Question 1207779
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
<pre>
{{{matrix(1,8,

Let, z, ""="", x+i*y, and, w,""="",u+i*v)}}}

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

{{{abs(z)=1}}}
{{{sqrt(x^2+y^2)=1}}}
{{{x^2+y^2=1}}}
{{{x^2-(-y^2)=1}}}
{{{x^2-(-1*y^2)=1}}}
{{{x^2-(i^2y^2)=1}}}
{{{(x-i*y)(x+i*y)=1}}}
{{{x-i*y=1/(x+i*y)}}}
</pre>
(b) Prove that ={z + w}/{zw + 1} is a real number.<pre>

{{{(z + w)/(z*w + 1)}}}

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

{{{(z + w)/(z*w + 1)}}}{{{""*""}}}{{{(1/z + 1/w)/(1/z + 1/w)}}}

{{{(1+w*expr(1/z)+z*expr(1/w)+1)/(w+z+1/z+1/w)}}}

{{{matrix(1,8,

Write, z, ""="", x+i*y, and, w,""="",u+i*v)}}}

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

{{{matrix(1,8,write, 1/z, ""="", x-i*y, and, 1/w,""="",u-i*v)}}}

{{{(1+(u+i*v)*(x-i*y)+(x+i*y)*(u-i*v)+1)/((u+i*v)+(x+i*y)+(x-i*y)+(u-i*v))}}}

{{{(2+u*x-i*u*y+i*v*x+x*u+i*xv+i*y*u-i^2*y*v)/(2*u+2*x)}}}

{{{(2+u*x-i*u*y+i*v*x+x*u+i*xv+i*y*u+y*v)/(2*u+2*x)}}}

The imaginary terms cancel, and we have:

{{{(2+u*v+x*u)/(2*u+2*x)}}}

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

Edwin</pre>