# geometric interpretation of quadratic equation with complex coefficients

When an equation has real coefficients and non-negative discriminant, the geometric meaning of it's roots is intersection of the function with the x-axis.

I know how to get roots of quadratic equation with complex coefficients, I just wonder if there is any geometric interpretation for that.

Try my paper, Graphing the Complex Roots of Quadratic Functions on a Three Dimensional Coordinate Space, detailing a geometric construction detailing how to derive the complex roots geometrically.

Say you have a polynomial $F(x_1,...,x_n)$ with real or complex coefficient. Then you can consider it as definining a function $$K^n\longrightarrow K\qquad\qquad(*)$$ where $K$ is either $\Bbb R$ or $\Bbb C$. Simply, the value associated to an $n$-ple $\vec a=(a_1,...,a_n)$ is $F$ evaluated at $\vec a$. In geometric terms, solving the equation $$F(x_1,...,x_n)=0$$ means finding the zero-locus of the function $(*)$ which is some subset $Z_F\subseteq K^n$. The case you are familiar with is that of $n=1$, where $Z_F$ is typically a set of isolated real or complex points.

Studying the geometry of the sets $Z_F$ is the foundational motivation of Algebraic Geometry

• Thanks for your response, but i'd like to know if it is possible to draw the roots and see how they look like. Wikipedia article doesn't answer my question. – koddo Feb 21 '13 at 14:49
• Since you talk about "roots", I'm assuming that you're looking at the one variable case. Sometimes the roots can be computed exactly, more often not. In any event, there techniques that allow to find approximations good enough to make an accurate drawing. – Andrea Mori Feb 21 '13 at 17:17
• Excuse me, but you do not need to assume anything since i explicitly wrote i'm interested in quadratic equation. – koddo Feb 21 '13 at 18:06
• quadratic polynomials in one variable always admit two, possibly equal, complex roots. In more variables their zero-locus are called quadrics which, in many respect, are analogous to the conic sections first studied by the Greeks, e.g. see en.wikipedia.org/wiki/Quadric – Andrea Mori Feb 21 '13 at 21:54

I have one that I actually just derived today. The following is equivalent to the quadratic formula and provides a geometric interpretation of the imaginary part of complex roots. Let's do an example first, and then below I will write out a more formal treatment. Let's say you have the equation $$x^2+4x+5$$ This equation can't be factored using two real roots, but that doesn't mean we can't try. Let's try some factors that only satisfy the x term without satisfying the constant term, and let's do our best to get as close to 5 as possible. We get 'closest' to our desired formula when we factor it like $$(x+2)(x+2)$$ which gives us $$x^2+4x+4$$ In fact, it is always true that if we can't satisfy the quadratic equation with real roots, we can get closest by choosing two real roots that are equal. We can draw an analogy (and it's not just an analogy) with the area of a rectangle. We maximize the area of a rectangle with sides $p1$ and $p2$ by choosing the two sides to be of equal length. Here, the area of our rectangle we are trying to get to is 5, but the best we can do is $4$, by choosing side lengths of $2$ and $2$. That's ok. How much area is left? So we only have $1$ unit of area remaining. Let's 'borrow' this area from the imaginary numbers, and let's create a 'rectangle' with side lengths of equal magnitude that, multiplied together, give us an area of $1$. We can do this with the numbers $i$ and $-i$, since $i*(-i)=1$ That's the answer. Our equation factors into $$[x+[2+i])(x+[2-i])$$ and indeed, this is the correct answer. So the geometric interpretation is that we are trying to 'generate' a certain amount of area (in this example $5$), but once we have our 'real square' (of area $4$) we need to borrow area from an 'imaginary square', which has area equal to the remainder of the area that we need.

Slightly more formal version:

If we have a quadratic equation with real coefficients $d$ and $e$ $$x^2+2 dx+e$$ (any quadratic equation with real coefficients can be put into this form by factoring out a constant, and the reason I put the two there will be clear later) then we know from the fundamental theorem of algebra that it can be factored into two terms $$(x+p1)(x+p2)$$ where p1 and p2 may be complex. From this, you can show that p1 and p2 are related to d and e by the two equations $$e=p1p2$$ $$d=p1+p2$$ Why, this looks awfully similar to the equations for the area and perimeter of a square. Indeed, we can interpret e as the area of a rectangle with sides p1 and $p2$, and $2d$ as the perimeter. We know from the quadratic formula that if a quadratic equation cannot be satisfied with real roots, it will be satisfied with a conjugate pair of roots. Then we can rewrite the equations above as $$e=p1 p1^*=pr^2+pi^2$$ $$2d=2(p1+p1^*)=2Re[p1]$$ By inspection then, we can see the real part of p1 is equal to d. Or, the real part is equal to the side length of the 'real square' needed to generate the area $e$. The imaginary part can be solved for and we see that $$pi=\pm\sqrt{e-pr^2}$$ The imaginary parts, then, are just the side lengths of the 'imaginary square' with area $e-pr^2$, or the total area that we want minus the area we are able to get from our real square. We are then left with the factors $$(x+(pr-i pi))(x+(pr+i pi))$$ which yields the roots of $$x=-(d\pm\sqrt{e-\frac{d^2}{4}})$$ From the quadratic formula, with a=1, we see the roots are $$x=\frac{-2d}{2}\pm\frac{\sqrt{(2d)^2-4e}}{2}$$ which is equal to our result above (you have to switch a sign inside the $\pm$ square root, but that's totally legal), and this interpretation is equivalent to the quadratic formula.

• As an aside, from a practical standpoint I have found this is MUCH faster than the quadratic formula for hand calculations in electrical engineering. – Jordan Nov 5 '17 at 18:15
• Let me simplify a little. So you are saying $x^2+2p+q = x^2 + 2p + (p^2 + h)$, where $h$ is height above the $x$ axis. Then the roots are $x = -p \pm \sqrt{-h}$. Interesting. – koddo Nov 7 '17 at 17:45
• Yeah, that's interesting, but the question is about quadratic equations with complex coefficients. – koddo Nov 7 '17 at 17:50
• I’m sure this could be extended to accommodate “complex area”, I’m just not immediately sure how to do it. – Jordan Nov 8 '17 at 19:07
• Yeah, to extend it to complex area we should imagine what $y = x^2$ looks like for complex $x$. – koddo Nov 9 '17 at 11:04