SOLUTION: Let z and w be complex numbers such that |z| = |w| = 1 and zw is not equal to -1. (a) Prove that conjugate {z} = 1/z and conjugate{w} = 1/w (b) Prove that ={z + w}/{zw + 1} i

It is well known fact that the product  . = |z|^2.


Indeed,  if z = a + bi, then   = a - bi,  and, therefore,


    . = (a+bi)*(a-bi) =  =  =  = .


But we are given that  |z| = 1;  hence,   = 1.


Thus . = 1.


Hence,   = .


It is what was required to prove.


For   =   the proof is the same.


It is, actually, the same statement as   = ,  but expressed using letter w instead of z.

z and w are unit complex numbers, in the sense that their modules are equal to 1 (given).


Hence, in geometric presentation in complex plane, z and w are adjacent side of a rhombus.

Then, according to the parallelogram rule of adding complex numbers, the sum z+w
is the diagonal of the rhombus on side z and w.

According to properties of a rhombus, its diagonal is a bisector of the angle 
between vectors z and w.


Therefore, for the argument of the complex number  z+w  we can write

    arg(z+w) =   mod .    (1)



Now consider complex number zw.  As the product of two complex unit numbers, it is a unit
complex number, too, i.e. has the modulus 1, and its argument is the sum of arguments z and w

    arg(zw) = arg(z) + arg(w)  mod  .    (2)



Also, notice that  "1"  is a complex unit number too, with the argument arg(1) = 0.

Applying the same proof as in (1),  we get

     arg(zw+1) =  =  =   mod .    (3)

    
Using (2), we can continue line (3) this way

     arg(zw+1) =   mod .   (4)


Comparing (1) with (4), we see that

    arg(z+w) = arg(zw+1)  mod .


It means that complex numbers  z+w  and  (zw+1)  are either co-directed vectors or oppositely directed vectors.


In any case, it implies that the number    has the argument 0 or   mod ,
i.e. is a real number.


At this point, the proof  (b)  is complete.

where x,y,u,v are real numbers.

Multiply numerator and denominator by the sum of the reciprocals of z and w:







and since we have proved in part (a) that their reciprocals are their
conjugates,









The imaginary terms cancel, and we have:



which is real since 2,u,v,x,y are all real.

Edwin

Here's another way to do the (a) part:

Since  |z| = |w| = 1

This problem is about points on the unit circle, where the horizontal axis
is the real axis and the vertical axis is the imaginary axis.

  


All complex numbers on the unit circle have modulus 1., the radius of the
unit circle.

The conjugate of any complex number on the unit circle is obviously its
reflection in the horizontal (real) axis.  It is also obvious that the 
argument (angle) of the conjugate is the negative of the argument (angle).

The reciprocal of any complex number is that complex number when multiplied by
the complex number, gives the result 1, which is the complex number 1+0i,
thought of as the point (1,0).

DeMoivre's theorem tell us that to multiply two complex numbers, we
multiply their moduli and add their arguments.  

So when we multiply the arguments of any two complex numbers on the unit circle,
we'll get 1.
 
It's also obvious to see that when you add the arguments, the
red and green arcs above, you will get 0 for the argument.   

So it's obvious that the reciprocal and conjugate of any complex number of the
unit circle are the same. 

This is what the (a) part asks you to prove:   

We can write z and w in trigonometric form:





We will show that the reciprocal equals the conjugate:







Since the denominator equals 1, 



So the reciprocal of a complex number on the unit circle is also its
conjugate.

Edwin