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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