Let `z_1=10+6i` and `z_2=4+6idot` If `z` is any complex number such that the argument of `((z-z_1))/((z-z_2))` is `pi/4,` then prove that `|z-7-9i|=3sqrt(2)` .

Explanation

`arg((z-z_(1))/(z-z_(2)))=(pi)/(4)`

Therefore, z lies on the major are whose centre is `z_(0)("say").`
Applying rotation at `z_(0)` , we have
`(z_(0)-(10+6i))/(z_(0)-(4+6i))=(|z_(0)-(10+6i)|)/(|z_(0)-(4+6i)|)e^(i(pi)/(2))`
`implies(z_(0)-(10+6i))/(z_(0)-(4+6i))=i`
`impliesz_(0)-10-6i=iz_(0)-4i+6`
`impliesz_(0)=7+9i`
Thus, centre is `7+9i` and z is any point on the arc.
Hence, for any complex number z on the arc, `|z-(7+9i)|=|10+6i-(7+9i)|=3sqrt(2)`