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

1 Answer
Jake M.
Mar 31, 2018

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.

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