Show that the equation of a circle passings through the origin and having intercepts a and b on real and imaginary axis, respectively, on the argand plane is `Re ((z-a)/(z-ib)) = 0`

Explanation

According to the question, we have circle as shown in the figure.

Clearly, complex numbers 'a' and 'ib' are and points of diameter. So, for any complex number z on the circle, we have
`arg((z-a)/(z-ib))=+-(pi)/(2)`
Thus, `(z-a)/(z-ib)` is purely imaginary.
`:." ""Re"((z-a)/(z-ib))=0`