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.

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
2162 views around the world
You can reuse this answer
Creative Commons License