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.
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:
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.