Z=2e^(ipi/4)
What will the conjugate be ?
-
Z=?
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
To find the conjugate of a complex number, we need to change the sign of the imaginary part.
So, for Z = 2e^(iπ/4), the conjugate will be Z* = 2e^(-iπ/4).
To find the conjugate of a complex number, we need to change the sign of the imaginary part. In this case, the expression Z = 2e^(iπ/4) is written in polar form, where the real part is 2 and the imaginary part is e^(iπ/4).
To express the complex number in rectangular form, we can use Euler's formula, which states that e^(ix) = cos(x) + i*sin(x).
So, in rectangular form, Z can be written as Z = 2 * (cos(π/4) + i*sin(π/4)).
Now, to find the conjugate, we change the sign of the imaginary part, giving Z = 2 * (cos(π/4) - i*sin(π/4)).
Therefore, the conjugate of Z = 2e^(iπ/4) is Z = 2(cos(π/4) - i*sin(π/4)).
2e^(-π/4 i) or 2e^(7π/4 i)
since sinθ = y/r, and sin(-θ) = -sinθ,
z* = x-iy = e^(-iθ)