Ooh too Complex

Let $Z=z$ , $Z_1=z_1$ , and $Z_2=z_2$ be three points on the complex plane.

Since $\text{arg}\left(\dfrac{z-z_1}{z_2-z}\right)=\dfrac{\pi}{2}$ , then $\angle Z_1ZZ_2=90^{\circ}$ .

Thus, the locus of $Z$ is on the circle with diameter $Z_1Z_2$ .

We are trying to find the distance from $5-i$ to $Z$ . However, note that $5-i = \dfrac{Z_1+Z_2}{2}$ . Thus, this is the center of the locus of $Z$ , and we are simply trying to find the radius of it.

Since $Z_1Z_2=\sqrt{2^2+4^2}=2\sqrt{5}$ , then the radius of the locus is $\sqrt{5}$ and our answer is $\boxed{5}$

Let:-

$z=x+iy$

Since $arg(\frac{z-z_1}{z_2-z})=0$ it must lie on imaginary axis hence $Re\frac{z-z_1}{z_2-z}=0......(1)$

Putting the values of $z,z_1,z_2$ in equation (1) gives:-

$Re\frac{(x-6)+i(y-1)}{(4-x)-i(y+3)}=0$

Rationalizing the above equation give and setting real part of above equation equal to zero yeilds:-

$\frac{(x-6)(4-x)-(y-1)(y+3)}{(4-x)^2+(y+3)^2}=0$

Multiply both sides by denominator and simplify:-

$x^2+y^2-10x+2y+21=0......(2)$

Also:-

$|z-5+i|=|(x-5)+i(y+1)|$

$=\sqrt{x^2+y^2-10x+2y+26} = \sqrt{x^2+y^2-10x+2y+21+5} = \sqrt{5}$

Let z = a + bi

$Z = \frac{z-z_1}{z_2-z} = \frac{a + bi - 6 - i}{4-3i-6-i} = \frac{(a-6) + (b-1)i}{-2-4i}=-\frac{1}{2}\frac{(a-6) + (b-1)i}{1+2i}$

Multiply with 1-2i above an below we get

$Z = -\frac{1}{2}\frac{[(a-6) + (b-1)i](1+2i)}{-3}$

$Z = \frac{1}{6}[(a+2b - 8) + (-2a+b+11)i]$

$arg Z = \frac{\pi}{2} = \theta$

$Z = r (\cos \theta + i\sin \theta)$ $Z = r (\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})$ $Z = r i$

We now have $a + 2b - 8 = 0$ $-2a + b + 11 = 1$

With solutions a = 6 , b= 1 and z = 6 + i

$| z - (5 - i)| = \sqrt{(a-5)^2 + (b+1)^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$

so m is $\boxed{5}$