Question

If `z=z_0+A( z -( z )_0), w h e r eA` is a constant, then prove that locus of `z` is a straight line.

Answer 1

`bar(z)=bar(z)_(0)+A(z-z_(0))`
`impliesAz-bar(z)-Az_(0)+bar(z)_(0)=0" "(1)`
`impliesbar(A)bar(z)-z-bar(A)bar(z)_(0)+z_(0)=0" "(2)`
Adding (1) and (2), we get
`(A-1)z+(bar(A)-1)bar(z)-(Az_(0)+bar(A)bar(z)_(0))+z_(0)+bar(z)_(0)=0`
This is of the form `bar(a)z+abar(z)+b=0," where "a=bar(A)-1" and "b=-1(Az_(0)+bar(A)bar(z)_(0))+z_(0)+bar(z)_(0)inR.` Hence, locus of z is a straight line.