# Do hyperbolas exist in the complex plane?

My question is somewhat long, but I will try to be concise and unambiguous in my explanation.

In my high school math class we were recently presented with the graph of a hyperbola, and given certain facts challenged to derive its equation.

My question only centers on the final part of the derivation, when the equation dealt with was:

$x^2(1 - e^2) + y^2 = a^2(1 - e^2)$

Where $e$ represents the eccentricity, and $a$ represents a positive constant.

Simply divide by $a^2(1 - e^2)$, and use the substitution $b^2 = a^2(1 - e^2)$, where $b$ is another positive constant, and one finds the equation for a ellipse:

$x^2/a^2 + y^2/b^2 = 1$

We were shown that all one had to do to get the equation for a hyperbola instead was multiply the original equation by $-1$, so that it became:

$x^2(e^2 - 1) - y^2 = a^2(e^2 - 1)$

The substitution also changes slightly, to become $b^2 = a^2(e^2 - 1)$. Then the equation for a hyperbola can be found.

My question is simply why this multiplication was necessary. It seems to me that multiplying both sides of an equation by some factor does not change anything about it, and therefore the only real change was when $b^2 = a^2(1 - e^2)$ became $b^2 = a^2(e^2 - 1)$. Why was this necessary, and what does it imply about hyperbolas?

Was it necessary because for hyperbolas the eccentricity $e$ is greater than one, and so $b^2 = a^2(1 - e^2)$ would imply $b$ is a complex number? And if so, can the operation be thought of as restricting the domain and range of a hyperbola to the set of all real numbers?

That is my primary question, but I do have another, which is a little more vague:

If hyperbolas do have a complex component, does a plot of one in the complex plane form a cross-section of a three-dimensional or four-dimensional shape, the way an ellipse forms a cross-section of a football? Phrased another way: ellipses are bounded in the set of all real numbers – are hyperbolas bounded in the set of all complex numbers?

• You're right that multiplying an entire equation by $-1$ does not change the solution set; are you sure the entire equation was multiplied by $-1$, and not merely the $y^{2}$ term? (Over the complex numbers, $y \mapsto iy$ is a type of "Wick rotation", and converts the equation of an ellipse into the equation of a hyperbola.) – Andrew D. Hwang Jan 22 '17 at 17:18
• Yes, I believe so. Because $1 - e^2$ became $e^2 - 1$. You can multiply both sides by $-1$ and expand to confirm. – Steffan Jan 22 '17 at 17:35
• Your equation as typed has certainly been multiplied throughout by $-1$. :) My question is, are you sure it's typed correctly? On the assumption the equation is indeed typed correctly, the point is that if $e^{2} < 1$ and $a$ is real, then $b$ is imaginary. Absorbing the imaginary unit into the variable $y$ amounts to making the stated Wick rotation. – Andrew D. Hwang Jan 22 '17 at 17:43

My question only centers on the final part of the derivation, when the equation dealt with was: $$x^2(1−e^2)+y^2=a^2(1−e^2)$$

Note in this equation that if $e < 0$ then the coefficients of $x^2$ and $y^2$ are both positive, indicating an ellipse. If $e > 0$, then the coefficient of $x^2$ is negative while that of $y^2$ (which is $1$) is still positive. Since they are of opposite signs, this would be a hyperbola.

In either case, dividing by the RHS gives $$\frac{x^2}{a^2}+\frac{y^2}{a^2(1−e^2)} = 1$$

If you multiply by $-1$ first, then divide it out, you get

$$\frac{x^2}{a^2}-\frac{y^2}{a^2(e^2-1)} = 1$$

These two equations are exactly the same. In both of them, if $e < 1$, then the coefficient of $y^2$ is positive, and if $e>1$, the coefficient is negative.

The reason for doing this is that we want to substitute $b^2$ for the denominator of the $y^2$ term, and $b^2$ must be positive. So when $e < 1$, we use the first form and let $b^2 = a^2(1-e^2)$. When $e > 1$, we use the second form and let $b^2 = a^2(e^2 - 1)$.

So multiplying by $-1$ didn't change the equation. It just allowed the $b^2$ substitution when $e > 1$.

Ellipses are bounded in the set of all real numbers – are hyperbolas bounded in the set of all complex numbers?

I assume by this you are talking about letting $x, y$ be complex values, not just real numbers. Then the answer is obviously not, as real numbers are also complex numbers. Since hyperbolas are unbounded in the reals, they are also unbounded in the complex numbers. Instead, what happens is that ellipses are unbounded as well. In fact these "complex ellipses" and "complex hyperbolas" are the same shape. Note that $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$ is equivalent to $$\frac{x^2}{a^2}-\frac{(iy)^2}{b^2} = 1$$

So $(x,y)$ lies on the ellipse if and only $(x,iy)$ lies on the hyperbola. Multiplying by $i$ rotates the complex plane by 90 degrees.

So the only difference between ellipses and hyperbolas is which plane you are slicing the 4D complex shape with to get the 2D real shape.

But you are already familiar with a relationship like that. After all, you chose the tag "conic sections". I assume you know where that term comes from. (Though the shape being sliced is different: cones are 3D, while this complex shape is 4D.)