8
$\begingroup$

So recently we've been doing the complex roots of quadratics, cubics and polynomials in general in school. But my question is, is there a way to see where these roots are, just like you can see where the real roots are by seeing where they intercept with the X-axis?

enter image description here

For example, in this cubic here, it is evident that there is a real root just under -1, but is there a way to visualise the complex roots? Is there another line (similar to the x-axis) which intercepts the equation in another dimension?

$\endgroup$
  • $\begingroup$ We can present complex roots to equation on the "complex plane" with one axis for the real part and the other for the imaginary part. You can play with, for instance, WolframAlpha, to give it a polynomial equation to solve and get a display of the complex roots. If you look up "DeMoivre's Theorem" online, you will find something interesting about the roots of equations $ \ z^n \ = \ c \ $ , with $ \ c \ $ being some complex (including real) number. $\endgroup$ – colormegone May 23 '16 at 21:16
  • $\begingroup$ +1 for the graph. You've now made this a very nice question. Similar questions have been asked, but those that I've seen so far are all pitched at a much higher level than your question, and thus not likely to be understandable to someone in an introductory (i.e. precalculus level) course. $\endgroup$ – Dave L. Renfro May 23 '16 at 21:36
11
$\begingroup$

Here is a way.

Suppose that we want to visualize the roots of the cubic equation $$ z^3+z+1=0 $$

Write $z=x+iy$ and expand: \begin{align} (x+iy)^3+(x+iy)+1&=0\\ x^3+3ix^2y-3xy^2-iy^3+x+iy+1&=0\\ \end{align}

Taking real and imaginary parts, we get \begin{align} x^3-3xy^2+x+1&=0\\ 3x^2y-y^3+y&=0 \end{align} Plotting the solution sets of these two equations gives us two curves in the $xy$-plane:

A plot of those curves

Now, the original expression is zero if and only if its real and imaginary parts are both zero. In other words, the roots of our original polynomial correspond to the points of intersection of these two curves.

This trick can be used to visualize the roots of any complex function $f$. Just write it in the form $f(x+iy)=u(x,y)+iv(x,y)$ and plot the solution sets to $u(x,y)=0$ and $v(x,y)=0$. Then the roots will correspond to the intersections of these two curves.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

I have developed a very clear method of visualizing where the complex roots of an equation are. The method involves drawing a graph of y = f(x) in the usual way on x, y plane but adding a 3rd axis to allow those special complex x values which also produce real y values. This means we have a normal y AXIS but a complex x PLANE. This is my first time on stackexchange so I cannot yet provide you with some excellent diagrams showing this method. I have however, just made a short video showing how the method works. I encouraging you to view it. Solutions of Cubics using Phantom Graphs http://screencast.com/t/dkAYxFDwH

Also, I have written a special section on this exact topic in my website: http://www.phantomgraphs.weebly.com just scroll right down to the last entry I have made especially for this question.

| 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.