What is #(sqrt(5)+i)^6# in the form #a+bi#?
2 Answers
Explanation:
#(sqrt(5)+i)^2 = (sqrt(5))^2+2sqrt(5)i+i^2 = 4+2sqrt(5)i#
#(4+2sqrt(5)i)^2 = 4^2+2(4)(2sqrt(5)i)+(2sqrt(5)i)^2#
#color(white)((4+2sqrt(5)i)^2) = 16+16sqrt(5)i-20#
#color(white)((4+2sqrt(5)i)^2) = -4+16sqrt(5)i#
So:
#(sqrt(5)+i)^6 = (4+2sqrt(5)i)(-4+16sqrt(5)i)#
#color(white)((sqrt(5)+i)^6) = -16+64sqrt(5)i-8sqrt(5)i-160#
#color(white)((sqrt(5)+i)^6) = -176+56sqrt(5)i#
Check
We can make a rudimentary check of the answer by making sure that:
#abs(-176+56sqrt(5)i) = abs(sqrt(5)+i)^6#
We find:
#abs(sqrt(5)+i) = sqrt((sqrt(5))^2+1^2) = sqrt(5+1) = sqrt(6)#
#abs(-176+56sqrt(5)i) = sqrt((-176)^2+(56sqrt(5))^2) = sqrt(30976+15680)#
#= sqrt(46656) = sqrt(6^6) = (sqrt(6))^6#