Let (z) represent the principal argument of the complex number z.

Question

Let $\arg(z)$ represent the principal argument of the complex number $z$. Then $|z|=3$ and $\arg(z-1)-\arg(z+1)=\dfrac\pi4$ intersect:

Accepted Answer

$|z|=3$ is the circle $x^2+y^2=9$ centred at the origin, radius 3.

For the second locus, let $z=x+iy$:
$$\tan^{-1}\frac y{x-1}-\tan^{-1}\frac y{x+1}=\frac\pi4 \implies \frac{2y}{x^2+y^2-1}=1 \implies x^2+(y-1)^2=2,$$

a circle centred $(0,1)$ with radius $\sqrt2$.

Distance between centres $=1$. Sum of radii $=3+\sqrt2\approx4.41$; difference of radii $=3-\sqrt2\approx1.59$. Since the distance between centres ($1$) is less than the difference of the radii, one circle lies entirely inside the other without touching --- so the two curves do not intersect anywhere.

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