How do you evaluate # e^( ( pi)/2 i) - e^( ( 23 pi)/12 i)# using trigonometric functions?

1 Answer
Binayaka C.
Sep 13, 2017

# e^(i pi/2) - e^(i (23*pi)/12) = -0.9659 + 1.2588i #

Explanation:

We know #e^(itheta) = cos theta +i sin theta#

#e^(i pi/2) = cos (pi/2) +i sin (pi/2) = 0 + i= i# and

#e^(i (23*pi)/12) = cos( (23pi)/12) +i sin ( (23pi)/12) # or

#e^(i (23*pi)/12) = cos 345 +i sin 345 = 0.9659 - 0.2588i #

# e^(i pi/2) - e^(i (23*pi)/12) = i- (0.9659 - 0.2588i)# or

# e^(i pi/2) - e^(i (23*pi)/12) = -0.9659 + 1.2588i # [Ans]

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