Locus corresponding to sum of two arguments in Argand diagram?

Question

I am attempting to determine the locus of z for $\arg(z^2+1)$. I do know that since $z^2+1 = (z+i)(z-i)$, I could write $\arg(z^2+1) = \arg(z+i) + \arg(z-i)$ but I don’t really know how to figure out the locus of the points from here. How should I approach this? Is there a relationship with the difference in two arguments (arc of a circle)? Thank you!

Answer 1

Let $\,z=x+iy\,$ with $\,x,y \in \mathbb{R}\,$, and let $\,\arg(z^2+1)=\alpha\,$.

• If $\,\alpha = \pm \pi/2\,$, then $\,z^2+1\,$ must lie on the imaginary axis, so $\,z^2+1=i b\,$ for some $\,b \in \mathbb{R}\,$, which in cartesian coodinates translates to $\,x^2-y^2+2ixy+1=ib\,$, so the locus is part of the hyperbola $\,x^2-y^2+1=0\,$.

• If $\,\alpha \ne \pm \pi/2\,$, then $\,a = \tan(\alpha) = \operatorname{Im}(z) / \operatorname{Re}(z)\,$ is well defined, and the equation becomes: $$ \frac{2xy}{x^2-y^2+1} = a \quad\iff\quad a x^2 -2 xy - a y^2 + a = 0 $$ This is a quadratic equation with discriminant $\,2^2 - 4 \cdot a \cdot (-a) \gt 0 \,$, so the locus is again part of a hyperbola.

In both cases, the actual locus is the part of the hyperbola lying in the quadrant determined by $\,\alpha\,$.