 # I am attempting to determine the locus of z for arg &#x2061;<!-- ⁡ --> ( z 2 junoonib89p4 2022-05-07 Answered
I am attempting to determine the locus of z for $\mathrm{arg}\left({z}^{2}+1\right)$. I do know that since ${z}^{2}+1=\left(z+i\right)\left(z-i\right)$, I could write $\mathrm{arg}\left({z}^{2}+1\right)=\mathrm{arg}\left(z+i\right)+\mathrm{arg}\left(z-i\right)$ 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!
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Let $\phantom{\rule{thinmathspace}{0ex}}z=x+iy\phantom{\rule{thinmathspace}{0ex}}$ with $\phantom{\rule{thinmathspace}{0ex}}x,y\in \mathbb{R}\phantom{\rule{thinmathspace}{0ex}}$, and let $\phantom{\rule{thinmathspace}{0ex}}\mathrm{arg}\left({z}^{2}+1\right)=\alpha \phantom{\rule{thinmathspace}{0ex}}$.

1. If $\phantom{\rule{thinmathspace}{0ex}}\alpha =±\pi /2\phantom{\rule{thinmathspace}{0ex}}$, then $\phantom{\rule{thinmathspace}{0ex}}{z}^{2}+1\phantom{\rule{thinmathspace}{0ex}}$ must lie on the imaginary axis, so $\phantom{\rule{thinmathspace}{0ex}}{z}^{2}+1=ib\phantom{\rule{thinmathspace}{0ex}}$ for some $\phantom{\rule{thinmathspace}{0ex}}b\in \mathbb{R}\phantom{\rule{thinmathspace}{0ex}}$, which in cartesian coodinates translates to $\phantom{\rule{thinmathspace}{0ex}}{x}^{2}-{y}^{2}+2ixy+1=ib\phantom{\rule{thinmathspace}{0ex}}$, so the locus is part of the hyperbola $\phantom{\rule{thinmathspace}{0ex}}{x}^{2}-{y}^{2}+1=0\phantom{\rule{thinmathspace}{0ex}}$.
2. If $\phantom{\rule{thinmathspace}{0ex}}\alpha \ne ±\pi /2\phantom{\rule{thinmathspace}{0ex}}$, then $\phantom{\rule{thinmathspace}{0ex}}a=\mathrm{tan}\left(\alpha \right)=\mathrm{Im}\left(z\right)/\mathrm{Re}\left(z\right)\phantom{\rule{thinmathspace}{0ex}}$ is well defined, and the equation becomes:
$\frac{2xy}{{x}^{2}-{y}^{2}+1}=a\phantom{\rule{1em}{0ex}}\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{1em}{0ex}}a{x}^{2}-2xy-a{y}^{2}+a=0$
This is a quadratic equation with discriminant $\phantom{\rule{thinmathspace}{0ex}}{2}^{2}-4\cdot a\cdot \left(-a\right)>0\phantom{\rule{thinmathspace}{0ex}}$, 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 $\phantom{\rule{thinmathspace}{0ex}}\alpha \phantom{\rule{thinmathspace}{0ex}}$.