## Tuesday, December 24, 2013

### Question of the week #3 - the answer

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 $z^{2}+1=0$.