SOLUTION: The complex numbers z and w satisfy |z| = |w| = 1 and zw is not equal to -1. (a) Prove that \overline{z} = {1}/{z} and \overline{w} = {1}/{w}. (b) Prove that {z + w}/{zw + 1} i

Since |z| = 1 and |w| = 1, it means that z and w are the unit vectors of the length 1: their endpoints lie on the unit circle.

To calculate (z+w), apply the parallelogram's rule.  Since the sides of the parallelogram on vectors z and w are equal,

the parallelogram is a rhombus.  The sum (z+w) is the diagonal of the parallelogram, and since parallelogram is a rhombus,

arg(z+w) is  EITHER   OR  .   Here arg() means the argument of complex number.



The first case   arg(z+w) =  is when the angle between vectors z and w is less than  :  |arg(z)-arg(w)| <= .

The second case  arg(z+w) =  is when the angle between vectors z and w is greater than  :  |arg(z)-arg(w)| > .

Notice that by the modulo of ,  arg(z+w) =    always.



Further, the product zw is the unit vector, again, so the same formulas are applicable to vectors zw and 1 = (1,0).

Notice that arg(zw) = arg(z) + arg(w), so arg(zw+1) is EITHER ,  or  , depending
on the angle between vectors zw and 1 = (1,0).

But in any case,  the vectors (z+w) and (zw+1) are EITHER parallel OR anti-parallel (opposite).

By the modulo of ,  arg(zw+1) =    always.



By the rule of argument of quotient for complex numbers, it means that the ratio   is real number.

This real number is EITHER positive (when the vectors (z+w) and (zw+1) are parallel), 

                      OR   negative (when the vectors (z+w) and (zw+1) are anti-parallel). 


At this point, the proof is completed.