# Thread: complex number polar form

1. ## complex number polar form

how do I write -8i in polar form?

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

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

Tonio

3. z^3 = -8i . Does the polar form change for z^3 = -8i

4. Solving $z^3 = -8i$...so u want to solve for z right?

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

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

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

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

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

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

Therfore $\theta = \frac{\pi}{2} + \frac{2k\pi}{3} , (k=0,1,2)$
Thanks, however i am just after the polar form for -8i

6. 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.