# find the cube roots

• Dec 11th 2008, 02:45 PM
robasc
find the cube roots
I am learning how to find cube roots of a complex number and I wanted to see if I am doing this right. could someone let me know?

find the cube roots of -8i:

$\displaystyle 8(cos (3\pi/2)+ i sin(3\pi/2))$

$\displaystyle (8)^1/3[cos((3\pi/2)+2\pi(k)/3) + i sin((3\pi/2)+2\pi(k)/3)]$

$\displaystyle k= 0, 1, 2$

$\displaystyle 2(cos(\pi/2) + i sin(\pi/2))$

Is this good so far? I just done the first one to see if I done it right.
• Dec 11th 2008, 03:06 PM
Skalkaz
Dear robasc,

http://www.mathhelpforum.com/math-he...9985f458-1.gif

It's not trully correct. Check it cubing for k=1 or 2.
• Dec 11th 2008, 03:09 PM
Soroban
Hello, robasc!

I'm pretty sure you've got it right . . . I'll tidy-up a bit just to make sure.

Quote:

Find the cube roots of $\displaystyle -8i$

$\displaystyle z\;=\;8\left(\cos\frac{3\pi}{2}+ i\sin\frac{3\pi}{2}\right)$

$\displaystyle z^{\frac{1}{3}} \;=\;8^{\frac{1}{3}}\bigg[\cos\left(\frac{\frac{3\pi}{2}+2\pi k}{3}\right) + i\sin\left(\frac{\frac{3\pi}{2}+2\pi k}{3}\right)\bigg] \;=$ .$\displaystyle 2\bigg[\cos\left(\frac{\pi}{2} + \frac{2\pi}{3}k\right) + i\sin\left(\frac{\pi}{2} + \frac{2\pi}{3}k\right) \bigg]$

. . for $\displaystyle k\:=\: 0, 1, 2$

$\displaystyle 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$

Is this good so far? I just done the first one to see if I done it right.

Looks good to me!

• Dec 11th 2008, 03:10 PM
robasc
do you mean check for the rest of them?

$\displaystyle 2(cos(7\pi/6)+ i sin(7\pi/6))$

$\displaystyle 2(cos(11\pi/6)+ i sin(11\pi/6))$
• Dec 11th 2008, 03:23 PM
Skalkaz
Okey, than you only misspeled the parenthesis but though right.
• Dec 11th 2008, 03:25 PM
robasc
Thanks you friends!