Question

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

Answer 1

`C_12=e^[((2 pi)/3 +(3pi)/2)] = e^ [(13pi)/6i]`

Explanation:

Given: Two complex numbers,
`C_1=e^((2 pi)/ 3 i), C_2= e^((3 pi)/2i)`

Required: The product, `C_1*C_2=e^((2 pi)/ 3 i)*e^((3 pi)/2i)`

Solution Strategy: Use complex multiplication mechanism using the general exponential product rule, `e^(ai)* e^(bi) = e^(ai+bi)`
Thus;
`C_(12)=C_1*C_2=e^((2 pi)/ 3 i)*e^((3 pi)/2i)=e^[((2 pi)/ 3 i)+((3pi)/2i)]`
`C_12=e^[((2 pi)/3 +(3pi)/2)] = e^ [(13pi)/6i]`