De Moivre's Theorem Power and Root

De Moivre's theorem is used to find powers of complex numbers and is also extended to find roots of complex numbers and solve equations. Several questions are presented along with their detailed step-by-step solutions hidden in collapsible dropdowns.

De Moivre's Theorem to Find Power of Complex Numbers

If \( z \) is a complex number in polar form written as:

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

then:

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

where \( n \) is an integer.

Example 1

Use De Moivre's theorem to simplify and write in standard form the following expressions:

View Solution

a) Simplify \( i^{23} \):

Write \( i \) in polar form:

\[ i = \cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right) \]

Use De Moivre's theorem to find \( i^{23} \):

\[ i^{23} = \left(\cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right)\right)^{23} = \cos\left(\frac{23\pi}{2}\right) + i \sin\left(\frac{23\pi}{2}\right) \] \[ = \cos\left(\frac{3\pi}{2} + 5(2\pi)\right) + i \sin\left(\frac{3\pi}{2} + 5(2\pi)\right) = \cos\left(\frac{3\pi}{2}\right) + i \sin\left(\frac{3\pi}{2}\right) = -i \]

b) Simplify \( (1 - i)^{12} \):

Write \( 1 - i \) in polar form:

\[ 1 - i = \sqrt{2}\left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right) \]

Use De Moivre's theorem:

\[ (1 - i)^{12} = \left(\sqrt{2}\right)^{12} \left(\cos\left(12 \times \frac{7\pi}{4}\right) + i \sin\left(12 \times \frac{7\pi}{4}\right)\right) \] \[ = 64(\cos(21\pi) + i \sin(21\pi)) = 64(-1 + 0) = -64 \]

c) Simplify \( \left(\dfrac{\sqrt{2}}{2} - i\dfrac{\sqrt{2}}{2}\right)^{400} \):

Write in polar form:

\[ \dfrac{\sqrt{2}}{2} - i\dfrac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}(1 - i) = \cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right) \]

Apply De Moivre's theorem:

\[ \left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right)^{400} = \cos\left(400 \times \frac{7\pi}{4}\right) + i \sin\left(400 \times \frac{7\pi}{4}\right) \] \[ = \cos(700\pi) + i \sin(700\pi) = 1 + 0i = 1 \]

De Moivre's Theorem to Find Roots of Complex Numbers

De Moivre's theorem can also be used to find the \( n \)-th roots of a complex number. If \( z \) is a complex number of the form:

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

then its \( n \)-th roots are given by:

\[ z_k = r^{1/n} \left( \cos\left( \dfrac{\theta + 2k\pi}{n} \right) + i \sin\left( \dfrac{\theta + 2k\pi}{n} \right) \right) \]

where \( k = 0, 1, \dots, (n - 1) \).

Example 2

Use De Moivre's theorem to find:

View Solution

a) Third roots of \( 1 \):

Write \( 1 = 1(\cos(0) + i \sin(0)) \). The formula for third roots (\( n = 3 \), \( r = 1 \), \( \theta = 0 \)) is:

\[ z_k = \cos\left(\frac{2k\pi}{3}\right) + i \sin\left(\frac{2k\pi}{3}\right), \quad k = 0, 1, 2 \]
  • \( z_0 = \cos(0) + i \sin(0) = 1 \)
  • \( z_1 = \cos\left(\frac{2\pi}{3}\right) + i \sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \)
  • \( z_2 = \cos\left(\frac{4\pi}{3}\right) + i \sin\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \)
third roots of z = 1 on complex plane
Figure 1. Third roots of 1 on the complex plane

b) Third roots of \( i \):

\( i = 1\left(\cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right)\right) \):

\[ z_k = \cos\left(\frac{\frac{\pi}{2} + 2k\pi}{3}\right) + i \sin\left(\frac{\frac{\pi}{2} + 2k\pi}{3}\right), \quad k = 0, 1, 2 \]
  • \( z_0 = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + i\frac{1}{2} \)
  • \( z_1 = \cos\left(\frac{5\pi}{6}\right) + i \sin\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + i\frac{1}{2} \)
  • \( z_2 = \cos\left(\frac{3\pi}{2}\right) + i \sin\left(\frac{3\pi}{2}\right) = -i \)

c) Sixth roots of \( 2 + 2i \):

\( 2 + 2i = 2\sqrt{2}\left(\cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right)\right) \), where \( r = 2\sqrt{2} = 2^{3/2} \), so \( r^{1/6} = (2^{3/2})^{1/6} = 2^{1/4} \):

\[ z_k = 2^{1/4}\left(\cos\left(\frac{\frac{\pi}{4} + 2k\pi}{6}\right) + i \sin\left(\frac{\frac{\pi}{4} + 2k\pi}{6}\right)\right), \quad k = 0, 1, 2, 3, 4, 5 \]
  • \( z_0 = 2^{1/4}\left(\cos\left(\frac{\pi}{24}\right) + i \sin\left(\frac{\pi}{24}\right)\right) \)
  • \( z_1 = 2^{1/4}\left(\cos\left(\frac{3\pi}{8}\right) + i \sin\left(\frac{3\pi}{8}\right)\right) \)
  • \( z_2 = 2^{1/4}\left(\cos\left(\frac{17\pi}{24}\right) + i \sin\left(\frac{17\pi}{24}\right)\right) \)
  • \( z_3 = 2^{1/4}\left(\cos\left(\frac{25\pi}{24}\right) + i \sin\left(\frac{25\pi}{24}\right)\right) \)
  • \( z_4 = 2^{1/4}\left(\cos\left(\frac{11\pi}{8}\right) + i \sin\left(\frac{11\pi}{8}\right)\right) \)
  • \( z_5 = 2^{1/4}\left(\cos\left(\frac{41\pi}{24}\right) + i \sin\left(\frac{41\pi}{24}\right)\right) \)
sixth roots of z = 2 + 2 i on complex plane
Figure 2. Sixth roots of \( 2 + 2i \) on the complex plane

d) Solve \( z^4 = \dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i \):

Write the right side in polar form: \( \frac{\sqrt{3}}{2} - \frac{1}{2}i = \cos\left(\frac{11\pi}{6}\right) + i \sin\left(\frac{11\pi}{6}\right) \).

The four roots are given by:

\[ z_k = \cos\left(\frac{\frac{11\pi}{6} + 2k\pi}{4}\right) + i \sin\left(\frac{\frac{11\pi}{6} + 2k\pi}{4}\right), \quad k = 0, 1, 2, 3 \]
  • \( z_0 = \cos\left(\frac{11\pi}{24}\right) + i \sin\left(\frac{11\pi}{24}\right) \)
  • \( z_1 = \cos\left(\frac{23\pi}{24}\right) + i \sin\left(\frac{23\pi}{24}\right) \)
  • \( z_2 = \cos\left(\frac{35\pi}{24}\right) + i \sin\left(\frac{35\pi}{24}\right) \)
  • \( z_3 = \cos\left(\frac{47\pi}{24}\right) + i \sin\left(\frac{47\pi}{24}\right) \)

Practice Questions

Questions

  1. Evaluate the expression: \( \left(\cos\left(-\frac{\pi}{3}\right) - i \sin\left(-\frac{\pi}{3}\right)\right)^{18} \)
  2. Show that \( 1 - i \) is one of the third roots of \( -2 - 2i \).
  3. Find all the fourth roots of \( i \).
  4. Solve for the complex number \( z \) the equation: \( (z - i)^2 = -1 \).

Solutions to Practice Questions

View Solutions

1) Evaluate \( \left(\cos\left(-\frac{\pi}{3}\right) - i \sin\left(-\frac{\pi}{3}\right)\right)^{18} \):

Using identities \( \cos(-x) = \cos(x) \) and \( \sin(-x) = -\sin(x) \):

\[ = \left(\cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right)\right)^{18} = \cos\left(\frac{18\pi}{3}\right) + i \sin\left(\frac{18\pi}{3}\right) = \cos(6\pi) + i \sin(6\pi) = 1 \]

2) Show that \( 1 - i \) is a third root of \( -2 - 2i \):

Evaluate \( (1 - i)^3 \):

\[ (1 - i)^3 = \left(\sqrt{2}\left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right)\right)^3 = 2\sqrt{2}\left(\cos\left(\frac{21\pi}{4}\right) + i \sin\left(\frac{21\pi}{4}\right)\right) \] \[ = 2\sqrt{2}\left(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = -2 - 2i \]

Since \( (1 - i)^3 = -2 - 2i \), \( 1 - i \) is indeed one of the third roots.

3) Fourth roots of \( i \):

\( i = \cos\left(\frac{\pi}{2}\right) + i \sin\left(\frac{\pi}{2}\right) \). The four roots are:

\[ z_k = \cos\left(\frac{\frac{\pi}{2} + 2k\pi}{4}\right) + i \sin\left(\frac{\frac{\pi}{2} + 2k\pi}{4}\right), \quad k = 0, 1, 2, 3 \]
  • \( z_0 = \cos\left(\frac{\pi}{8}\right) + i \sin\left(\frac{\pi}{8}\right) \)
  • \( z_1 = \cos\left(\frac{5\pi}{8}\right) + i \sin\left(\frac{5\pi}{8}\right) \)
  • \( z_2 = \cos\left(\frac{9\pi}{8}\right) + i \sin\left(\frac{9\pi}{8}\right) \)
  • \( z_3 = \cos\left(\frac{13\pi}{8}\right) + i \sin\left(\frac{13\pi}{8}\right) \)

4) Solve \( (z - i)^2 = -1 \):

Let \( w = z - i \), so \( w^2 = -1 \). The solutions are the second roots of \( -1 = \cos(\pi) + i \sin(\pi) \) (or using standard square roots of \(-1\)):

\[ w = \pm i \]

Solving for \( z \) (\( z = w + i \)):

  • If \( w = i \): \( z = i + i = 2i \)
  • If \( w = - i \): \( z = -i + i = 0 \)
    • Hence the solutions to the given equation
  • \( z_0 = 2i \)
  • \( z_1 = 0 \)

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