Complex Number Problem! (1 Viewer)

[Lindurr]

Show that Re(1-z/1+z) = 0 for any complex z with |z|=1

Thanks heaps for the help!

 

[nightweaver066]

 

[deswa1]

<a href="http://www.codecogs.com/eqnedit.php?latex=z=x@plus;iy\\ Re(\frac{1-z}{1@plus;z})=Re(\frac{(1-x)-iy}{(1@plus;x)@plus;iy})\\ Re(\frac{1-z}{1@plus;z})=Re(\frac{(1-x)(1@plus;x)-iy(1@plus;x)-iy(1-x)-iy(-iy)}{(1@plus;x)^2@plus;y^2})\\ Re(\frac{1-z}{1@plus;z})=Re(\frac{1-x^2-iy-iyx-iy@plus;iyx-y^2}{(1@plus;x)^2@plus;y^2})\\ Re(\frac{1-z}{1@plus;z})=\frac{1-x^2-y^2}{(1@plus;x)^2@plus;y^2)}\\ Re(\frac{1-z}{1@plus;z})=0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?z=x+iy\\ Re(\frac{1-z}{1+z})=Re(\frac{(1-x)-iy}{(1+x)+iy})\\ Re(\frac{1-z}{1+z})=Re(\frac{(1-x)(1+x)-iy(1+x)-iy(1-x)-iy(-iy)}{(1+x)^2+y^2})\\ Re(\frac{1-z}{1+z})=Re(\frac{1-x^2-iy-iyx-iy+iyx-y^2}{(1+x)^2+y^2})\\ Re(\frac{1-z}{1+z})=\frac{1-x^2-y^2}{(1+x)^2+y^2)}\\ Re(\frac{1-z}{1+z})=0" title="z=x+iy\\ Re(\frac{1-z}{1+z})=Re(\frac{(1-x)-iy}{(1+x)+iy})\\ Re(\frac{1-z}{1+z})=Re(\frac{(1-x)(1+x)-iy(1+x)-iy(1-x)-iy(-iy)}{(1+x)^2+y^2})\\ Re(\frac{1-z}{1+z})=Re(\frac{1-x^2-iy-iyx-iy+iyx-y^2}{(1+x)^2+y^2})\\ Re(\frac{1-z}{1+z})=\frac{1-x^2-y^2}{(1+x)^2+y^2)}\\ Re(\frac{1-z}{1+z})=0" /></a>

Since |z|=1