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

Question of the week #2 - the answer

Question of the week #2

   Last week's question was dealing with two different standard forms of the hyperbola equation.
The answer to the question, is that these two different forms are equivalent descriptions and this can be shown by a counterclockwise rotation $(x,y)\rightarrow(x',y')$ of the planar coordinate system, through an angle $φ=π/4 (rad)$.
   Here are some more details (in the example that follows $a > 0$):

Question of the week #2

This week's question comes form $2d$ analytic geometry, and deals more specifically with the coordinate equations of the hyperbola:

"In a given coordinate system $(x,y)$ the equation $y=\frac{a}{x}$, $a \in \mathbb{R}$ represents an hyperbola. Show that under a suitable change of coordinates i.e. under a suitable transformation $(x,y)\rightarrow(x',y')$ the same hyperbola becomes $x'^{2} - y'^{2}=2a$"

check out the pdf version here.

Waiting again for your thoughts, ideas and answers till next week!