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:
\( z_0 = 1^{1/3} ( \cos( 0 ) + i \sin ( 0 )) = 1 \)
\(z_1 = 1^{1/3} (\cos(2\pi/3)+ i \sin (2\pi/3 )) = -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2} \)
\( z_2 = 1^{1/3} ( \cos ( 4\pi/3 ) + i \sin (4\pi/3) = -\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} ) \)
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
argumernts
are equal.
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:
\( z_0 = 1^{1/3} ( \cos( \pi/6 ) + i \sin ( \pi/6 )) = \dfrac{\sqrt{3}}{2}+i\dfrac{1}{2} \)
\( z_1 = 1^{1/3} ( \cos ( \dfrac{ \pi/2 + 2\pi}{3} ) + i \sin ( \dfrac{\pi/2 + 2\pi}{3} )) \)
\( = ( \cos ( 5\pi/6 ) + i \sin ( 5\pi/6 ) ) = -\dfrac{\sqrt{3}}{2}+i\dfrac{1}{2} \)
\( z_2 = 1^{1/3} ( \cos ( \dfrac{ \pi/2 + 4\pi}{3} ) + i \sin ( \dfrac{\pi/2 + 4\pi}{3} ) ) \)
\( = \cos ( 3\pi/2 ) + i \sin ( 3\pi/2 ) = - i \)
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
modulii are equal and they are equally spaced.
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.
\(w_0 = 1^{1/2} ( \cos ( \dfrac{3\pi/2}{2} ) + i \sin ( \dfrac{3\pi/2}{2} ) ) = \cos ( 3\pi/4 ) + i \sin ( 3\pi/4 ) = -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2} \)
\(w_1 = 1^{1/2} ( \cos ( \dfrac{3\pi/2 + 2\pi}{2} ) + i \sin ( \dfrac{3\pi/2 + 2\pi}{2} ) ) = \cos ( 7\pi/4 ) + i \sin ( 7\pi/4 ) =\dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2} \)
We now solve for z
\( z_0 = W_0 + i = -\dfrac{\sqrt{2}}{2}+i\dfrac{\sqrt{2}}{2} + i = -\dfrac{\sqrt{2}}{2}+i\dfrac{2+\sqrt{2}}{2} \)
\( z_1 = W_1 + i = \dfrac{\sqrt{2}}{2}-i\dfrac{\sqrt{2}}{2} + i = \dfrac{\sqrt{2}}{2}+i\dfrac{2-\sqrt{2}}{2} \)