# complex number polar form

• Apr 28th 2010, 03:36 AM
tim_mannire
complex number polar form
how do I write -8i in polar form?
• Apr 28th 2010, 03:56 AM
tonio
Quote:

Originally Posted by tim_mannire
how do I write -8i in polar form?

As $\displaystyle -i=cis(3\pi/2)=e^{3\pi i/2}$ , $\displaystyle -8i=8e^{3\pi i/2}$

Tonio
• Apr 28th 2010, 04:20 AM
tim_mannire
z^3 = -8i . Does the polar form change for z^3 = -8i
• Apr 28th 2010, 04:57 AM
raheel88
Solving $\displaystyle z^3 = -8i$...so u want to solve for z right?

If you write z as $\displaystyle z = re^{i\theta}$ and substitute this into $\displaystyle z^3 = -8i$ and solve.

You get r = 2 and $\displaystyle 3\theta = \frac{3\pi}{2} + 2k\pi , (k=0,1)$.

Therfore $\displaystyle \theta = \frac{\pi}{2} + \frac{2k\pi}{3} , (k=0,1,2)$
• Apr 28th 2010, 05:04 AM
tim_mannire
Quote:

Originally Posted by raheel88
Solving $\displaystyle z^3 = -8i$...so u want to solve for z right?

If you write z as $\displaystyle z = re^{i\theta}$ and substitute this into $\displaystyle z^3 = -8i$ and solve.

You get r = 2 and $\displaystyle 3\theta = \frac{3\pi}{2} + 2k\pi , (k=0,1)$.

Therfore $\displaystyle \theta = \frac{\pi}{2} + \frac{2k\pi}{3} , (k=0,1,2)$

Thanks, however i am just after the polar form for -8i
• Apr 28th 2010, 06:04 AM
raheel88
Quote:

Originally Posted by tim_mannire
Thanks, however i am just after the polar form for -8i

that's what tonio has posted.

the polar form doesn't change for z^3 but there are more solutions for z.