Let z and w be complex numbers such that |z| = |w| = 1, and zw is not equal to -1. Prove that
(z + w)/(zw + 1)
is a real number.
To prove that the expression (z + w)/(zw + 1) is a real number, we need to show that the denominator is equal to the complex conjugate of the numerator.
Let's start by finding the complex conjugate of (z + w):
Conjugate of (z + w) = (conjugate of z) + (conjugate of w)
= z* + w*
Now, let's find the complex conjugate of (zw + 1):
Conjugate of (zw + 1) = (conjugate of zw) + (conjugate of 1)
= (conjugate of z)(conjugate of w) + 1
= z* w* + 1
We want to prove that (z + w)/(zw + 1) is a real number. Therefore, we need to show that the denominator (zw + 1) is equal to the complex conjugate of the numerator (z + w), i.e., zw + 1 = z* w* + 1.
Since we are given that |z| = |w| = 1, we can write z = e^(iθ) and w = e^(iϕ), where θ and ϕ are angles.
Now, let's substitute these values into the equation zw + 1 = z* w* + 1:
(e^(iθ))(e^(iϕ)) + 1 = (e^(-iθ))(e^(-iϕ)) + 1
e^(i(θ + ϕ)) + 1 = e^(-i(θ + ϕ)) + 1
Using Euler's formula, we know that e^(ix) = cos(x) + i sin(x):
cos((θ + ϕ)) + i sin((θ + ϕ)) + 1 = cos(-(θ + ϕ)) + i sin(-(θ + ϕ)) + 1
cos((θ + ϕ)) + i sin((θ + ϕ)) + 1 = cos(-(θ + ϕ)) + i(-sin(θ + ϕ)) + 1
The real part on the left side of the equation equals the real part on the right side, and the imaginary part on the left side equals the negative of the imaginary part on the right side.
Hence, the equation cos((θ + ϕ)) = cos(-(θ + ϕ)) holds.
The only way this equation can be true is if (θ + ϕ) = -(θ + ϕ) + 2πm, where m is an integer.
Simplifying this equation, we get:
2θ + 2ϕ = 2πm
θ + ϕ = πm
Since θ and ϕ are angles of the complex numbers z and w respectively, θ + ϕ is also an angle of the complex number zw.
Therefore, we have zw = e^(i(θ + ϕ)) = e^(iπm), where m is an integer.
Recall that we are given that zw is not equal to -1.
If zw = e^(iπm) = -1, then m must be an odd integer.
But since zw is not equal to -1, m must be an even integer.
Therefore, we have proved that zw = e^(iπm) is a real number.
Now let's substitute this result into the expression (z + w)/(zw + 1):
(z + w)/(zw + 1) = (z + w)/(e^(iπm) + 1)
= (z + w)/(cos(πm) + i sin(πm) + 1)
= (z + w)/(cos(πm) + 1 + i sin(πm))
The denominator cos(πm) + 1 + i sin(πm) is the complex conjugate of the numerator (z + w). So, the expression (z + w)/(zw + 1) is indeed a real number.
Therefore, we have proved that if |z| = |w| = 1 and zw is not equal to -1, then the expression (z + w)/(zw + 1) is a real number.
To prove that (z + w)/(zw + 1) is a real number, we need to show that its imaginary part is equal to zero.
Let's start by expressing z and w in polar form. Since |z| = |w| = 1, we can write z = e^(iθ1) and w = e^(iθ2), where θ1 and θ2 are the arguments (angles) of z and w respectively.
Now, we can express zw as:
zw = e^(iθ1) * e^(iθ2) = e^(i(θ1 + θ2))
We know that zw is not equal to -1, so e^(i(θ1 + θ2)) is not equal to -1.
Next, let's simplify (z + w)/(zw + 1):
(z + w)/(zw + 1) = (e^(iθ1) + e^(iθ2))/(e^(i(θ1 + θ2)) + 1)
To find the imaginary part of the expression above, we can multiply both the numerator and the denominator by the complex conjugate of the denominator, e^(-i(θ1 + θ2)), which will eliminate the complex parts in the denominator:
(z + w)/(zw + 1) = [ (e^(iθ1) + e^(iθ2))/(e^(i(θ1 + θ2)) + 1) ] * [ e^(-i(θ1 + θ2))/e^(-i(θ1 + θ2)) ]
= [ (e^(iθ1) + e^(iθ2))*e^(-i(θ1 + θ2)) ] / [ (e^(i(θ1 + θ2)) + 1) * e^(-i(θ1 + θ2)) ]
= (e^(iθ1 - i(θ1 + θ2)) + e^(iθ2 - i(θ1 + θ2))) / (e^(i(θ1 + θ2) - i(θ1 + θ2)) + e^(-i(θ1 + θ2)))
Simplifying further, we have:
= (e^(-iθ2) + e^(-iθ1)) / (e^0 + e^(-i2(θ1 + θ2)))
= (e^(-iθ2) + e^(-iθ1)) / (1 + e^(-i2(θ1 + θ2)))
The denominator, 1 + e^(-i2(θ1 + θ2)), is real since the complex exponential term has an argument of -2(θ1 + θ2), which means the real part is 1 and the imaginary part is 0.
The numerator, e^(-iθ2) + e^(-iθ1), is also real since the complex conjugates of e^(-iθ2) and e^(-iθ1) are e^(iθ2) and e^(iθ1) respectively, and the sum of these conjugates is real.
Therefore, (z + w)/(zw + 1) is a real number since its imaginary part is 0.