what are the real and imaginary part of ie^(i theta)??
2007-11-21 07:26:25 · 5 answers · asked by leyzeebum 1 in Science & Mathematics ➔ Mathematics
ie^(iθ) = i(cos θ + i sin θ) = -sin θ + i cos θ
So assuming that θ is real, we have:
Re(ie^(iθ)) = -sin θ
Im(ie^iθ) = cos θ
2007-11-21 07:35:13 · answer #1 · answered by Pascal 7 · 4⤊ 0⤋
Using DeMoivers Theorem:
ie^(i theta) = i[cos(theta) + i*sin(theta)]
using fact that i^2 = -1:
ie^(i theta)= i*cos(theta) - sin(theta)
so the real part would be -sin(theta)
2007-11-21 15:38:45 · answer #2 · answered by KEYNARDO 5 · 0⤊ 0⤋
real part = - sin (theta)
imaginary part = cos (theta)
2007-11-21 15:36:07 · answer #3 · answered by best-doctor 2 · 0⤊ 0⤋
overall it is imaginary
the coefficient is imaginary
the e is real and so is the theta
but the i coefficient in front of the theta is imaginary...
2007-11-21 15:33:04 · answer #4 · answered by sayamiam 6 · 0⤊ 1⤋
ie^(i theta) = i ( cos(theta) + i sin(theta)) =
-sin(theta) + i cos(theta) :
real part = -sin(theta)
imaginary part = cos(theta)
2007-11-21 15:36:53 · answer #5 · answered by Any day 6 · 0⤊ 0⤋