how do I write -8i in polar form?
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Originally Posted by tim_mannire how do I write -8i in polar form? As , Tonio
z^3 = -8i . Does the polar form change for z^3 = -8i
Solving ...so u want to solve for z right? If you write z as and substitute this into and solve. You get r = 2 and . Therfore
Originally Posted by raheel88 Solving ...so u want to solve for z right? If you write z as and substitute this into and solve. You get r = 2 and . Therfore Thanks, however i am just after the polar form for -8i
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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