If z = 3 + 1, and w = 3-(root(3)) + 2i, determine (z-w)^10 in the form a+bi

if z = 3 + 1, and w = 3-(square root of 3) + 2i,
determine (z-w)^10 in the form a+bi .

×−
z = 3 + i
w = 3−√3 + 2i
z−w={(3 + i )−(3−√3 + 2i)}
z−w={(3 + i −3+√3 − 2i)}
(z−w)=(√3 −i)} =
Le √3 =Rcosθ and 1= Rsinθ
Hence (Rcosθ)²+( Rsinθ)² =(√3)²+(1)²
→ R²(cos²θ+sin²θ) =3+1=4
→R²=4 and R= 2
2cosθ=√3 →cosθ=√3/2 Hence θ =30°­ =π/3
Hence (z−w)=(√3 −i)} =2{cos(π/3)−isin(π/3)} =2 e^(−iπ/3)
and (z−w)^10 =[2 e^(−iπ/3)]^10
=2^10 e^(−i10π/3)=2^10{ cos(10π/3) −isin(10π/3)}=
= 2^10{ cos(2π+π +π/3) −isin(2π+π +π/3)}=
=2^10{ cos(π +π/3) −isin(π +π/3)}=
=2^10{ −cos(π/3) +isin(π/3)}= 2^10{ −cos60° +isin60°}
= 2^10{ −1/2 +i√3/2} == 2^9{ −1 +i√3}
= −512 +512√3 i