Question

(1+ i square root 3)^8?
in standard form?

Answer 1

i think the answer is 82..i like did this in my head

Answer 2

(1+i√3)^8=(((1+i√3)^2)^2)^2

Answer 3

because i^8 is 1 and radical(square root) 3 to the 8th is 81 so 81 plus 1 is 82

Answer 4

Continuation of my previous post:
(1+i√3)^8
(((1+i√3)^2)^2)^2
(1+i√3)^2=1+2i√3-3=2i√3-2=2(i√3-1)
(2(i√3-1))^2=4(-3-2i√3+1)=4(-2-2i√3)=-8(1+i√3)
(-8(1+i√3))^2=64(1+2i√3-3)=64(2i√3-2)=128(i√3-1)
In standard form:
-128+128i√3

Answer 5

i would change first to trig form, then use demoivre's theorem

Answer 6

if we plot the number inside the expresion, ( 1 + i sqrt(3)) , we get ( 1 , sqrt(3))
change this to trig form.
r = sqrt( 1 ^2 + sqrt(3)^2) = sqrt( 1 + 3) + sqrt( 4) = 2
and tan(theta) = sqrt(3)/1 , theta = arctan sqrt(3)
theta = 60 degrees