complex number polar form

**tim_mannire** · April 28th 2010, 03:36 AM

how do I write -8i in polar form?

**tonio** · April 28th 2010, 03:56 AM

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

**tim_mannire** · April 28th 2010, 04:20 AM

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

**raheel88** · April 28th 2010, 04:57 AM

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

**tim_mannire** · April 28th 2010, 05:04 AM

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

**raheel88** · April 28th 2010, 06:04 AM

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.

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