35
$\begingroup$

Consider the function $f(x)$ such that $f(x) = x^2-4x+13$. By considering the discriminant, it can immediately be seen that the function has no real roots, since $b^2-4ac = (-4)^2-4(13) = -36$ and $-36 < 0 $, and hence the graph looks as follows

enter image description here

With no real roots, and hence the $x$-axis is never intercepted.

However, with knowledge of complex numbers, it can be seen that the roots are $$x=\frac{4\pm\sqrt{-36}}{2}$$ $$x_1=2+3i$$ $$x_2=2-3i$$ $$\implies (x-2-3i)(x-2+3i)=0$$

and the question on the table is, are there any means of visualising this fact? I know the closest we generally come to visualising complex numbers is argand diagrams, but I've thus far failed to recognise whether or not this can be used to visualise solutions such as these.

An example of a visualisation such as this enquires looking at real solutions, so consider when $$x^2-4x+13=\frac{1}{2}x+10$$

Here, the graph of $0=2x^2-9x+6$ indicates the solutions to the system of equations, but is there any way this example can be applied when the roots are complex?

enter image description here

Any help is appreciated. Thank you.

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ For quadratics with real coefficients, $x^2-bx+c=0$, with $b^2<4c$, you have that the real part of the complex number is $\frac{b}{2}$ and the absolute value of the complex number is $\sqrt{c}$. (Note that $c>0$ since $b^2<4c$.) But the general case is harder. $\endgroup$ – Thomas Andrews Oct 19 '17 at 15:13
  • 3
    $\begingroup$ A related answer Graphically solving for complex roots - how to visualize $\endgroup$ – StephenG Oct 19 '17 at 15:15
  • $\begingroup$ From the graph you can read off the axis of symmetry ($x=2$) and the y intercept ($13$). The axis of symmetry is the real part of the complex roots; the imaginary part can be found by subtracting the square of the axis (here, $4$) from the intercept ($13-4=9$) and then taking the square root ($3$). This assumes the roots come in conjugate pairs (so the coefficients of your quadratic are real numbers). $\endgroup$ – symplectomorphic Oct 19 '17 at 15:15
  • $\begingroup$ There are lots of ways to visualize roots of complex functions because there are lots of ways to visualize complex functions. StephenG's comment points to an which presents one technique and LarryB's answer provides another technique. $\endgroup$ – Mark McClure Oct 19 '17 at 18:31
  • $\begingroup$ You might want to check out this short YouTube playlist on imaginary numbers. It's pretty good at ironing out the relevant intuitions if you're not entirely accustomed to the topic. $\endgroup$ – Denis de Bernardy Oct 20 '17 at 13:01
31
$\begingroup$

If your question is related to quadratic equations then here is the solution. Let's take your example.

enter image description here

Now, do the following in the case of a parabolic curve not intersecting the $x$ axis. Mirror the curve about the straight line tangent to it at its minimum. This mirror image will certainly cross the $x$ axis. (Assume that $a$, the coefficient belonging to $x^2$, is $1$. Otherwise, by dividing both sides by this coefficient the new one will be one, and the roots will be the same.)

Take the $x_m$ value at which the minimum occurs. That value will be the common real part of the two solutions. Then take the differences between this $x_m$ and the values of the crossing points of the mirrored curve.

In your case, the minimum is taken at $2$. This is the real part. Then the differences are: $-3$ and $3$. Then $-i3$ and $i3$ are the imaginary parts.

EDIT

Let's prove that the method described above works in general. (not only in the case of the example given)

Consider the quadratic equation $x^2+bx+c=0$ and assume that $b^2-4c<0$. Then the graph of $$f(x)=x^2+bx+c$$ will not intersect the $x$ axis. The complex roots are

$$x_{1,2}=-\frac b2\pm i\frac{\sqrt{-(b^2-4c)}}2.$$

Now, consider another function: $$h(x)=-x^2-bx-\frac{b^2}2+c.$$

This is the mirror image of $f$ (mirrored about the tangent at the minimum point).

Let's calculate the roots of $h(x)=0$.

$$x_{1,2}=-\frac {b}{2}\pm\frac{\sqrt{-(b^2-4c)}}2.$$ Since $b^2-4c<0$ ($f(x)=0$ did no have real roots), we have now a positive discriminant and two real roots. And, exactly, the imaginary parts, as real numbers of the complex roots of $f(x)=0$ are now added and subtracted to/from $x_m$ at which $f$ takes its minimum.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm assuming this isn't mathematically rigorous, but it definitely is an easy way to visualise the scenario without relying on more than 3 dimensions! I'm curious, however, has this worked for every situation like this that you've encountered? $\endgroup$ – joshuaheckroodt Oct 19 '17 at 15:32
  • $\begingroup$ Do yo need the proof that this is always true? Or rather, will you try to prove it on your own? $\endgroup$ – zoli Oct 19 '17 at 15:34
  • $\begingroup$ I admire that you presume I'll be trying such a thing, but almost certainly not, haha! In fact the whole reason for my question is sheer curiosity, so no I won't be attempting to prove this, I'm was just asking. $\endgroup$ – joshuaheckroodt Oct 19 '17 at 15:37
  • 1
    $\begingroup$ @Ooker: I don't thik so. This is so special an accidetal. $\endgroup$ – zoli Oct 19 '17 at 21:06
  • 1
    $\begingroup$ A great merit of this approach is that you don't have to be able to sketch 3d plots for this. $\endgroup$ – Ruslan Oct 20 '17 at 10:15
17
$\begingroup$

What you can do is use either a 3D plot or a color scheme, and have your input be a complex number and your output be $|f(x + iy)|$. This would give the following picture :

$|f(x + iy)|$

From there you can see where the two roots are ; they are in the center of the two red spots.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Was this done on WolframAlpha? $\endgroup$ – joshuaheckroodt Oct 19 '17 at 18:03
  • $\begingroup$ Yes, this is a 3D contour plot in fact. $\endgroup$ – Matrefeytontias Oct 19 '17 at 19:03
  • $\begingroup$ Why do the solutions lower than their neighborhoods? The plot doesn't seem to be a generalized form of the 2D parabola. $\endgroup$ – Ooker Oct 19 '17 at 21:15
  • 4
    $\begingroup$ @Ooker Note that this is a plot of the absolute value, not of the function itself (which would result in a 4-dimensional plot). Ploting $|x^2-1|$ as a normal real function plot gives a similar not-a-parabola result. $\endgroup$ – tomsmeding Oct 19 '17 at 22:08
6
$\begingroup$

Try entering $z^2-4z+13$ into this tool. Make sure to zoom out, and you'll see something that looks like this:

a plot of the complex plane for z squared minus 4 z + 13

There's not a very good indicator of result coordinates on this tool. But you can clearly see the two roots, with reflection symmetry across the real line.

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ It seems the deeper you dive into imaginary maths the more the prefix of "complex numbers" starts to make sense... $\endgroup$ – joshuaheckroodt Oct 19 '17 at 15:22
5
$\begingroup$

If we define a polynomial $f(z) = az^2 +bz +c$ (where $z \in \mathbb C$), then $$f(z) = a(z - p)^2 - \frac{\Delta}{4a},$$ where $p = -\frac{b}{2a}$ and $\Delta = b^2 - 4ac.$

If $a,$ $b,$ and $c$ are all real, then $p$ and $\Delta$ are real. It follows that $f(x)$ is real if and only if $(z-p)^2$ is real. This can happen in either of two ways: $z-p$ is real, or $z - p$ is purely imaginary.

In the case where $z-p$ is real, we can set $z = x$ where $x\in\mathbb R,$ and you can plot $y = f(x)$ over $x\in\mathbb R$ as the question has done for $a=1,$ $b=-4,$ and $c=13.$ We get a parabola with a vertex at the coordinates $\left(p, - \frac{\Delta}{4a}\right).$

In the case where $z - p$ is purely imaginary, we can set $z = p + ix$ where $i$ is the imaginary unit and $x\in\mathbb R.$ Then \begin{align} f(z) &= a(p + ix)^2 + b(p + ix) + c \\ &= ap^2 + i2apx + i^2 ax^2 + bp + ibx + c \\ &= \frac{b^2}{4a} - ibx - ax^2 - \frac{b^2}{2a} + ibx + c \\ &= - ax^2 - \frac{b^2}{4a} + c \\ &= - ax^2 - \frac{\Delta}{4a} \\ \end{align}

If we set $g(x) = - ax^2 - \frac{\Delta}{4a},$ we can plot $y = g(x)$ in a Cartesian plane; it is a parabola with a vertex at $\left(0, - \frac{\Delta}{4a}\right),$ oriented in an opposite direction to the parabola we got for real values of $z.$

So imagine that we want to plot $y = f(z)$ over the complex plane, but only for the values of $f(z)$ that are purely real. We can visualize this in just three dimensions: two dimensions to plot the complex input value, $z,$ and one dimension to plot the real output value $y.$ If we write $z = u + iv$ we can describe this three-dimensional space by the coordinates $(u,v,y).$

The resulting plot consists of just two parabolas: the parabola for $y = f(x)$ where $x$ is real, presented as the plot $y = f(u)$ in the $u,y$ plane; and the parabola for $y = g(x),$ translated $p$ units in the $u$ direction and rotated $90$ degrees around the line $u = p,$ $v=0$ (a line through $(p,0,0)$ parallel to the $y$ axis) so that it lies in the plane $u = p.$

That is, the parabolas have the same vertex, $\left(p, 0, - \frac{\Delta}{4a}\right),$ and the same axis, the line $u=p,$ $v=0,$ but they are in perpendicular planes. In another answer to this question, this same plot is "flattened" into two dimensions by rotating the "imaginary" parabola by $90$ degrees around its axis.

Now, depending on whether $\frac{\Delta}{4a}$ is positive or negative, we have the common vertex of the parabolas either below or above the $u,v$ plane, and we have either two zeros on the upper parabola or two zeros on the lower parabola. That translates to two zeros on the line $v=0,y=0$ (two real solutions of $f(z) = 0$) or two zeros on the line $u=p,y=0$ (two conjugate complex solutions of $f(z) = 0$). Which of these pairs corresponds to the upper parabola and which to the lower parabola depends on the sign of $a.$ Of course if $\frac{\Delta}{4a} = 0$ then we get the double root $z = p.$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.