Question

show that if the point z,-iz,I are collinear then Z lies on a circle​

Answer 1

Solution:

Let, z=x + i y,

then, - i z=-i(x+i y)=-i x -i²y=y - ix

1= 1 + i 0

So, if three points are collinear, it means area of triangle must be equal to zero.

$\begin{vmatrix}x &y &1 \\ y &-x &1\\ 1 &0  &1 \end{vmatrix}=0$

Interchanging first row and third row

$-\begin{vmatrix}1 &0  &1\\ y &-x  &1\\ x &y  &1 \end{vmatrix}=0$

→1 [-x-y]+1(y²+x²)=0

→ x²+y²-x-y=0

Which is the equation of circle $(x-\frac{1}{2})^2+(y-\frac{1}{2})^2=[\frac{1}{\sqrt{2}}]^2$ , having center $(\frac{1}{2},\frac{1}{2}) {\text{radius is}}\frac{1}{\sqrt{2}}$.

Which shows z lies on a circle.