Polar Form Of Complex Numbers

Complex numbers in polar form

Let P(x, y) be a point representing a non-zero complex number $z = x + iy$ in the Argand diagram. If OP makes an angle $\theta$ with the positive x-axis, then $z = r(\cos\theta + i\sin\theta)$ is called the polar form of $z$, where $r = |z| = \sqrt{x^2 + y^2}$, and $\theta = \arctan\left(\frac{y}{x}\right)$ is the argument of $z$, written as arg($z$).

Represents the Cartesian coordinates (x, y) and the polar coordinates $(r, \theta)$ of the point P(x, y).

From Figure

$x = r\cos\theta$

$y = r\sin\theta$

$\tan\theta = \frac{y}{x}$

Product and quotient identities for complex numbers in polar form:

1. arg($z_1z_2$) = arg($z_1$) + arg($z_2$)
2. arg($\frac{z_1}{z_2}$) = arg($z_1$) - arg($z_2$)
3. $z_1z_2 = r_1r_2[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]$
4. $\frac{z_1}{z_2} = \frac{r_1}{r_2}[\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)]$

Example 1

Express $z = \frac{7}{4 - 3i}$ in polar form and represent it on the Argand diagram.

Solution:

$z = \frac{7}{4 - 3i} = \frac{7(4 + 3i)}{(4 - 3i)(4 + 3i)} = \frac{28 + 21i}{16 + 9} = \frac{28 + 21i}{25} = \frac{28}{25} + \frac{21}{25}i$

$r = |z| = \sqrt{\left(\frac{28}{25}\right)^2 + \left(\frac{21}{25}\right)^2} = \sqrt{\frac{784 + 441}{625}} = \sqrt{\frac{1225}{625}} = \frac{35}{25} = \frac{7}{5}$

$\theta = \arctan\left(\frac{21/25}{28/25}\right) = \arctan\left(\frac{21}{28}\right) = \arctan\left(\frac{3}{4}\right) \approx 36.87^\circ$

The polar form of $z$ is $z = \frac{7}{5}(\cos(36.87^\circ) + i\sin(36.87^\circ))$.

Example 2

Let $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$ be two complex numbers in polar form. Show that $z_1z_2 = r_1r_2[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]$.

Solution:

$z_1z_2 = r_1(\cos\theta_1 + i\sin\theta_1) \times r_2(\cos\theta_2 + i\sin\theta_2)$

$= r_1r_2(\cos\theta_1\cos\theta_2 + i\cos\theta_1\sin\theta_2 + i\sin\theta_1\cos\theta_2 + i^2\sin\theta_1\sin\theta_2)$

$= r_1r_2(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2))$

Using the trigonometric identities for the sum of angles:

$z_1z_2 = r_1r_2[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]$

Example 3

If $z_1 = 3(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3})$ and $z_2 = 12(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4})$, find:

(a) arg($z_1z_2$) (b) arg($\frac{z_1}{z_2}$) (c) $z_1z_2$ (d) $\frac{z_1}{z_2}$

Solution:

(a) arg($z_1z_2$) = arg($z_1$) + arg($z_2$) = $\frac{2\pi}{3} + \frac{\pi}{4} = \frac{8\pi + 3\pi}{12} = \frac{11\pi}{12}$

(b) arg($\frac{z_1}{z_2}$) = arg($z_1$) - arg($z_2$) = $\frac{2\pi}{3} - \frac{\pi}{4} = \frac{8\pi - 3\pi}{12} = \frac{5\pi}{12}$

(c) $z_1z_2 = (3)(12)[\cos(\frac{11\pi}{12}) + i\sin(\frac{11\pi}{12})] = 36[\cos(\frac{11\pi}{12}) + i\sin(\frac{11\pi}{12})]$

(d) $\frac{z_1}{z_2} = \frac{3}{12}[\cos(\frac{5\pi}{12}) + i\sin(\frac{5\pi}{12})] = \frac{1}{4}[\cos(\frac{5\pi}{12}) + i\sin(\frac{5\pi}{12})]$

Example 4

Simplify:

(a) $\frac{\cos4\theta + i\sin4\theta}{\sin\theta + i\cos\theta}$ (b) $\frac{(\cos\theta + i\sin\theta)(\cos2\theta + i\sin2\theta)}{(\cos2\theta + i\sin2\theta)(\cos3\theta + i\sin3\theta)}$

Solution:

(a) $\frac{\cos4\theta + i\sin4\theta}{\sin\theta + i\cos\theta} = \frac{\cos4\theta + i\sin4\theta}{i(\cos\theta - i\sin\theta)} = \frac{\cos4\theta + i\sin4\theta}{i(\cos(-\theta) + i\sin(-\theta))}$

Multiplying the numerator and denominator by -i:

$\frac{-i(\cos4\theta + i\sin4\theta)}{1(\cos(-\theta) + i\sin(-\theta))}$

$-i\cos4\theta + \sin4\theta$ * $\cos\theta - i\sin\theta$

$\sin4\theta\cos\theta - i\sin4\theta\sin\theta$

Loci in complex numbers

A locus of a complex number is the path traced out by a point on an Argand diagram satisfying a given condition. It is usually defined in terms of modulus or argument.

For instance, $|z - a| = k$, where $k$ is any positive real number, represents a circle. If $z = x + iy$, then:

$|x + iy - a| = k$

$|(x - a) + iy| = k$

$\sqrt{(x - a)^2 + y^2} = k$

$(x - a)^2 + y^2 = k^2$

This is the equation of a circle with center at $(a, 0)$ and radius $k$.

Similarly, an equation of the form arg$\left(\frac{z - a}{z - b}\right) = \mu$, where $\mu$ is an angle, defines the locus of a complex number in terms of an argument.

From the relation:

arg$\left(\frac{z - a}{z - b}\right)$ = arg($z - a$) - arg($z - b$) = $\mu$

Substituting $z = x + iy$:

arg$(x - a + iy)$ - arg$(x - b + iy)$ = $\mu$

Let $\alpha = \arctan\left(\frac{y}{x - a}\right)$ and $\beta = \arctan\left(\frac{y}{x - b}\right)$. Then:

$\alpha - \beta = \mu$

$\tan(\alpha - \beta) = \tan(\mu)$

$\frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} = \tan\mu$

Substituting the expressions for $\alpha$ and $\beta$:

$\frac{\frac{y}{x - a} - \frac{y}{x - b}}{1 + \frac{y}{x - a}\frac{y}{x - b}} = \tan\mu$

Let $c = \tan\mu$. Then:

$\frac{y(x - b) - y(x - a)}{(x - a)(x - b) + y^2} = c$

$y(x - b - x + a) = c(x^2 - ax - bx + ab + y^2)$

$y(a - b) = c(x^2 - (a + b)x + ab + y^2)$

$ay - by = cx^2 - c(a + b)x + cab + cy^2$

$cx^2 + cy^2 - c(a + b)x - (a - b)y + cab = 0$

This is the general equation of a circle.

Example 5

Show on an Argand diagram the region satisfying a complex number $z$ such that $|z + 2 - i| = 5$.

Solution:

Let $z = x + iy$. Then $|x + iy + 2 - i| = 5$.

$|(x + 2) + i(y - 1)| = 5$

$\sqrt{(x + 2)^2 + (y - 1)^2} = 5$

$(x + 2)^2 + (y - 1)^2 = 25$

The region satisfying $z$ is a circle with radius 5 and center at the point (-2, 1).

Example 6

Determine the locus defined by arg$\left(\frac{z - 1}{z + 1}\right) = \frac{\pi}{4}$.

Solution:

Let $z = x + iy$. Then:

arg$\left(\frac{x + iy - 1}{x + iy + 1}\right) = \frac{\pi}{4}$

arg$\left(\frac{(x - 1) + iy}{(x + 1) + iy}\right) = \frac{\pi}{4}$

Rationalizing the denominator:

arg$\left(\frac{((x - 1) + iy)((x + 1) - iy)}{((x + 1) + iy)((x + 1) - iy)}\right) = \frac{\pi}{4}$

arg$\left(\frac{x^2 - 1 - ixy + iy + ixy + y^2}{(x + 1)^2 + y^2}\right) = \frac{\pi}{4}$

arg$\left(\frac{x^2 + y^2 - 1 + 2iy}{(x + 1)^2 + y^2}\right) = \frac{\pi}{4}$

arg$\left(\frac{x^2 + y^2 - 1}{(x + 1)^2 + y^2} + i\frac{2y}{(x + 1)^2 + y^2}\right) = \frac{\pi}{4}$

Using the definition of the argument:

$\tan^{-1}\left(\frac{\frac{2y}{(x + 1)^2 + y^2}}{\frac{x^2 + y^2 - 1}{(x + 1)^2 + y^2}}\right) = \frac{\pi}{4}$

$\tan^{-1}\left(\frac{2y}{x^2 + y^2 - 1}\right) = \frac{\pi}{4}$

$\frac{2y}{x^2 + y^2 - 1} = \tan\frac{\pi}{4} = 1$

$2y = x^2 + y^2 - 1$

$x^2 + y^2 - 2y - 1 = 0$

$x^2 + (y - 1)^2 - 1 - 1 = 0$

$x^2 + (y - 1)^2 = 2$

This is a circle with center (0, 1) and radius $\sqrt{2}$.

Example 7

Show that the locus defined by $\left|\frac{z - 3 + 2i}{z + 3 - 2i}\right| = 5$ is a circle, and find its center and radius.

Solution:

Let $z = x + iy$. Then:

$\left|\frac{x + iy - 3 + 2i}{x + iy + 3 - 2i}\right| = 5$

$\left|\frac{(x - 3) + i(y + 2)}{(x + 3) + i(y - 2)}\right| = 5$

$\frac{|(x - 3) + i(y + 2)|}{|(x + 3) + i(y - 2)|} = 5$

$\sqrt{(x - 3)^2 + (y + 2)^2} = 5\sqrt{(x + 3)^2 + (y - 2)^2}$

$(x - 3)^2 + (y + 2)^2 = 25[(x + 3)^2 + (y - 2)^2]$

$x^2 - 6x + 9 + y^2 + 4y + 4 = 25(x^2 + 6x + 9 + y^2 - 4y + 4)$

$x^2 - 6x + y^2 + 4y + 13 = 25x^2 + 150x + 225 + 25y^2 - 100y + 100$

$24x^2 + 24y^2 + 156x - 104y + 312 = 0$

$x^2 + y^2 + \frac{13}{2}x - \frac{13}{3}y + 13 = 0$

Completing the square:

$(x + \frac{13}{4})^2 - \frac{169}{16} + (y - \frac{13}{6})^2 - \frac{169}{36} + 13 = 0$

$(x + \frac{13}{4})^2 + (y - \frac{13}{6})^2 = \frac{169}{16} + \frac{169}{36} - 13$

$(x + \frac{13}{4})^2 + (y - \frac{13}{6})^2 = \frac{1521 + 676 - 702}{144}$

$(x + \frac{13}{4})^2 + (y - \frac{13}{6})^2 = \frac{1495}{144}$

This is a circle with center $(-\frac{13}{4}, \frac{13}{6})$ and radius $\sqrt{\frac{1495}{144}} = \frac{\sqrt{1495}}{12}$.