Question

If $a>0,|z|=a$, then find the real part of $\left(\frac{z-a}{z+a}\right)$.

Answer 1

Given $|z|=a$ let $z=x+iy\Rightarrow \sqrt{x^2+y^2}=a...\left(1\right)$

$\frac{z-a}{z+a}=\frac{(x-a)+iy}{(x+a)+iy}=\frac{\left[\right(x-a)+iy]\left[\right(x+a)+iy(-1\left)\right]}{\left[\right(x+a)+iy]\left[\right(x+a)-iy]}$ [rationalizing]

$=\frac{(x^2-a^2)-iy(x-a)+iy(x+a)+y^2}{(x+a{)}^2+y^2}$

$=\frac{iy\left[\right(x+a)-(x-a\left)\right]}{y^2+(a+x{)}^2}$ $[\because x^2+y^2=a^2]$

$=\frac{2ayi}{y^2+(x+a{)}^2}$

$\therefore Re\left(\frac{z-a}{z+a}\right)=0$