How do you evaluate e^( ( 7 pi)/4 i) - e^( ( 23 pi)/12 i)e7π4ie23π12i using trigonometric functions?

1 Answer
Jul 18, 2016

e^((7pi/4)i)-e^((23pi/12)i= -0.259 - 0.448ie(7π4)ie(23π12)i=0.2590.448i

Explanation:

e^((7pi/4)i) = cos7pi/4+isin7pi/4 =cos315+isin315=0.707+(-0.707)i=0.707-0.707ie(7π4)i=cos7π4+isin7π4=cos315+isin315=0.707+(0.707)i=0.7070.707i
e^((23pi/12)i) = cos23pi/12+isin23pi/12 =cos345+isin345=0.966+(-0.259)i=0.966-0.259i :.
e^((7pi/4)i)-e^((23pi/12)i = (0.707-0.966)-(.707+.259)i = -0.259-0.448i[Ans]