If z=costheta +isintheta, prove that
a) z = 1/z
b) z+1/z = 2costheta
c) z-1/z=2isintheta
Hence ,deduce that z^2-1/z^2 +1=itantheta
z = cosθ + isinθ , then
a)
1/z = 1/(cosθ + isinθ) = 1/(cosθ + isinθ) * (cosθ - isinθ)/(cosθ - isinθ)
= (cosθ - isinθ)/(cos^2 θ - i^2 sin^2 θ)
= (cosθ - isinθ)/1 = cosθ - isinθ
b) z + 1/z = cosθ + isinθ + cosθ - isinθ = 2cosθ
c) z - 1/z = (cosθ + isinθ) - (cosθ - isinθ) = 2i sinθ
d) (z-1/z) / (z+1/z)
= ( (z^2 - 1)/z ) / ( (z^2 + 1)/z )
= ( (z^2 - 1)/z )*( z/(z^2 + 1) )
= (z^2 - 1)/(z^2 + 1) = 2isinθ/(2cosθ) = i tanθ
btw, did you notice that z ≠ 1/z ??
but rather, a), b), and c) are true for my results.
I think he had trouble with z* or something for z-conjugate.
overbars are hard to get right.
To prove the given statements, we will use the properties of complex numbers and trigonometric functions.
a) To prove that z = 1/z, we need to show that the left side of the equation is equal to the right side. Let's start by calculating the reciprocal of z:
(1/z) = (1/(costheta + isintheta))
To simplify this expression, we need to multiply the numerator and denominator by the conjugate of the denominator, which is (costheta - isintheta):
(1/z) = (1/(costheta + isintheta)) * ((costheta - isintheta)/(costheta - isintheta))
= (costheta - isintheta)/(costheta^2 - (isintheta)^2)
= (costheta - isintheta)/(costheta^2 + sin^2theta) [using the equation (i) i^2 = -1]
= (costheta - isintheta)/1
= costheta - isintheta
Now, we can see that z = costheta + isintheta and 1/z = costheta - isintheta. Since z and 1/z are equal, we have proven a) z = 1/z.
b) To prove that z + 1/z = 2costheta, let's substitute the values of z and 1/z:
z + 1/z = (costheta + isintheta) + (costheta - isintheta)
= 2costheta
Thus, we have proven b) z + 1/z = 2costheta.
c) To prove that z - 1/z = 2isintheta, let's substitute the values of z and 1/z:
z - 1/z = (costheta + isintheta) - (costheta - isintheta)
= 2isintheta
Thus, we have proven c) z - 1/z = 2isintheta.
Finally, let's deduce z^2 - 1/z^2 + 1 = itantheta:
We know that z^2 = (costheta + isintheta)^2 = costheta^2 + 2isintheta * costheta - sin^2theta.
Using the results from part a), we have 1/z^2 = (costheta - isintheta)^2 = costheta^2 - 2isintheta * costheta - sin^2theta.
Now, substituting these values into the expression z^2 - 1/z^2 + 1:
z^2 - 1/z^2 + 1 = (costheta^2 + 2isintheta * costheta - sin^2theta) - (costheta^2 - 2isintheta * costheta - sin^2theta) + 1
= 4isintheta * costheta + 1
We know that itantheta = sintheta / costheta, so 4isintheta * costheta + 1 = 4itantheta * costheta + 1.
Therefore, we can deduce that z^2 - 1/z^2 + 1 = itantheta.