Question #561d4

1 Answer
Scott F.
Feb 8, 2018

#z^8=-128+128isqrt3#

Explanation:

First put #z# into polar form. #a=1#, #b=sqrt3#, so #r=sqrt(a^2+b^2)=sqrt(1+3)=sqrt4=2# and #theta=arctan(b/a)=arctan(sqrt3/1)=arctansqrt3=pi/3# (this is fine since #a>0 and b>0#). So putting #z# into the polar form #z=r"cis"theta# where #"cis"theta=costheta+isintheta=e^(itheta)#, #z=2"cis"(pi/3)#.

De Moivre's theorem states that a complex number #z^n=r^n"cis"ntheta#. Therefore #z^8=2^8"cis"(8pi/3)=256"cis"((2pi)/3)#

Putting this back into rectangular form, #z^8=256cos((2pi)/3)+256isin((2pi)/3)=-128+128isqrt3#

Related questions
Impact of this question
835 views around the world
You can reuse this answer
Creative Commons License