What real function is (e^(ix)-e^(-ix))/(ie^(ix)+ie^(-ix)) to?

1 Answer
May 11, 2018

tan x

Explanation:

Using

e^{ix} = cos x + i sin x

and its conjugate

e^{-ix} = cos x-i sin x

we get

e^{ix}+e^{-ix} = 2 cos x
and
e^{ix}-e^{-ix} = 2i sin x

Thus

(e^(ix)-e^(-ix))/(ie^(ix)+ie^(-ix)) = (2i sin x)/(i 2 cos x) = tan x