Interpreting the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ as the unit circle stretched horizontally by $a$ and vertically by $-bi$ A hyperbola can be written in the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$
But this is same as $$\left(\frac{x}{a}\right)^2 + \left(i\frac{y}{b}\right)^2 = 1$$
The motivation is as follows. I thought of ellipse with equation $$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1$$ as being a circle stretched by a factor of $a$ in the direction of $x$ and by a factor of $b$ in the direction of $y$.
But what does it really mean, in the context of a hyperbola, to stretch by $$\frac{b}{i} \quad(= -bi) \ ?$$
 A: I guess that when you are considering the ellipse as a stretched circle you think of both objects as point sets in the real plane $\mathbb{R}^2$. To interpret the "stretching factor" $-bi$ you have to change your point of view and more generally consider the complex plane $\mathbb{C}^2$. From a "real point of view" $\mathbb{C}^2$ equals 4-dimensional real space $\mathbb{R}^4$: every coordinate is detemined by a real and an imaginary part. Geometric intuition is challenged ...
Lets look at the "circle" $x^2+y^2=1$ in $\mathbb{C}^2$ and split the coordinates into real and imaginary part: $x=x_1+ix_2$, $y=y_1+iy_2$. Splitting the circle equation accordingly leads to:
\begin{array}{rcl}
x_1^2-x_2^2+y_1^2-y_2^2 &=& 1 \\
x_1x_2+y_1y_2 &=& 0
\end{array}
These equations determine a 2-dimensional point set (a surface) $S$ in $\mathbb{R}^4$ that intersects the real plane $\{(x_1,0,y_1,0) : x_1,y_1\in\mathbb{R}\}$ in the ordinary circle.
Multiplying the complex coordinate $(x,y)$ with the factors $\frac{1}{a}$ and $\frac{1}{b}$ amounts to multiplying the real coordinates $x_1,x_2$ with the first factor and the real coordinates $y_1,y_2$ with the second. Nothing particularly new happens here: the surface $S$ is "stretched" giving a new surface $S^\prime$ that intersects the plane $\{(x_1,0,y_1,0) : x_1,y_1\in\mathbb{R}\}$  in the ordinary ellipse.
Now multiply the complex coordinates by $\frac{1}{a}$ and $\frac{i}{b}$. For the real coordinates $x_1,x_2$ nothing new happens, just stretching. For $y=y_1+iy_2$ however one gets
$$
\frac{i}{b}y=-\frac{1}{b}y_2+\frac{i}{b}y_1
$$
which means that the points in the plane $\{(0,0,y_1,y_2) : y_1,y_2\in\mathbb{R}\}$ are not only stretched by the factor $b$ but rotated by $90$ degrees.
This whole operation transforms $S$ into a new surface $T$ given by the equation
$$
(\frac{x}{a})^2+(\frac{y}{-bi})^2=1.
$$
What is the intersection of $T$ with the real plane $\{(x_1,0,y_1,0) : x_1,y_1\in\mathbb{R}\}$? Precisely the hyperbola.
Addition to the original answer: your question concerning how a rotation of $S$ leads to a non-closed shape in the real plane actually hints at the problem of understanding the shape of $S$ as a whole. This could be achieved by determining an injective differentiable mapping $f:S\rightarrow\mathbb{R}^3$ such that the left inverse of $f$ is differentiable too (significantly weaker: differentiability could be replaced by continuity). I have no such mapping at hand and don't even know whether it exists.
However one can easily embed parts of $S$ into $\mathbb{R}^3$: lets consider the subset $U:=\{(x_1,x_2,y_1,y_2)\in S : x_1\neq 0\}$. For points in this subset the second of the defining equations of $S$ can be resolved for $x_2$ and plugged in the first equation thus getting the single equation
$$
x_1^2(x_1^2+y_1^2-y_2^2-1)-y_1^2y_2^2=0
$$
for $U$. Note that $U$ is a subset of $\mathbb{R}^3$ now, since the variable $x_2$ is eliminated from the equations for $S$. Its still not easy to extract the shape of $U$ from this equation. Using a plotter software one gets the following beautiful image:
$U$" />
Since the software I used is not flexible enough I had to rename coordinates: $x,y,z$ correspond to $x_1,y_1,y_2$ in the same order.
Looking along the $y_2$-axis one can see the circle in the $(x_1,y_1)$-plane:
$U$ looked at along $y_2$" />
On the other hand looking along the $y_1$-axis you can see a hyperbola in the $(x_1,y_2)$-plane:
$U$ looked at along $y_1$" />
To come back to the question: one can see now how a rotation of $S$ and therefore $U$ could bring this hyperbola into the $(x_1,y_1)$-plane. You can even see that a rotation angle of $90$ degrees is necessary.
Thinking a bit more about the situation it seems that I found an embedding $f:S\rightarrow\mathbb{R}^3$, so that one can visualize $S$ as a whole: first observe that there is no point $(x_1,x_2,y_1,y_2)\in S$ such that $x_1=y_1=0$ because of the first of the equations defining $S$. The second of these equations has the following interpretation: in $\mathbb{R}^2$ the vector $(x_2,y_2)$ is perpenticular to $(x_1,y_1)$. Hence given $(x_1,y_1)$ there exists a unique $u\in\mathbb{R}$ such that $x_2=-uy_1$ and $y_2=ux_1$. We define
$$
f(x_1,x_2,y_1,y_2):=(u,x_1,y_1).
$$
One easily checks that $f$ is injective. Moreover a straighforward calculation shows that every point $(u,x_1,y_1)\in f(S)$ satisfies the equation
$$
(1-u^2)(x_1^2+y_1^2)=1.
$$
Let $T\subset\mathbb{R}^3$ be the set of all solutions of this equation. Then $f$ maps $S$ surjectively hence bijectively onto $T$. The inverse map being given by
$$
f^{-1}:T\rightarrow S, (u,x_1,y_1)\mapsto (x_1,ux_1,y_1,uy_1)
$$
is differentiable with Jacoabian matrix
$$
J=\left(
\begin{array}{ccc}
0 & 1 & 0 \\
x_1 & u & 0 \\
0 & 0 & 1 \\
y_1 & 0 & u
\end{array}
\right).
$$
$J$ has full rank $3$ everywhere on $T$, hence $f$ is also differentiable.
Here is an image of $T$:
$T$" />
