# Complex Numbers ( Roots)

• Jun 8th 2009, 12:04 PM
Daniels4691
Complex Numbers ( Roots)
Find the three cubic roots of -1

a> cos π/3 + i sin π/3, cos 3π + i sin 3π , cos 2π/3 + i sin 2π/3
b> cos π/3 + i sin π/3, cos π + i sin π, cos 5π/3 + i sin 5π/3
c> cos π/2 + i sin π/2, cos π + i sin π, cos 5π/2 + i sin 5π/2
d> cos π/4 + i sin π/4, cos π + i sin π, cos 5π/4 + i sin 5π/4

Any help would be greatly appreciated
• Jun 8th 2009, 12:21 PM
CaptainBlack
Quote:

Originally Posted by Daniels4691
Find the three cubic roots of -1

a> cos π/3 + i sin π/3, cos 3π + i sin 3π , cos 2π/3 + i sin 2π/3
b> cos π/3 + i sin π/3, cos π + i sin π, cos 5π/3 + i sin 5π/3
c> cos π/2 + i sin π/2, cos π + i sin π, cos 5π/2 + i sin 5π/2
d> cos π/4 + i sin π/4, cos π + i sin π, cos 5π/4 + i sin 5π/4

Any help would be greatly appreciated

Hint:

$\displaystyle -1=e^{i(2n+1)\pi}, \ n\in \mathbb{Z}$

CB
• Jun 8th 2009, 12:23 PM
Amer
let

$\displaystyle z^3=-1$

write -1 in exponential

$\displaystyle z^3=e^{i(\pi+2n\pi)}$ n take the values (1,2,3) since you want the cube square

take the cube square for the both sides ......... the rest for you
• Jun 8th 2009, 02:26 PM
yeongil
$\displaystyle -1 = 1(\cos \pi + i \sin \pi)$

If $\displaystyle z = r(\cos \theta + i \sin \theta)$, then the nth roots of a complex number would be in the form of

$\displaystyle \sqrt[n]{z} = \sqrt[n]{r}\left[\cos \left(\frac{\theta + 2\pi k}{n}\right) + i \sin \left(\frac{\theta + 2\pi k}{n}\right)\right]$ ,
where k = 0, 1, 2, ... n - 1.

In our case, n = 3, r = 1, and k will go from 0 to 2. The angle simplifies to
$\displaystyle \frac{\theta + 2\pi k}{n} = \frac{\pi + 2\pi k}{3} = \frac{\pi}{3} + \frac{2\pi k}{3}$

The three cube roots of $\displaystyle -1 = 1(\cos \pi + i \sin \pi)$ are:
\displaystyle \begin{aligned} z_1 &= \sqrt[3]{1}\left[\cos \left(\frac{\pi}{3} + 0\right) + i \sin \left(\frac{\pi}{3} + 0\right)\right] \\ &= \cos\left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right) \\ &= \frac{1}{2} + \frac{\sqrt{3}}{2}i \end{aligned}

\displaystyle \begin{aligned} z_2 &= \sqrt[3]{1}\left[\cos \left(\frac{\pi}{3} + \frac{2\pi}{3}\right) + i \sin \left(\frac{\pi}{3} + \frac{2\pi}{3}\right)\right] \\ &= \cos \pi + i \sin \pi \\ &= -1 \end{aligned}

\displaystyle \begin{aligned} z_3 &= \sqrt[3]{1}\left[\cos \left(\frac{\pi}{3} + \frac{4\pi}{3}\right) + i \sin \left(\frac{\pi}{3} + \frac{4\pi}{3}\right)\right] \\ &= \cos\left(\frac{5\pi}{3}\right) + i \sin \left(\frac{5\pi}{3}\right) \\ &= \frac{1}{2} - \frac{\sqrt{3}}{2}i \end{aligned}