Please Please Please help*************????

Question

let w1 = -2 (sqrt of 3) + 2i and w2 = -2 / (sqrt 2) - 2 / (sqrt 2)i

a) Rewrite w1 and w2 in trigonometric form.

b) Hence find z = w1^5 / w2^2 in simplest a+ bi form

Answer 1

w1 = sqrt(12+4) w2=sqrt(4/2+4/2) w1^5 = 4^5 <25pi/6 +10kpi =4^5 w2^2= 2^2 w1/w/2 = 4^4 <(pi/6-pi/3)+2kpi = 4^4 <(-pi/6)
= 4^4(-1/2sqrt3 -1/2i)

Answer 2

For w1
r=sqr((12+4)=4
sinθ=1/2
cosθ=-sqr(3)/2
Hence θ=5pi/6
w1=4(cos(5pi/6)+isin(5pi/6))

For w2
w2=-sqr(2)-sqr(2)i
r=2
sinθ=-sqr(2)/2=cosθ
θ=5pi/4
w2=2(cos(5pi/4)+isin(5pi/4))

w1^5/w2^2=
4^4(cos(5pi/3)+isin(5pi/3))
=256(cos(-pi/3)+isin(-pi/3))
=256(1/2-isqr(3)/2))
=128-i128sqr(3)