Complex Numbers in Exponential Form

Complex numbers can be written in exponential form. The multiplication, division, and power of complex numbers in exponential form are explained through examples and reinforced through questions with detailed step-by-step solutions hidden in collapsible dropdowns.

Exponential Form of Complex Numbers

A complex number in standard form \( z = a + bi \) is written in polar form as:

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

where \( r = \sqrt{a^2 + b^2} \) is called the modulus of \( z \), and \( \tan(\theta) = \dfrac{b}{a} \) such that \( 0 \le \theta < 2\pi \), where \( \theta \) is called the argument of \( z \).

complex plane
Figure 1. Complex plane representation

We now use Euler's formula, given by \( e^{i\theta} = \cos(\theta) + i \sin(\theta) \), to write the complex number \( z \) in exponential form as:

\[ z = r e^{i\theta} \]

where \( r \) and \( \theta \) are defined as above.

Example 1

Plot the complex number \( z = -1 + i \) on the complex plane and write it in exponential form.

View Solution

The complex number \( z = -1 + i = a + bi \) has real part \( a = -1 \) and imaginary part \( b = 1 \). It is plotted as a vector on the complex plane shown below.

plot of complex number z = -1 + i on complex plane
Figure 2. Plot of \( z = -1 + i \)

Calculate the modulus \( r \):

\[ r = \sqrt{a^2 + b^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

Find the reference angle \( \theta_r \):

\[ \theta_r = \tan^{-1}\left|\dfrac{b}{a}\right| = \tan^{-1}\left|\dfrac{1}{-1}\right| = \tan^{-1}(1) = \dfrac{\pi}{4} \]

Since the real part is negative and the imaginary part is positive, the terminal side of \( \theta \) lies in quadrant II. Thus:

\[ \theta = \pi - \theta_r = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4} \]

Therefore, \( z \) in exponential form is:

\[ z = \sqrt{2} e^{i\dfrac{3\pi}{4}} \]

Example 2

a) Plot the complex numbers \( i, -2, -i, -1 - 2i \) and \( 1 - i \) on the same complex plane.
b) Plot in separate complex planes and write each complex number in exponential form.

View Solution

a) Combined Plot:

plot of complex numbers in example 2 on complex plane
Figure 3. Combined plot of complex numbers

b) Separate Exponential Forms:

  • \( z = i \): \( a = 0, b = 1 \). Modulus \( r = 1 \). From the plot, argument \( \theta = \dfrac{\pi}{2} \).
    Exponential form: \( z = e^{i\pi/2} \)
    plot of complex number i
    Plot of \( z = i \)
  • \( z = -2 \): \( a = -2, b = 0 \). Modulus \( r = 2 \). Argument \( \theta = \pi \).
    Exponential form: \( z = 2 e^{i\pi} \)
    plot of complex number -2
    Plot of \( z = -2 \)
  • \( z = -i \): \( a = 0, b = -1 \). Modulus \( r = 1 \). Argument \( \theta = \dfrac{3\pi}{2} \).
    Exponential form: \( z = e^{i3\pi/2} \)
    plot of complex number -i
    Plot of \( z = -i \)
  • \( z = -1 - 2i \): \( a = -1, b = -2 \). Modulus \( r = \sqrt{(-1)^2 + (-2)^2} = \sqrt{5} \). Reference angle \( \theta_r = \tan^{-1}(2) \). In quadrant III, \( \theta = \pi + \tan^{-1}(2) \approx 4.25 \).
    Exponential form: \( z = \sqrt{5} e^{i(\pi + \arctan 2)} \)
    plot of complex number -1-2i
    Plot of \( z = -1 - 2i \)
  • \( z = 1 - i \): \( a = 1, b = -1 \). Modulus \( r = \sqrt{2} \). Reference angle \( \theta_r = \pi/4 \). In quadrant IV, \( \theta = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \).
    Exponential form: \( z = \sqrt{2} e^{i 7\pi/4} \)
    plot of complex number 1 - i
    Plot of \( z = 1 - i \)

Example 3

Write the complex number \( z = \sqrt{2} e^{i \dfrac{5\pi}{4}} \) in standard form.

View Solution

Using Euler's formula:

\[ z = \sqrt{2}\left(\cos\left(\dfrac{5\pi}{4}\right) + i \sin\left(\dfrac{5\pi}{4}\right)\right) \] \[ = \sqrt{2}\left(-\dfrac{\sqrt{2}}{2} - i \dfrac{\sqrt{2}}{2}\right) = -1 - i \]

Multiplication and Division of Complex Numbers in Exponential Forms

Complex numbers in exponential form are easily multiplied and divided. The power and root of complex numbers in exponential form are also easily computed.

Multiplication

Let \( z_1 = r_1 e^{i\theta_1} \) and \( z_2 = r_2 e^{i\theta_2} \) be complex numbers in exponential form. Their product is given by:

\[ z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)} \]

Example 4

Given \( z_1 = 3 e^{i\pi/4} \) and \( z_2 = 5 e^{i3\pi/4} \), find \( z_1 z_2 \) and write it in standard form.

View Solution
\[ z_1 z_2 = (3 e^{i\pi/4})(5 e^{i3\pi/4}) = (3 \times 5) e^{i(\pi/4 + 3\pi/4)} = 15 e^{i\pi} \]

Rewrite in polar/standard form:

\[ 15(\cos(\pi) + i \sin(\pi)) = 15(-1 + 0) = -15 \]

Division

Let \( z_1 = r_1 e^{i\theta_1} \) and \( z_2 = r_2 e^{i\theta_2} \) be complex numbers in exponential form. Their ratio (or division) is given by:

\[ \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2} e^{i(\theta_1 - \theta_2)} \]

Example 5

Given \( z_1 = 10 e^{i\pi/3} \) and \( z_2 = 2 e^{i2\pi/3} \), find \( \dfrac{z_1}{z_2} \) and write it in standard form.

View Solution
\[ \dfrac{z_1}{z_2} = \dfrac{10 e^{i\pi/3}}{2 e^{i2\pi/3}} = \dfrac{10}{2} e^{i(\pi/3 - 2\pi/3)} = 5 e^{-i\pi/3} \]

Rewrite in polar/standard form:

\[ 5\left(\cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right)\right) = 5\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = \dfrac{5}{2} - \dfrac{5\sqrt{3}}{2}i \]

You may also review De Moivre's Theorem Power and Root of Complex Numbers.

Practice Questions

Question 1

Write the following complex numbers in exponential form:

  1. \( z_1 = -1 \)
  2. \( z_2 = -2i \)
  3. \( z_3 = -\sqrt{3} - i \)
  4. \( z_4 = -3 + 3\sqrt{3}i \)
  5. \( z_5 = 7 - 7i \)

Question 2

Use the results in Question 1 above to evaluate the following expressions and write them in exponential form:

  1. \( z_4 z_5 \)
  2. \( \dfrac{z_3 z_5}{z_4} \)

Solutions to Practice Questions

Solutions to Question 1

  1. \( z_1 = -1 = e^{i\pi} \)
  2. \( z_2 = -2i = 2 e^{i3\pi/2} \)
  3. \( z_3 = -\sqrt{3} - i = 2 e^{i7\pi/6} \)
  4. \( z_4 = -3 + 3\sqrt{3}i = 6 e^{i2\pi/3} \)
  5. \( z_5 = 7 - 7i = 7\sqrt{2} e^{i7\pi/4} \)

Solutions to Question 2

  1. \( z_4 z_5 = (6 e^{i2\pi/3})(7\sqrt{2} e^{i7\pi/4}) = 42\sqrt{2} e^{i(2\pi/3 + 7\pi/4)} = 42\sqrt{2} e^{i 29\pi/12} \).
    Subtracting \( 2\pi \) (\( 24\pi/12 \)) from the argument gives: \( 42\sqrt{2} e^{i 5\pi/12} \)
  2. \( \dfrac{z_3 z_5}{z_4} = \dfrac{(2 e^{i7\pi/6})(7\sqrt{2} e^{i7\pi/4})}{6 e^{i2\pi/3}} = \dfrac{14\sqrt{2}}{6} e^{i(7\pi/6 + 7\pi/4 - 2\pi/3)} = \dfrac{7\sqrt{2}}{3} e^{i 9\pi/4} \).
    Subtracting \( 2\pi \) (\( 8\pi/4 \)) from the argument gives: \( \dfrac{7\sqrt{2}}{3} e^{i \pi/4} \)

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