At the moment I only have time to say the following:
This is not a circle. It's the major arc of a circle.
The arc joins z = 1 and z = -1 (these points are NOT included in the locus) and lies on a circle of radius and centre located on the perpendicular bisector of the chord joining z = 1 and z = -1 at a distance of 1 from this chord.
This will give the circle upon which the locus lies. But the locus itelf is only a subset of this circle.
In fact, since cannot have an argument of unless its real and imaginary parts are BOTH positive, the solution has to satisfy the following two implied restrictions:
.... (1)
.... (2)
The locus is therefore the major arc of the circle I have previously described and which can be found by solving . As well as this algebraic approach suggested by Prove It, the parameters of this circle can also be found from geometric considerations:
is the locus of the intersection point P of the rays and under the restriction .
It's not hard to show that these two rays only intersect when , in which case the angle between these two rays is always .
Using the the well known theorem that all angles subtended by the same chord of a circle at the circumference are equal, it follows that the locus is the major arc joining but not including z = -1 and z = 1. Using simple trigonometry on the right triangle AOC (where a is the point representing , O is the origin and C is the centre of the circle) it is easily seen that the arc lies on a circle with the following parameters:
1. Radius = .
2. Centre C located on the Imaginary axis at a distance from the origin.
Several other approaches to finding the locus are possible, including a vector approach (using the dot product) and a parametric method.