Question

How do you multiply `e^(( 5 pi )/ 3 i) * e^( 3 pi/2 i )` in trigonometric form?

Answer 1

`e^(i(7pi)/6)`

Explanation:

By Euler's formula:
`e^(ix) = cosx + i sinx`

Hence, the above becomes a FOIL problem:

`e^(i(5pi)/3) * e^(i(3pi)/2)`
`= (cos((5pi)/3) + i sin((5pi)/3))(cos((3pi)/2) + i sin((3pi)/2) )`
`= cos((5pi)/3)cos((3pi)/2) + i cos((5pi)/3)sin((3pi)/2) + isin((5pi)/3)cos((3pi)/2) - sin((5pi)/3) sin((3pi)/2)`

Let's plug in the values, which we know from the unit circle:
`= 1/2 * 0 + i 1/2 * -1 + i * -sqrt(3)/2 * 0 - (-1) * (-sqrt(3)/2)`
`= -1/2 i - sqrt(3)/2`
Which is clearly on the unit circle! With a little thinking, we realize this is at `theta = (7pi)/6`.

We can now realize that since this is an exponential, we could have used an exponential rule that we know:
`e^a * e^b = e^(a+b)`
`e^(i(5pi)/3) * e^(i(3pi)/2) = e^(i(5pi)/3 + i (3pi)/2) = exp(i [(5pi)/3 + (3pi)/2]) = exp(i[19pi]/6)`
Subtracting off a full rotation (`2pi`) from the argument leaves us with
`= e^(i (19 - 12)pi / 6) = e^(i (7pi) / 6)`
as we found trigonometrically.