How do you evaluate #e^( ( 23 pi)/12 i) - e^( ( 17 pi)/12 i)# using trigonometric functions?

1 Answer
Roy E.
Jan 20, 2017

Possibly it is #(cos alpha+sin alpha)+i(cos alpha -sin alpha)# where #alpha= pi/12#

Explanation:

Noting that #e^((24pi i)/12)=1# and #e^((18pi i)/12)=-i# we take out the common factor #e^((-pi i)/12)#:

#e^((23pi)/12i)-e^((27pi)/12 i)#
#=e^((24-1)/12 pi i)-e^((18-1)/12 pi i)#
#=e^(-pi/12 i)(e^(24/12pi i) - e^(18/12pi i))#
#=e^(-pi/12 i)(1-i))#
#=(cos (pi/12) + sin (pi/12)) + i (cos (pi/12) - sin (pi /12))#
or something like that.

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
1423 views around the world
You can reuse this answer
Creative Commons License