Question of the week #3: Given a complex number $z$, determine its locus, given that $w=\frac{i}{z^{2}+1}$ belongs on the real axis (i.e. $w$ is a real number)
Solution: Substituting $z=x+iy$ with $x,y \in \mathbb{R}$ and denoting by $\bar{z}=x-iy$ the complex conjugate, we have:
$$
\begin{array}{c}
w \in \mathbb{R} \Leftrightarrow w=\bar{w} \Leftrightarrow \frac{i}{z^{2}+1}=\overline{\frac{i}{z^{2}+1}} \Leftrightarrow \\
\\
\Leftrightarrow \frac{i}{z^{2}+1}=-\frac{i}{\bar{z}^{2}+1} \Leftrightarrow \bar{z}^{2}+1= -z^{2}-1 \Leftrightarrow \\
\\
\Leftrightarrow \bar{z}^{2}+z^{2}+2=0 \Leftrightarrow \\
\\
\Leftrightarrow x^{2}-y^{2}-2ixy+x^{2}-y^{2}+2ixy+2=0 \Leftrightarrow \\
\\
\Leftrightarrow 2x^{2}-2y^{2}+2=0 \Leftrightarrow y^{2}-x^{2}=1
\end{array}
$$
which is an isosceles hyperbola:
with the vertices $A(0,1)$ and $B(0,-1)$ excluded, because for these values we have $z=i$ and $z=-i$ respectively, thus $z^{2}+1=0$.
Solution: Substituting $z=x+iy$ with $x,y \in \mathbb{R}$ and denoting by $\bar{z}=x-iy$ the complex conjugate, we have:
$$
\begin{array}{c}
w \in \mathbb{R} \Leftrightarrow w=\bar{w} \Leftrightarrow \frac{i}{z^{2}+1}=\overline{\frac{i}{z^{2}+1}} \Leftrightarrow \\
\\
\Leftrightarrow \frac{i}{z^{2}+1}=-\frac{i}{\bar{z}^{2}+1} \Leftrightarrow \bar{z}^{2}+1= -z^{2}-1 \Leftrightarrow \\
\\
\Leftrightarrow \bar{z}^{2}+z^{2}+2=0 \Leftrightarrow \\
\\
\Leftrightarrow x^{2}-y^{2}-2ixy+x^{2}-y^{2}+2ixy+2=0 \Leftrightarrow \\
\\
\Leftrightarrow 2x^{2}-2y^{2}+2=0 \Leftrightarrow y^{2}-x^{2}=1
\end{array}
$$
which is an isosceles hyperbola:
with the vertices $A(0,1)$ and $B(0,-1)$ excluded, because for these values we have $z=i$ and $z=-i$ respectively, thus $z^{2}+1=0$.
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