Rewrite as arg(z-5) - arg(z-3) = π/6 and let arg(z-5) = b and arg(z-3) = a, so that b-a = π/6Since we know that each argument makes a half line (starting at (3,0) for angle a, (5,0) for angle b) the half lines must intersect at a point P which is on the locus of z. The angle formed by this intersection must be equal to b-a = π/6 since the exterior angle in a triangle (in this case b) is equal to the sum of the interior angles (in this case a and π/6).We know from circle theorems that the angles subtended at the circumference in the same segment are always equal. Hence we can deduce that since the angle formed by the intersection is constant (equal to π/6) as b and a both vary, the locus of z must be an arc of a circle from x=3 to x=5 for y>0 (since the angle is positive).
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