate can only be given up to adding a multiple of $$2\pi$$. Turning through $$2\pi$$ keeps a point unchanged.

Every complex number can be written as $$$ z = r(\cos\theta +i \sin\theta)$$$ with $$r\ge0$$. $$r$$ is the modulus and $$\theta$$ is the argument. This is the polar representation of $$z$$, and is sometimes called the modulus-argument form. It corresponds to the polar coordinates $$(r, \theta)$$, and is an alternative to the Cartesian representation $$z=x+iy$$.

The polar angle $$\theta$$ is the argument of $$z$$. It is characterised by $$$\cos\theta = \frac{x}{\sqrt{x^2+y^2}},\quad \sin\theta = \frac{y}{\sqrt{x^2+y^2}}.$$$

The argument, denoted $$\arg z$$, is unique up to multiples of $$2\pi$$. The principal argument, denoted $$\Arg z$$, is the argument in the interval $$(-\pi, \pi ]$$.

We often use the exponential notation for the polar representation of complex numbers. If we have a number $$z$$ with modulus $$r$$ and argument $$\theta$$, we can write $$$ z = r e^{i\theta}$$$

Exponential representations of some complex numbers are: $$$ 1 = e^{i0},\; i = e^{i\pi/2},\; -1 = e^{i\pi},\; -i = e^{-i\pi/2},\; 1+i = \sqrt{2}e^{i\pi/4}$$$

In particular, $$e^{i\pi} + 1 = 0$$ is known as Euler identity.

Conjugation, and multiplicative operations (including inverse, division, root extraction, etc.) are simple using the polar representation.

The **complex exponential** $$e^{i\theta} = (\cos\theta +i \sin\theta)$$ has the usual properties of the exponential. For instance, $$$e^{i(\theta_1+\theta_2)} = e^{i\theta_1} e^{i\theta_2}. $$$ This arises from the trigonometric identities \begin{gather*} \cos(\theta_1+\theta_2) = \cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2,\\ \sin(\theta_1+\theta_2) = \cos\theta_1\sin\theta_2+\sin\theta_1\cos\theta_2. \end{gather*}

If $$z=re^{i\theta}$$, using the properties of the exponential, we find $$$\zo = r e^{-i\theta},\quad z^{-1}= r^{-1}e^{-i\theta},\quad z_1z_2 = r_1r_2 e^{i(\theta_1+\theta_2)}.$$$

This is equivalent to \begin{gather*} \vert\zo\vert = \vert z\vert,\; \arg\zo = -\arg z, \; \vert z^{-1}\vert = \vert z\vert^{-1},\; \arg z^{-1} = -\arg z\\ \vert z_1z_2\vert = \vert z_1\vert\,\vert z_2\vert,\; \arg(z_1z_2) = \arg z_1 + \arg z_2 \end{gather*}

De Moivre's formula states that for all $$\theta\in\R$$ and all integers $$n$$, we have $$$\cos(n\theta)+i\sin(n\theta) = (\cos\theta + i \sin\theta)^n. $$$

The formula can be seen using the property of the exponential $$$ e^{i n\theta} = (e^{i\theta})^n.$$$

We can also use proof by induction and trigonometric formulae.

Sometimes the expression $$\cos\theta + i \sin\theta$$ is abbreviated to cis.

De Moivre's formula and the binomial formula are often used to express polynomials in $$\cos\th$$ or $$\sin\th$$ of degree $$n$$ as a linear combination of $$\cos(k\th)$$ and $$\sin(k\th)$$ (for $$0\le k\le n$$) and vice versa.

To express $$\cos(n\theta)$$ and $$\sin(n\theta)$$ in powers of $$\cos\theta$$ and $$\sin\theta$$, we use \begin{gather*} \cos(n\th) = \Real(e^{in\th}) = \Real\big((\cos\th + i\sin\th)^n\big),\\ \sin(n\th) = \Imag(e^{in\th}) = \Imag\big((\cos\th + i\sin\th)^n\big). \end{gather*}

Writing $$c=\cos\theta$$ and $$s=\sin\theta$$, we can expand $$\cos(4\th)$$. \begin{gather*}\cos(4\theta) = \Real(c+is)^4 = c^4 + 6i^2c^2s^2+i^4s^4\\ =c^4-6c^2s^2+s^4. \end{gather*}

To express polynomials of $$\sin\theta$$ and $$\cos\theta$$ using sums of $$\cos(n\theta)$$ and $$\sin(n\theta)$$, we use $$$\cos^n(\th) = \frac{(e^{i\th}+e^{-i\th})^n}{2^n} ,\quad \sin^n(\th) = \frac{(e^{i\th}-e^{-i\th})^n}{(2i)^n} $$$

We can express $$\sin^2\th$$ in terms of $$\cos(2\th)$$. $$$ \sin^2\theta = -\frac{(e^{i\th}-e^{-i\th})^2}{4} =-\frac{e^{i2\th}-2 + e^{-i2\th}}{4} = \frac{1-\cos(2\th)}{2}. $$$

The solutions of $$z^n=1$$, called the $$n^{\text{th}}$$ roots of unity, are given by $$$z_k = e^{i 2k\pi/n},\quad k\in\{0,1,\dots,n-1\}. $$$ To see this, we note that $$(z_k)^n = e^{i 2kn\pi} = 1$$. Since the equation has $$n$$ roots, we have them all.

The $$n^{\text{th}}$$ roots of unity sum to $$0$$. They are often denoted $$\omega$$. All except $$1$$ satisfy the equation $$$\omega^{n-1} + \omega^{n-2} + \dots + \omega + 1 =\frac{\omega^n - 1}{\omega -1} = 0.$$$

The roots of unity lie at the vertices of a regular polygon.

More generally, the solutions of $$ z^n = Z = R e^{i\Theta}$$, called the $$n^{\text{th}}$$ roots of a complex number, are given by $$$ z_k = R^{1/n}e^{i(\Theta/n + 2k\pi/n)},\quad k\in\{0,1,\dots,n-1\}. $$$