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.
you've been posting this over and over. Did you think of google? Many problems posted here have appeared elsewhere. My first hit on google discusses this problem and one similar to it. Scroll down about halfway and you will find what you need.
http://math.stackexchange.com/questions/427663/prove-if-z-w-1-and-1zw-neq-0-then-zw-over-1zw-is-a-real
To prove that the expression (z + w)/(zw + 1) is a real number, we need to show that the imaginary part of the expression is equal to zero.
Let's assume z = x1 + y1*i and w = x2 + y2*i, where x1, x2, y1, and y2 are real numbers.
Using the properties of complex conjugates, we can write the expression as:
(z + w)/(zw + 1) = (x1 + y1*i + x2 + y2*i)/((x1 + y1*i)*(x2 + y2*i) + 1)
Expanding the denominator, we get:
(z + w)/(zw + 1) = (x1 + y1*i + x2 + y2*i)/(x1*x2 + x1*y2*i + x2*y1*i - y1*y2 + 1)
Now, we need to prove that the imaginary part of this expression is equal to zero.
The imaginary part of a complex number is given by the coefficient of 'i', which is y1*x2 + y2*x1.
So, we need to show that y1*x2 + y2*x1 equals zero.
Since |z| = |w| = 1, we have x1^2 + y1^2 = 1 and x2^2 + y2^2 = 1.
We can rewrite the equation y1*x2 + y2*x1 = 0 as:
(y1*x2 + y2*x1) = (y1*x2 + y2*x1)(x1^2 + y1^2)(x2^2 + y2^2)
Expanding and simplifying, we get:
(y1*x2 + y2*x1) = (x1^2*y2 + x2^2*y1 + x1*y1*(y2 - x2) + x2*y2*(x1 - y1))
Since x1^2 + y1^2 = 1 and x2^2 + y2^2 = 1, we can replace them in the equation:
(y1*x2 + y2*x1) = x1*y1*(y2 - x2) + x2*y2*(x1 - y1)
Since we know that zw is not equal to -1, we can conclude that z and w are not complex conjugates of each other. Therefore, x1, y1, x2, and y2 cannot be equal to zero at the same time.
Since x1, y1, x2, and y2 are real numbers, the imaginary part of the expression (z + w)/(zw + 1) is equal to zero.
Hence, we have proved that (z + w)/(zw + 1) is a real number.