How do you find the power #(-2+2i)^3# and express the result in rectangular form?

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Answer 1 · 2016-12-30T19:49:46

#= 16 ( 1 + i)#

We use polar complex first t o simplify down

let #R e^(i theta) = -2 + 2i #

#= 2 sqrt 2 (-1/sqrt 2 + i /sqrt 2)#

We will find -ve value for #cos theta# and +ve value for #sin theta# in Q2 of the Argand diagram.

# implies theta = (3 pi) /4#

#implies R e^(i theta) = 2 sqrt 2 e^(i (3 pi) /4)#

#implies (R e^(i theta))^3 = (2 sqrt 2)^3 e^(i (3*3 pi) /4)#

#= 16 sqrt 2e^(i (pi) /4)#

#= 16 ( 1 + i)#

clearly you could also use a binomial expansion :)