Prove that if $w,z$ are complex numbers such that $|w|=|z|=1$ and $wz\ne -1$, then $\frac{w+z}{1+wz}$ is a real number.

michaelcai
Dec 14, 2017

#1**+3 **

Let \(\displaystyle w = \cos\theta+ \imath \sin\theta \text{ and } z = \cos\phi + \imath\sin\phi\).

The bottom line will be \(\displaystyle 1+\cos(\theta+\phi)+\imath\sin(\theta + \phi)\).

Multiply top and bottom by the conjugate of this number and then pick out the imaginary part of the top line, showing that it's equal to zero, (therefore showing that the fraction is a real number).

You will need to use the expansion for sin(A + B) in the simplification.

Tiggsy

Guest Dec 15, 2017

#1**+3 **

Best Answer

Let \(\displaystyle w = \cos\theta+ \imath \sin\theta \text{ and } z = \cos\phi + \imath\sin\phi\).

The bottom line will be \(\displaystyle 1+\cos(\theta+\phi)+\imath\sin(\theta + \phi)\).

Multiply top and bottom by the conjugate of this number and then pick out the imaginary part of the top line, showing that it's equal to zero, (therefore showing that the fraction is a real number).

You will need to use the expansion for sin(A + B) in the simplification.

Tiggsy

Guest Dec 15, 2017

#3**+1 **

If you are aware of the way in which complex numbers are added multiplied and divided on an Argand diagram then yes it is possible to avoid much of the algebra.

Suppose that P and Q are points representing the complex numbers w and z, that O is the origin, then the complex number w+z will be at the point which is the fourth vertex of the parallelogram defined by OP and OQ. Because, in this case, w and z both lie on the unit circle, the parallelogram will be a rhombus and in that case its diagonal will bisect the angle POQ.

So, if arg(w) = theta and arg(z) = phi (and if, for convenience, phi > theta)

then arg(w + z) = theta + (phi - theta)/2 = (phi + theta)/2 ................... (1)

With multiplication, the moduli of the numbers are multiplied and the arguments added, that means that |wz| = 1

and arg(wz)= (phi + theta).

Applying the rule for addition again, (arg(1) = 0), so arg(1 + wz) = 0 + (phi + theta - 0)/2 = (phi + theta)/2 ................... (2)

With division, one modulus is divided by the other and one argument subracted from the other (the reverse of the operation for multiplication), so, from (1) and (2), arg[(w + z)/(1 + wz)] = 0. That means that the number lies on the real axis and so is real.

If you are going to solve this problem algebraically though, involving the circular functions sine and cosine is the best, easiest, way to go.

Tiggsy

Guest Dec 17, 2017

#4**0 **

I should be more specific. I need to find a way to solve it using the complex plane

michaelcai
Dec 17, 2017