Jul 272012

For today’s #myfavfriday I am presenting an idea that has been percolating in my head for a while. If you want to know what a #myfavfriday is, then see Druinok’s blog here.


Learners have a devil of a time with quadratics. Afterall, there can be 2 solutions, 1 solution, or no solutions in Algebra 1, and then in Algebra 2, we come at them with the fact that those equations with no solution really do have 2 solutions after all, they are just “imaginary” (could there be a worse name for them, really? Thanks a lot Descarte.”)

But I came across a picture on some site one day, and it has stuck with me. I never bookmarked it, or wrote down the site, so it is lost to me (and I have searched hard for it) but the work blew my mind, and as I have shown it to learners, they have at least gotten a sense that the “imaginary” really does have meaning.

Let’s begin with 2 equations and graphs that are simple, straight-forward and make sense. [all images are clickable to see full size]

graph1and graph2

The equations are y = x^2 – 4x + 3 and y = x^2 – 4x + 4.

A simple change of one number changes the number of solutions from 2 distinct to 2 repeating solutions, and learners don’t have a problem with that idea, generally. Then comes this bad boy.

graph3 y = x^2 –4x + 6

Now they have to do the whole Quadratic formula on it to get the solution, and the solution has those i thingies in it, which makes them all confused and irritated until they wrap their heads around it. And why does it still have 2 solutions? It doesn’t touch at all!

But wait! We can play a game with this quadratic function. What if we reflect the parabola around the vertex in the downward direction? Then we end up with something that looks like this:

graph4a To do this reflection, we first had to complete the square on the original equation to get y = (x-2)^2 + 2. Now, with this equation, we can put the – sign into the equation and get the reflection, y = -(x-2)^2 + 2.

But hold on, see those 2 points where it crosses the X-axis? And see the Axis of Symmetry that goes through both equations? If we use those three points as definitions for a circle, we get the following graph and equation.

graph5 (x-2)^2 + y^2 = 2

Guess what the solution to the quadratic equation y = x^2 –4x + 6 is. If you guessed 2 + root(2)i and 2 – root(2)i  then you are absolutely correct.

The real number part of the complex solution of a quadratic with two imaginary roots is the X value of the Axis of Symmetry, and the imaginary part of the solution is the radius of the circle created by the center and endpoints created when the inverted parabola crosses the X-Axis!

Okay, mind blown. Why? How could I prove this?

Aha! now come into play the hours I spend on a motorcycle every summer. How could I PROVE that this will always work? I have the proof. I am working it up, but it is a pain to type. That, I think, will be the focus of a future, #myfavfriday!

[And I really need to look in to a LaTex module for my blog if I am going to do math. The equations look horrible.]

Edit: 29 July 2012: I proved this assertion, at least to my satisfaction in a followup post: http://blog.mrwaddell.net/archives/348

Edit: 4 August 2012: I found, stuffed in the bottom of my backpack, a rumpled piece of paper with this link on it. I think I did this page justice with my treatment. I wish I had found the page before I spent hours thinking about how to prove it, it gives the suggestion right there at the bottom!

Edit: 18 Dec 2012: @Mythagon posted this picture on Twitter. It is a great visual of what is discussed above, and clearly shows why the rotation is so important.

From: Teaching Mathematics, 2nd edition by m. Sobel and E. Maletsky

Edit: 27 Sep 2015: Wow, a long time since the original post, however I still come back to this every year. Love it. Now, Luke Walsh, aka @LukeSelfwalker added this to the mix. Love it. Click it for the live Desmos file.

  5 Responses to “A visual representation of Imaginary solutions”

  1. different visualization at http://en.wikipedia.org/wiki/Geometry_of_roots_of_real_polynomials … I love telling students that the name ‘imaginary’ was coined as an insult and stuck; especially when you point out that it was Descartes’ invention of the coordinate plane that allows you to easily visualize complex numbers (Real Axis, Imaginary Axis, Complex Number as Ordered Pair) … the way that I think about visualizations of parabolas with complex roots is more like the 3-D graphs on http://www.bikinfo.com/HTML/quadratic.html (random google image that matches my mental image) – of course, you have to get kids thinking about 3-D graphs before you can begin to approach this in class, but is a somewhat natural dimensional extension of a video I like to use to introduce imaginary numbers http://www.youtube.com/watch?v=T-c8hvMXENo this is a little long/slow for some students, but there is a lot of really good math in there if you’re paying attention!

  2. Wow, Awesome links Aaron. Thank you for the additional links and insight. This will help me complete the proof I am working on.

  3. […] comes via Glenn Waddell: But hold on, see those 2 points where it crosses the X-axis? And see the Axis of Symmetry that […]

  4. I was driving home from Delaware yesterday and, for no discernible reason, started to think about representing the solutions to complex functions. Inspired by this brilliance I was imagining functions from C to C as transformations of the plane. The reason to think of it this way, for me, was to be able to think of transformations of 2D numbers to 2D numbers. The visualizations that only take into account the real part of real part of complex curves by graphing them on 3D planes (like in Ashli’s nifty picture or the links in the first comment) never quite made sense to me — you’re leaving something out! What are you leaving out? How do you know it doesn’t matter?

    So I wanted to think about quadratic functions as mapping points to other points in some kind of predictable way, and to think about what maps to (0,0) — in other words, finding complex roots of parabolas.

    I started with looking at transforming the complex plane with a quadratic: https://www.desmos.com/calculator/bwmfemi03r

    The square centered at 0 with side length 8 got turned into this lovely, spirally swoosh. But most interestingly, I noticed that every point on the new swoosh was mapped there from 2 different input points — so when I tried to find what maps to (0,0) I would need to find two different points that got there. It happened because of the squaring — squaring the complex plane doubles the rotation angles of each point, and so the plane kinda wraps around itself again, as everything with angles from 0 to pi now covers the whole plane and everything with angles pi to 2pi gets mapped on top of what’s already there.

    I did happen to find the roots of that polynomial (the points that got mapped to (0,0)) but not in any systematic way.

    Next was the desire to get systematic about finding roots. I realized I needed to think about undoing all the pieces of a quadratic — the squaring, the shifting… it took a while but I realized I needed to think about the function in “vertex form” because that’s the best form for undoing. So what’s a quadratic from a transformations perspective? I think it’s easiest if I define a generic quadratic in the form y = a(x-h)^2 + k. If (and only if) a and k are both positive, then there are 2 complex roots. So we’re looking at quadratics where a and k are positive (that turned out to be useful). Oh, important note: even though I’m using complex inputs and outputs I’m only using real coefficients. So a, h, and k are real numbers (and a and k are positive).

    Here are the transformations, in the order they happen:
    First: -h shifts the whole plane (to the left if h is positive and to the right if h is negative)
    Second: squaring doubles the angle (argument) of each point and squares its distance from the center.
    Third: multiplying by a dilates points away from the center by a factor of a.
    Fourth: adding k shifts everything to the right k units. [This one is weird — why does it shift everything to the right, not up? Because k is a positive real number, and real numbers on the complex planes have no vertical component. So it’s translating by a vector pointing to the right].

    To find what mapped to (0, 0) I had to undo the transformations in backwards order:
    First, shift everything to the left k units. So (0,0) goes back to (-k, 0)
    Second, un-dialate by a factor of a, so (0, 0) → (-k, 0) → (-k/a, 0)
    Third, un-double-the angle and un-square the distance from the center. Remember that there are 2 angles that double to pi, so 2 points are created: at angles of pi/2 and 3pi/2 (0, 0) → (-k, 0) → (-k/a, 0) → (0,sqrt(k/a)) and (0, -sqrt(k/a))
    Fourth, shift everything h units: (0, 0) → (-k, 0) → (-k/a, 0) → (0, -sqrt(k/a)) and (0, sqrt(k/a)) → (h,sqrt(k/a)) and (h, -sqrt(k/a))

    A couple things to check-up on:
    Are h + isqrt(k/a) and h – isqrt(k/a) the roots of the quadratic y = a(x-h)^2 + k. Yep!
    What happens when you run those two points through the transformations defined by y = a(x-h)^2 + k?

    First shift the points by -h: (h,sqrt(k/a)) and (h,-sqrt(k/a)) → (0,sqrt(k/a)) and (0,-sqrt(k/a)).
    Second, square by doubling the angle and squaring the distance. Since (0,sqrt(k/a)) and (0,-sqrt(k/a)) are on the y-axis, doubling their angles will take both to the negative x-axis, and squaring their distance gets rid of the square root: (h,sqrt(k/a)) and (h,-sqrt(k/a)) → (0,sqrt(k/a)) and (0,-sqrt(k/a)) → (k/a, 0)
    Third, dilate by a factor of a: (h,sqrt(k/a)) and (h,-sqrt(k/a)) → (0,sqrt(k/a)) and (0,-sqrt(k/a)) → (-k/a, 0) → (-k, 0)
    Fourth, translate by k: (h,sqrt(k/a)) and (h,-sqrt(k/a)) → (0,sqrt(k/a)) and (0,-sqrt(k/a)) → (-k/a, 0) → (-k, 0) → (0, 0).

    Interestingly, as you follow the points through the transformations, you see how they’re related to your circle and to the x-intercepts of the reflected parabola!

    Here’s the other Desmos file I made: https://www.desmos.com/calculator/d6wo91oc0v

    Update: I also made a file to actually show the quadratic transformations! https://www.desmos.com/calculator/sb4eyeqh4h

  5. Hey Glenn,
    I came across this post a while ago and have kept it in the back of my mind waiting for the opportunity to use it. Last Friday, a student in my intermediate algebra class asked me “Is there a way for us to see the imaginary solutions to a quadratic by looking at the graph?” BOOM!!!! I didn’t completely give them the answer but offered them just enough to peak their curiosity. Your post has helped me to organize my lesson for Monday. Thanks Bud!!!!

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