Z=2e^(ipi/4)

Question

What will the conjugate be ?
-
Z=?

Answer 1

e^ix= cos x + i sin x
so what we have here is
2 cos pi/4 + 2 i sin pi/4
conjugate is
2 cos pi/4 - 2 i sin pi/4
but
cos pi/4 = .5 sqrt 2
and
sin pi/4 = .5 sqrt 2
so
sqrt 2 - i sqrt 2

Answer 2

2e^(-π/4 i) or 2e^(7π/4 i)

since sinθ = y/r, and sin(-θ) = -sinθ,

z* = x-iy = e^(-iθ)