How do you convert #(2-2i)^5# to polar form?

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Answer 1 · 2016-10-24T07:18:56

For #2-i2#, the polar form #re^(itheta)# is given by

#(2, -2)=r(cos theta, sin theta)#.

Here,

#r = sqrt(2^2+(-2)^2)=2sqrt2#,

#cos theta =2/r=1/sqrt 2>0# and

#sin theta =-2/r=-1/sqrt2<0#.

The #Q_4# #theta# is #-pi/4#

So, the given expression

#(2-i2)^5#

#=(2sqrt2 e^(i(-pi/4)))^5#

#=64sqrt2e^(5(-ipi/4))#

#-64sqrt2e^(i(-2pi+3/4pi))#

#=64sqrt2e^(-i2pi)e^(i(3/4pi))#

#=64sqrt2(cos(3/4pi)+isin(3/4pi))#, using #e^(-i2pi)=1#

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