How do you simplify #(2+2i)^6#?
1 Answer
Mar 23, 2016
Explanation:
Using polar form
#2+2i = 2sqrt(2)(cos (pi/4) + i sin (pi/4))#
So:
#(2+2i)^6 = (2sqrt(2))^6(cos (pi/4) + i sin (pi/4))^6#
#=(2^(3/2))^6 (cos 6(pi/4) + i sin 6(pi/4))#
#=2^9 (cos ((3pi)/2) + i sin ((3pi)/2))#
#=2^9 * (-i)#
#=-512i#
Using rectangular form
#(2+2i)^2 = (2(1+i))^2 = 4(1+i)(1+i) = 4(1+2i+i^2) = 4(2i) = 8i#
So:
#(2+2i)^6 = ((2+2i)^2)^3 = (8i)^3 = 8^3i^3 = -512i#