Radius of circle with complex equation

The expression $\frac{z-i}{z-1}$ represents the following operation in complex space:

Consider two vectors $z-i$ and $z-1$ . Suppose the magnitudes are the same. Then, the operation of division is subtracting their angles. If the angles are 90 degrees apart then the result is purely complex. One place where this occurs is at the origin.

Since the circle is centered at $(0.5, 0.5i)$ we simply need to find the distance between the origin and that point.

Letting $z=x+iy$ , we can develop easily $\mathrm{Re}\left(\frac{z-i}{z-1}\right)=\frac{x^2+y^2-x-y}{(x-1)^2+y^2}$ Thus such points $z$ satisfy the equation $x^2+y^2-x-y=0$ , and this is the equation of a circle of radius $\frac{\sqrt{2}}{2}$ , centered at the point $(\frac{1}{2},\frac{1}{2})$ , so we obtain that the solution is $\frac{\sqrt{2}}{2}\approx 0.707$ .

In fact, the hypothesis that the center is $\frac{1}{2}(1+i)$ is not necessary.

If radius of the circle is r

| z - 1/2(1+i) | = r

As ( z - i) / ( z - 1) is purely imaginary ,

( z - i) / ( z - 1) = a i

Let's take z = x + y i , x and y are non zero

( z - i) / ( z - 1) = a i

[ x + ( y - 1) i ] / [ ( x - 1 ) + y i ] = a i

[ x ( x - 1 ) + y ( y - 1 ) ] / [ ( x -1 ) ^ 2 + y ^ 2 ] = 0 as ( z - i) / ( z - 1) is purely imaginary

x ( x - 1 ) + y ( y - 1 ) = 0

As x and y are non zero

x = 1 and y = 1

z = 1 + i

| z - 1/2(1+i) | = r

| ( 1 + i) - 1/2 ( 1 + i ) | = r

1/2 | 1+ i | = r

r = 0 . 707

If above expression is purely imaginary, then , real part of z must be 1 ( see the denominator) and imaginary part must be i (see the numerator).

If any real number in denominator is left, then it will surely give a real part.

Similarly, if there is another imaginary part except i, then in numerator one real number is surely left, which will give a real part.