Complex Numbers in Polar Form

Let us represent the complex number \( z = a + bi \), where \( i = \sqrt{-1} \), in the complex plane (a system of rectangular axes) such that the real part \( a \) is the coordinate on the horizontal axis and the imaginary part \( b \) is the coordinate on the vertical axis, as shown below.

Where:

\( r = \sqrt{a^2 + b^2} \), where \( r \) is called the modulus of \( z \).
\( \tan(\theta) = \dfrac{b}{a} \), such that \( 0 \le \theta < 2\pi \), where \( \theta \) is called the argument of \( z \).
The reference angle \( \theta_r \) to the argument \( \theta \) is the acute angle made by the horizontal axis and the terminal side of \( \theta \), defined as: \[ \theta_r = \tan^{-1}\left|\dfrac{b}{a}\right| \]

If \( r \) and \( \theta \) are known, then \( a \) and \( b \) are given by:

\[ a = r \cos(\theta) \quad \text{and} \quad b = r \sin(\theta) \]

Thus, the complex number \( z = a + bi \) may be expressed in polar form involving \( r \) and \( \theta \) as:

\[ z = r (\cos(\theta) + i \sin(\theta)) \]

Example 1

a) Plot the complex number \( z = 2\sqrt{3} - 2i \) on the complex plane and write it in polar form.

View Solution

plot of complex number z on complex plane — Figure 2. Plot of \( z = 2\sqrt{3} - 2i \)

Calculate the modulus \( r \):

\[ r = \sqrt{(2\sqrt{3})^2 + (-2)^2} = \sqrt{12 + 4} = \sqrt{16} = 4 \]

Find the reference angle \( \theta_r \):

\[ \theta_r = \tan^{-1}\left|\dfrac{b}{a}\right| = \tan^{-1}\left|\dfrac{-2}{2\sqrt{3}}\right| = \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right) = \tan^{-1}\left(\dfrac{\sqrt{3}}{3}\right) = \dfrac{\pi}{6} \]

Since the real part of \( z \) is positive and its imaginary part is negative, the terminal side of the argument \( \theta \) lies in quadrant IV. Therefore:

\[ \theta = 2\pi - \theta_r = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6} \]

Thus, \( z \) in polar form is:

\[ z = 4 \left(\cos\left(\dfrac{11\pi}{6}\right) + i \sin\left(\dfrac{11\pi}{6}\right)\right) \]

Example 2

Write the complex number \( z = \sqrt{2}\left(\cos\left(\dfrac{3\pi}{4}\right) + i \sin\left(\dfrac{3\pi}{4}\right)\right) \) in standard form.

View Solution

\[ z = \sqrt{2}\left(\cos\left(\dfrac{3\pi}{4}\right) + i \sin\left(\dfrac{3\pi}{4}\right)\right) \] \[ = \sqrt{2}\left(-\dfrac{\sqrt{2}}{2} + i \dfrac{\sqrt{2}}{2}\right) \] \[ = -\dfrac{2}{2} + i \dfrac{2}{2} = -1 + i \]

Multiplication and Division of Complex Numbers in Polar Forms

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 complex numbers in polar form. Using sum and difference trigonometric formulas for sine and cosine, their product and quotient are given by:

\[ z_1 z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)) \] \[ \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2} (\cos(\theta_1 - \theta_2) + i \sin(\theta_1 - \theta_2)) \]

Example 3

Given \( z_1 = 3 \left(\cos\left(\dfrac{5\pi}{4}\right) + i \sin\left(\dfrac{5\pi}{4}\right)\right) \) and \( z_2 = \cos\left(\dfrac{\pi}{4}\right) + i \sin\left(\dfrac{\pi}{4}\right) \), find \( z_1 z_2 \) and \( \dfrac{z_1}{z_2} \).

View Solution

Product (\( z_1 z_2 \)):

\[ z_1 z_2 = 3 \left(\cos\left(\dfrac{5\pi}{4}\right) + i \sin\left(\dfrac{5\pi}{4}\right)\right) \left(\cos\left(\dfrac{\pi}{4}\right) + i \sin\left(\dfrac{\pi}{4}\right)\right) \] \[ = 3 \left(\cos\left(\dfrac{5\pi}{4} + \dfrac{\pi}{4}\right) + i \sin\left(\dfrac{5\pi}{4} + \dfrac{\pi}{4}\right)\right) \] \[ = 3 \left(\cos\left(\dfrac{6\pi}{4}\right) + i \sin\left(\dfrac{6\pi}{4}\right)\right) = 3 \left(\cos\left(\dfrac{3\pi}{2}\right) + i \sin\left(\dfrac{3\pi}{2}\right)\right) = 3(0 - i) = -3i \]

Quotient (\( \dfrac{z_1}{z_2} \)):

\[ \dfrac{z_1}{z_2} = \dfrac{3 \left(\cos\left(\dfrac{5\pi}{4}\right) + i \sin\left(\dfrac{5\pi}{4}\right)\right)}{\cos\left(\dfrac{\pi}{4}\right) + i \sin\left(\dfrac{\pi}{4}\right)} \] \[ = 3 \left(\cos\left(\dfrac{5\pi}{4} - \dfrac{\pi}{4}\right) + i \sin\left(\dfrac{5\pi}{4} - \dfrac{\pi}{4}\right)\right) \] \[ = 3 \left(\cos\left(\dfrac{4\pi}{4}\right) + i \sin\left(\dfrac{4\pi}{4}\right)\right) = 3 (\cos(\pi) + i \sin(\pi)) = 3(-1 + 0) = -3 \]

Solutions to the Practice Questions

Solutions to Question 1

plot of complex numbers z_1, z_2, z_3 and z_4 on the complex plane — Figure 3. Plots for parts a, b, c, and d

\( z_1 = i \):
\( r = \sqrt{0^2 + 1^2} = 1 \). The point is at \((0, 1)\) on the positive vertical axis, so \( \theta = \frac{\pi}{2} \).

Polar form: \( z_1 = \cos\left(\dfrac{\pi}{2}\right) + i \sin\left(\dfrac{\pi}{2}\right) \)
\( z_2 = 2 \):
\( r = \sqrt{2^2 + 0^2} = 2 \). The point is at \((2, 0)\) on the positive horizontal axis, so \( \theta = 0 \).

Polar form: \( z_2 = 2\left(\cos(0) + i \sin(0)\right) \)
\( z_3 = -2i \):
\( r = 2 \), \( \theta = \dfrac{3\pi}{2} \).

Polar form: \( z_3 = 2\left(\cos\left(\dfrac{3\pi}{2}\right) + i \sin\left(\dfrac{3\pi}{2}\right)\right) \)
\( z_4 = -3 \):
\( r = 3 \), \( \theta = \pi \).

Polar form: \( z_4 = 3\left(\cos(\pi) + i \sin(\pi)\right) \)
\( z_5 = \sqrt{3} + i \):
\( r = \sqrt{(\sqrt{3})^2 + 1^2} = 2 \). Reference angle \( \theta_r = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \). Both parts are positive (Quadrant I), so \( \theta = \frac{\pi}{6} \).

Figure 4. Plot of \( z_5 \)

Polar form: \( z_5 = 2\left(\cos\left(\dfrac{\pi}{6}\right) + i \sin\left(\dfrac{\pi}{6}\right)\right) \)
\( z_6 = -3 - 3\sqrt{3}i \):
\( r = \sqrt{(-3)^2 + (-3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6 \). Reference angle \( \theta_r = \tan^{-1}\left(\frac{3\sqrt{3}}{3}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \). Both parts are negative (Quadrant III), so \( \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} \).

Figure 5. Plot of \( z_6 \)

Polar form: \( z_6 = 6\left(\cos\left(\dfrac{4\pi}{3}\right) + i \sin\left(\dfrac{4\pi}{3}\right)\right) \)
\( z_7 = -5 + 5i \):
\( r = \sqrt{(-5)^2 + 5^2} = \sqrt{50} = 5\sqrt{2} \). Reference angle \( \theta_r = \tan^{-1}(1) = \frac{\pi}{4} \). Real part negative, imaginary part positive (Quadrant II), so \( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \).

Figure 6. Plot of \( z_7 \)

Polar form: \( z_7 = 5\sqrt{2}\left(\cos\left(\dfrac{3\pi}{4}\right) + i \sin\left(\dfrac{3\pi}{4}\right)\right) \)

Solutions to Question 2

\( z_5 z_6 \):
Using the product formula:
\[ z_5 z_6 = 2\left(\cos\left(\dfrac{\pi}{6}\right) + i \sin\left(\dfrac{\pi}{6}\right)\right) \cdot 6\left(\cos\left(\dfrac{4\pi}{3}\right) + i \sin\left(\dfrac{4\pi}{3}\right)\right) \] \[ = (2 \times 6)\left(\cos\left(\dfrac{\pi}{6} + \dfrac{4\pi}{3}\right) + i \sin\left(\dfrac{\pi}{6} + \dfrac{4\pi}{3}\right)\right) \] \[ = 12\left(\cos\left(\dfrac{9\pi}{6}\right) + i \sin\left(\dfrac{9\pi}{6}\right)\right) = 12\left(\cos\left(\dfrac{3\pi}{2}\right) + i \sin\left(\dfrac{3\pi}{2}\right)\right) = 12(0 - i) = -12i \]
\( \dfrac{z_6 z_7}{z_5} \):
Combine multiplication in the numerator and division:
\[ \dfrac{z_6 z_7}{z_5} = \dfrac{6\left(\cos\left(\dfrac{4\pi}{3}\right) + i \sin\left(\dfrac{4\pi}{3}\right)\right) \cdot 5\sqrt{2}\left(\cos\left(\dfrac{3\pi}{4}\right) + i \sin\left(\dfrac{3\pi}{4}\right)\right)}{2\left(\cos\left(\dfrac{\pi}{6}\right) + i \sin\left(\dfrac{\pi}{6}\right)\right)} \] \[ = \dfrac{30\sqrt{2}\left(\cos\left(\dfrac{4\pi}{3} + \dfrac{3\pi}{4}\right) + i \sin\left(\dfrac{4\pi}{3} + \dfrac{3\pi}{4}\right)\right)}{2\left(\cos\left(\dfrac{\pi}{6}\right) + i \sin\left(\dfrac{\pi}{6}\right)\right)} \] \[ = 15\sqrt{2}\left(\cos\left(\dfrac{4\pi}{3} + \dfrac{3\pi}{4} - \dfrac{\pi}{6}\right) + i \sin\left(\dfrac{4\pi}{3} + \dfrac{3\pi}{4} - \dfrac{\pi}{6}\right)\right) \]
Finding a common denominator for the angles (\(12\)):
\[ \dfrac{16\pi}{12} + \dfrac{9\pi}{12} - \dfrac{2\pi}{12} = \dfrac{23\pi}{12} \]
Result: \( 15\sqrt{2}\left(\cos\left(\dfrac{23\pi}{12}\right) + i \sin\left(\dfrac{23\pi}{12}\right)\right) \)

Complex Numbers in Polar Form