
Bobosharif S. answered 04/27/18
Tutor
4.4
(32)
Mathematics/Statistics Tutor
1.
(2-2i)/(-√3 + i)=((2-2i)(-√3 - i))/((-√3 + i)(-√3 - i))
=(-2(1+√3)+ i2(√3-1))/4=-(1+√3)/2+i(√3-1)/2
r=√[((1+√3)/2)2+((√3-1)/2)2]=√2
θ=arctan((√3-1)/(√3+1))=11π/12
(2-2i)/(-√3 + i)=√2e11π/12
2)
(2i)/(3e(8+i) =(2/3e8)ie-i/ei)(e-i)=(2/3e8)ie-i
=(2/3e8)i[(cos(1)-isin(1)]=(2/3e8)[(sin(1)+icos(1)]
r=(2/3e8)
θ=π/4.
So
(2i)/(3e(8+i))=(2/3e8)eπ/4=(2/3)e8+π/4.