Question

How do you multiply `e^(( 2 pi )/ 3 i) * e^( 3 pi/2 i )` in trigonometric form?

Answer 1

`e^((2pi)/3i)*e^((3pi)/2i)=sqrt3/2+1/2i`

Explanation:

`a+ib` can be written in trigonometric form `re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)`,
where `r=sqrt(a^2+b^2)`, but here in given question `r=1`

Hence `e^((2pi)/3i)*e^((3pi)/2i)=e^(((2pi)/3+(3pi)/2)i)`

= `e^(((4pi)/6+(9pi)/6)i)=e^(13pi/6i)`

= `(cos(13pi/6)+isin(13pi/6))`

= `(cos(2pi+pi/6)+isin(2pi+pi/6))`

= `(cos(pi/6)+isin(pi/6))`

= `sqrt3/2+1/2i`