find the cube roots

**robasc** · December 11th 2008, 02:45 PM

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:

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

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

$ k= 0, 1, 2 $

$ 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.

**Skalkaz** · December 11th 2008, 03:06 PM

Dear robasc,

It's not trully correct. Check it cubing for k=1 or 2.

**Soroban** · December 11th 2008, 03:09 PM

Hello, robasc!

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

Find the cube roots of $-8i$

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

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

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

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

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

Looks good to me!

**robasc** · December 11th 2008, 03:10 PM

do you mean check for the rest of them?

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

$ 2(cos(11\pi/6)+ i sin(11\pi/6)) $

**Skalkaz** · December 11th 2008, 03:23 PM

Okey, than you only misspeled the parenthesis but though right.

**robasc** · December 11th 2008, 03:25 PM

Thanks you friends!

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