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.

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