locus

why nobody is replying

Can u please post the question again ?
The statements seem to be quite jumbled up.....

Let z = x + iy.
(x + iy)(1 + i) = x + ix + iy - y
= x - y + (x + y)i

The argument of this would be tan^-1((x + y)/(x - y)).

I'll solve the rest of this when I get back from walk...others can try after applying the modulus condition to x + iy.
|z| = √x² + y²
|z - 2| = √(x - 2)² + y²

Use the above two equations as your conditions(or region equations) to apply on the argument = 3pi/4 or -pi/4. You'll get your answer.
Square both sides of inequality to make it easier.

Ok back now...first let's find the region conditions from the inequalities.
The first region is |z| ≤ |z - 2|
=> x² + y² ≤ x² + y² + 4 - 2x
=> 2x - 4 ≤ 0
=> x ≤ 2
Similarly the second region is x > 2.

So our expression arg[z(1 + i)] = 3pi/4 if x ≤ 2
and -pi/4 if x > 2.

Now for x ≤ 2,
tan^-1[x + yx - y] = 3pi/4
=> x + y = y - x as tan(3pi/4) = -1.
=> x = 0 or the Y-axis.

For x > 2,
tan^-1[x + yx - y] = -pi/4
Which is again x = 0.
So for all regions, x = 0 or the Y-axis is the locus.

I guess the answer is (C) then?