Question

If $z=2+k+i\surd (3-k^2)$ where $k$ is real such that $k^2<3$, prove that $\left|\frac{z+1}{z-1}\right|$ is independent of $k$. Also prove that the locus of the point $z$ for different values of $k$ is a part of the circle. What is its centre and radius?

Answer 1

$\frac{z+1}{z-1}=\frac{3+k+i\sqrt{(3-k^2)}}{1+k+i\sqrt{(3-k^2)}}$
$\therefore \left|\frac{z+1}{z-1}\right|=\left|\frac{z+1}{z-1}\right|=\frac{{(3+k)}^2+3-k^2}{{(1+k)}^2+3-k^2}$
$=\frac{6(k+2)}{2(k+2)}=3$
Form given value of $z$
$x=2+k,y=\sqrt{3-k^2}$
$(x-2{)}^2=k^2,y^2=3-k^2$
$\therefore (x-2{)}^2+(y-0{)}^2=3$
or $(x-2{)}^2+y^2={\left(\sqrt{3}\right)}^2\therefore$ $(2,0)$, $\sqrt{3}$