Prove that if w,z are complex numbers such that |w|=|z|=1 and wz\=-1, then w+z/1+wz is a real number
I so far i have determined that w and z both have to equal less than one, but i cannot figure out how to prove this. if wz does not equal one it must mean that w and z both equal less than one. i'm not sure anything i have done is right if someone could look at this it would be amazing, thank you.
To prove that the expression (w + z)/(1 + wz) is a real number, we need to show that its imaginary part is zero.
Given that |w| = |z| = 1, we can express w and z in trigonometric form as follows:
w = cos(a) + i sin(a)
z = cos(b) + i sin(b)
where a and b are real numbers.
Now, let's compute wz:
wz = (cos(a) + i sin(a))(cos(b) + i sin(b))
= cos(a)cos(b) + i sin(a)cos(b) + i cos(a)sin(b) - sin(a)sin(b)
= cos(a)cos(b) - sin(a)sin(b) + i(sin(a)cos(b) + cos(a)sin(b))
= cos(a + b) + i sin(a + b)
Since wz ≠ -1, we have cos(a + b) + i sin(a + b) ≠ -1.
From the equation (w + z)/(1 + wz), let's work with the numerator first:
(w + z) = (cos(a) + i sin(a)) + (cos(b) + i sin(b))
= cos(a) + cos(b) + i(sin(a) + sin(b))
Now, we can rewrite the expression as:
(w + z)/(1 + wz) = (cos(a) + cos(b) + i(sin(a) + sin(b))) / (1 + cos(a + b) + i sin(a + b))
To prove that this expression is real, we need the imaginary part to be zero:
Im((w + z)/(1 + wz)) = Im((cos(a) + cos(b) + i(sin(a) + sin(b))) / (1 + cos(a + b) + i sin(a + b)))
To simplify the expression, we multiply both the numerator and the denominator by the conjugate of the denominator:
(w + z)/(1 + wz) = ((cos(a) + cos(b) + i(sin(a) + sin(b))) / (1 + cos(a + b) + i sin(a + b))) * ((1 + cos(a + b) - i sin(a + b)) / (1 + cos(a + b) - i sin(a + b)))
Expanding and simplifying the expression, we get:
(w + z)/(1 + wz) = [(cos(a) + cos(b) + i(sin(a) + sin(b)))(1 + cos(a + b) - i sin(a + b))] / [(1 + cos(a + b))^2 + sin^2(a + b)]
Using the distributive property and simplifying further, we have:
(w + z)/(1 + wz) = [cos(a) + cos(b) + cos(a + b) + cos(a)cos(b) + sin(a)sin(b) + i(sin(a) + sin(b) + sin(a + b) + cos(a)sin(b) - sin(a)cos(b))] / [(1 + cos(a + b))^2 + sin^2(a + b)]
Now, let's focus on the imaginary part:
Im((w + z)/(1 + wz)) = sin(a) + sin(b) + sin(a + b) + cos(a)sin(b) - sin(a)cos(b)
Since sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and sin(-x) = -sin(x), we can simplify further:
Im((w + z)/(1 + wz)) = sin(a) + sin(b) + sin(a + b) + cos(a)sin(b) - sin(a)cos(b)
= sin(a) + sin(b) + sin(a)cos(b) + cos(a)sin(b) - sin(a)cos(b)
= sin(a) + sin(b)
Since sin(x) is a real number for any real number x, we have proved that the expression (w + z)/(1 + wz) is a real number.
Note: In this proof, we used trigonometric form to express complex numbers. However, this could also be proven using algebraic manipulations.
The mean of four consecutive even numbers is 15.
The greatest of these numbers is a1.
The least of these numbers is