how do I write -8i in polar form?

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- Apr 28th 2010, 03:36 AMtim_mannirecomplex number polar form
how do I write -8i in polar form?

- Apr 28th 2010, 03:56 AMtonio
- Apr 28th 2010, 04:20 AMtim_mannire
z^3 = -8i . Does the polar form change for z^3 = -8i

- Apr 28th 2010, 04:57 AMraheel88
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 AMtim_mannire
- Apr 28th 2010, 06:04 AMraheel88