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 solutions.

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 forms the following expressions.
a) \( i^{23} \)
b) \( (1 - i)^{12} \)
c) \( (\sqrt2 / 2 - i\sqrt2 / 2)^{400} \)

Solution to Example 1
a) Write \( i \) in polar form
\( i = \cos(\pi/2) + i \sin(\pi/2) \)
Use De Moivre's theorem to find \( i^{23} \)
\( i^{23} = (\cos(\pi/2) + i \sin(\pi/2))^{23} \)
\( = (\cos(23\pi/2) + i \sin(23\pi/2)) \)
Simplify
\( = \cos(3\pi/2 + 5(2\pi)) + i \sin(3\pi/2 + 5(2\pi))\)
\(= \cos(3\pi/2) + i \sin(3\pi/2)\)
\( = - i \)

b) Write \( (1 - i)\) in polar form
\( (1 - i) = \sqrt2 (cos(7\pi/4) + i sin(7\pi/4)) \)
Use De Moivre's theorem to find \( (1 - i)^{12} \)
\( (1 - i)^{12} = (\sqrt2 (cos(7\pi/4) + i sin(7\pi/4)))^{12} \)
\( = (\sqrt2)^12 (cos(12 \times 7\pi/4) + i sin(12 \times 7\pi/4)) \)
Simplify
\( = 64 (cos(21\pi) + i sin(21\pi)) \)
\( = 64(-1 + 0) \)
\( = - 64 \)

c) Write \( (\sqrt2 / 2 - i\sqrt2 / 2) \) in polar form
\( (\sqrt2 / 2 - i\sqrt2 / 2) = \sqrt2 / 2 (1 - i) = (cos(7\pi/4) + i sin(7\pi/4)) \)
Use De Moivre's theorem to find \( (\sqrt2 / 2 - i\sqrt2 / 2)^{400}\)
\( (\sqrt2 / 2 - i\sqrt2 / 2)^{400} = (cos(7\pi/4) + i sin(7\pi/4))^{400} \)
\( = (cos(400 \times 7\pi/4) + i sin(400 \times 7\pi/4)) \)
Simplify
\( = (cos( 700 \pi) + i sin(700 \pi)) \)
\( = 1 \)



De Moivre's Theorem to Find Roots of Complex Numbers

De Moivre's theorem can also be used to find the nth roots of a complex number as follows
If z is a complex number of the form
\[ z = r (\cos(\theta)+ i \sin(\theta)) \]
then the nth 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, ... , (n - 1).
Example 2
Use De Moivre's theorem to find
a) all the third roots of \( 1 \)
b) all the third roots of \( i \)
c) all the sixth roots of \( 2 + 2i \)
d) to solve over the complex numbers the equation : \( z^4 = \sqrt3 / 2 - (1/2) i\)

Solution to Example 2
a) Write \( 1 \) in polar form
\( 1 = 1(\cos(0)+ i \sin(0)) \)
Use the De Moivre's theorem to find all third roots
\( z_k = 1^{1/3} \left ( \cos \left( \dfrac{ 0 + 2k\pi}{3} \right ) + i \sin \left ( \dfrac{0 + 2k\pi}{3} \right) \right ) \) , k = 0, 1,2.
The three roots are
Set k = 0, 1 and 2 in the formula above to obtain the roots:
\( z_0 = 1^{1/3} ( \cos( 0 ) + i \sin ( 0 )) = 1 \)
\(z_1 = 1^{1/3} (\cos(2\pi/3)+ i \sin (2\pi/3 )) = -\frac{1}{2}+i\frac{\sqrt{3}}{2} \)
\( z_2 = 1^{1/3} ( \cos ( 4\pi/3 ) + i \sin (4\pi/3) = -\frac{1}{2}-i\frac{\sqrt{3}}{2} ) \)
The three third roots are shown on the complex plane below. They are located on the same circle because their moduli are equal and they are equally spaced because the difference between their arguments are equal.

third roots of z = 1 on complex plane



b) Write \( i \) in polar form
\( i = 1(\cos(\pi/2) + i \sin(\pi/2)) \)
Use the De Moivre's theorem to find all third roots
\( z_k = 1^{1/3} ( \cos ( \dfrac{ \pi/2 + 2k\pi}{3} ) + i \sin ( \dfrac{\pi/2 + 2k\pi}{3} ) ) \) , k = 0, 1,2.

Set k = 0, 1 and 2 in the formula above to obtain the roots:

\( z_0 = 1^{1/3} ( \cos( \pi/6 ) + i \sin ( \pi/6 )) = \frac{\sqrt{3}}{2}+i\frac{1}{2} \)

\( z_1 = 1^{1/3} ( \cos ( \dfrac{ \pi/2 + 2\pi}{3} ) + i \sin ( \dfrac{\pi/2 + 2\pi}{3} )) \)
\( = ( \cos ( 5\pi/6 ) + i \sin ( 5\pi/6 ) ) = -\frac{\sqrt{3}}{2}+i\frac{1}{2} \)

\( z_2 = 1^{1/3} ( \cos ( \dfrac{ \pi/2 + 4\pi}{3} ) + i \sin ( \dfrac{\pi/2 + 4\pi}{3} ) ) \)
\( = \cos ( 3\pi/2 ) + i \sin ( 3\pi/2 ) = - i \)

c) Write \( 2 + 2i \) in polar form
\( 2 + 2i = 2\sqrt2 (\cos(\pi/4) + i\sin(\pi/4) \)
Use De Moivre's theorem to find all six roots.
\( z_k = (2\sqrt2)^{1/6} ( \cos ( \dfrac{ \pi/4 + 2k\pi}{6} ) + i \sin ( \dfrac{\pi/4 + 2k\pi}{6} ) ) \) , k = 0,1,2,3,4,5.

Set k = 0, 1, 2, 3, 4 and 5 in the formula above to obtain the roots:
\( z_0 = (2\sqrt2)^{1/6} ( \cos ( \dfrac{ \pi/4}{6} ) + i \sin ( \dfrac{\pi/4}{6} ) ) \)
\( = 2^{1/4}( \cos (\pi/24 ) + i \sin (\pi/24) ) \)

\( z_1 = (2\sqrt2)^{1/6} ( \cos ( \dfrac{ \pi/4 + 2\pi}{6} ) + i \sin ( \dfrac{\pi/4 + 2\pi}{6} ) ) \)
\( = 2^{1/4} ( \cos ( 3\pi/8 ) + i \sin ( 3\pi/8 )) \)

\( z_2 = (2\sqrt2)^{1/6} ( \cos ( \dfrac{ \pi/4 + 4\pi}{6} ) + i \sin ( \dfrac{\pi/4 + 4\pi}{6} ) ) \)
\( = 2^{1/4} ( \cos ( 17\pi/24 ) + i \sin ( 17\pi/24 )) \)

\( z_3 = (2\sqrt2)^{1/6} ( \cos ( \dfrac{ \pi/4 + 6\pi}{6} ) + i \sin ( \dfrac{\pi/4 + 6\pi}{6} ) ) \)
\( = 2^{1/4} ( \cos ( 25\pi/24 ) + i \sin ( 25\pi/24 )) \)

\( z_4 = (2\sqrt2)^{1/6} ( \cos ( \dfrac{ \pi/4 + 8\pi}{6} ) + i \sin ( \dfrac{\pi/4 + 8\pi}{6} ) ) \)
\( = 2^{1/4} ( \cos ( 11\pi/8 ) + i \sin ( 11\pi/8 )) \)

\( z_5 = (2\sqrt2)^{1/6} ( \cos ( \dfrac{ \pi/4 + 10\pi}{6} ) + i \sin ( \dfrac{\pi/4 + 10\pi}{6} ) ) \)
\( = 2^{1/4} ( \cos ( 41\pi/24 ) + i \sin ( 41\pi/24 )) \)

d) The solutions to the given equation are the fourth roots to the complex number \(\sqrt3 / 2 - (1/2) i \)
Write \(\sqrt3 / 2 - (1/2) i \) in polar for
\(\sqrt3 / 2 - (1/2) i = \cos (11\pi/6) + i \sin(11\pi/6)\)
The four roots are given by De Moivre's theorem as follows
\( z_k = 1^{1/4} ( \cos ( \dfrac{11\pi/6 + 2k\pi}{4} ) + i \sin ( \dfrac{11\pi/6 + 2k\pi}{4} ) ) \)
where k = 0, 1,2,3.
The solutions (or roots) to the given equation are:

\( z_0 = 1^{1/4} ( \cos ( \dfrac{11\pi/6}{4} ) + i \sin ( \dfrac{11\pi/6}{4} ) ) = \cos ( 11\pi/24 ) + i \sin ( 11\pi/24 ) \)

\( z_1 = 1^{1/4} ( \cos ( \dfrac{11\pi/6 + 2\pi}{4} ) + i \sin { 2\pi/6 + 2\pi}{4} ) = \cos ( 23\pi/24) + i \sin ( 23\pi/24) \)

\( z_2 = 1^{1/4} ( \cos ( \dfrac{11\pi/6 + 4\pi}{4} ) + i \sin ( \dfrac{11\pi/6 + 4\pi}{4} ) ) = \cos ( 35\pi/24) + i \sin ( 35\pi/24) \)

\( z_3 = 1^{1/4} ( \cos ( \dfrac{11\pi/6 + 6\pi}{4} ) + i \sin ( \dfrac{11\pi/6 + 6\pi}{4} ) ) = \cos ( 47\pi/24) + i \sin ( 47\pi/24) \)
The six sixth roots are shown on the complex plane below; their moduli are equal and they are equally spaced.

sixth roots of z = 2 + 2 i on complex plane



Questions

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



Solutions to the Above Questions

1)
Use the trigonometric identities: cos(-x) = cos(x) and sin(-x) = - sin(x) to rewrite the given expressions as
\( = (cos(\pi/3) + i sin(\pi/3))^{18} \)
which is in polar form, hence the use of De Moivre's theorem
\( = (cos(18\pi/3) + i sin(18\pi/3)) = (cos(6\pi) + i sin(6\pi)) = 1 \)

2)
Rewrite \((1 - i)\) in polar form and evaluate \( (1 - i)^3 \)
\( (1 - i)^3 = (\sqrt(2) (\cos(7\pi/4) + i \sin(7\pi/4)))^3 \)
\( = 2\sqrt(2) (\cos(21\pi/4) + i \sin(21\pi/4)) \)
\( = 2\sqrt(2) ( - \sqrt(2) / 2 - i \sqrt(2) / 2) = - 2 - 2 i \)
Since \( (1 - i)^3 = - 2 - 2 i \) , \( 1 - i\) is one of the third roots of \( - 2 - 2 i \).

3)
Rewrite \( i \) in polar form and use De Moivre's theorem to find the roots.
\( i = ( \cos(\pi/2) + i\sin(\pi/2)) \)
roots are
\(z_k = 1^{1/4} ( \cos ( \dfrac{\pi/2 + 2k\pi}{4} ) + i \sin ( \dfrac{\pi/2 + 2k\pi}{4} ) ) \)
where k = 0,1,2,3.
\(z_0 = 1^{1/4} ( \cos ( \dfrac{\pi/2}{4} ) + i \sin ( \dfrac{\pi/2}{4}) ) = \cos ( \pi/8 ) + i \sin ( \pi/8 )) \)
\(z_1 = 1^{1/4} ( \cos ( \dfrac{\pi/2 + 2\pi}{4} ) + i \sin ( \dfrac{\pi/2 + 2\pi}{4}) ) = \cos(5\pi/8) + i \sin(5\pi/8) \)
\(z_2 = 1^{1/4} ( \cos ( \dfrac{\pi/2 + 4\pi}{4} ) + i \sin ( \dfrac{\pi/2 + 4\pi}{4}) ) = \cos(9\pi/8) + i \sin(9\pi/8) \)
\(z_3 = 1^{1/4} ( \cos ( \dfrac{\pi/2 + 6\pi}{4} ) + i \sin ( \dfrac{\pi/2 + 6\pi}{4}) ) = \cos(13\pi/8) + i \sin(13\pi/8) \)

4)
Let \( w = z - i \) and rewrite the equation as
\( w^2 = - 1 \)
The solutions to the above equation are the second roots of \( - 1 \). Rewrite \( - 1 \) in polar form and find the roots.
\( - 1 = \cos(3\pi/2) + i \sin(3\pi/2) \)
roots are
\(w_k = 1^{1/2} ( \cos ( \dfrac{3\pi/2 + 2k\pi}{2} ) + i \sin ( \dfrac{3\pi/2 + 2k\pi}{2} ) ) \)
where k = 0,1.
\(w_0 = 1^{1/2} ( \cos ( \dfrac{3\pi/2}{2} ) + i \sin ( \dfrac{3\pi/2}{2} ) ) = \cos ( 3\pi/4 ) + i \sin ( 3\pi/4 ) = -\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \)
\(w_1 = 1^{1/2} ( \cos ( \dfrac{3\pi/2 + 2\pi}{2} ) + i \sin ( \dfrac{3\pi/2 + 2\pi}{2} ) ) = \cos ( 7\pi/4 ) + i \sin ( 7\pi/4 ) =\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} \)
We now solve for z
\( z_0 = W_0 + i = -\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} + i = -\frac{\sqrt{2}}{2}+i\frac{2+\sqrt{2}}{2} \)
\( z_1 = W_1 + i = \frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} + i = \frac{\sqrt{2}}{2}+i\frac{2-\sqrt{2}}{2} \)

More References and Links

Complex Numbers in Polar Form
Complex Numbers - Basic Operations
Cube Root Calculator
Power Calculator .

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