I was reading a graphical explanation of complex roots, and between Figures 7 and 8 I became confused.

enter image description here

The roots appear in the imaginary plane, but I don't understand why the original function must be inverted before the graphical representation fits. I realize that inversion is necessary for the graphical representation of imaginary roots to fit, but the lack of explanation bothers me. I would like to understand this inverstion better than "we do it because it fits the data."

  • $\begingroup$ Please take screenshots or copy down the information yourself into the question, so that your question will still make sense if external content is removed. $\endgroup$ May 25, 2013 at 4:31

1 Answer 1


Comment: In general, a nonzero quadratic has the form $y = A(x-a)^2+b^2$ where $A,a,b$ are real constants, but I set $A=1$ for the sake of clarity.

The graph is given to extend $y=f(x) = (x-a)^2+b^2$ to $y=Re(f(z)) = Re((z-a)^2+b^2)$ where $z = x+it$ is a complex variable. The graph given indicates all points in $(t,x,y)$ space such that the equation $y=Re(f(z))$ is satisfied. In particular, $$ y = Re((x+it-a)^2+b^2) = (x-a)^2-t^2+b^2 $$ Considering $x=a$ which is a vertical plane in the pictures, $$ y = -t^2+b^2 $$ This is a parabola which opens downward in the $x=a$ plane. It has $y=0$ when $t = \pm b$. This is why the graph opens downward. On the other hand, the plane $t=0$ (otherwise known as the $xy$-plane) we have $y = (x-a)^2+b^2$ which is a parabola with vertex $(a,b^2)$ which opens upward with no $x$-intercept.

The function is not inverted, it is extended to a complex domain, in this larger context the three-dimensional graph looks as pictured. However, beware the real picture casts $y = Re(f(x+it))$ as a surface, it is one-equation in three variables. This gives two parameters at most points on its solution set. The picture you post focuses on two interesting curves on this surface (it has parametrization $X(u,v) = (u,v,Re(f(u+iv)))$ )

Also, I was tempted to say it was the graph of $y = f(x+it)$, but that is more difficult to graph directly... think about it.

  • 1
    $\begingroup$ +1 for "The function is not inverted, it is extended to a complex domain." $\endgroup$
    – User3910
    May 25, 2013 at 9:33
  • $\begingroup$ Do you know of any software that will graph a two variable function along with its complex domain like the picture in the question? I tried wolfram, which does them separately. $\endgroup$
    – User3910
    May 27, 2013 at 0:40
  • $\begingroup$ @JoeHobbit I don't know any particular site which is designed for this purpose. I suppose you can calculate the real part of the complex extension and graph in 3D as they did here. For that you could use Maple or Mathematica etc... there are a few decent Java-based 3D graphers. I've used web.monroecc.edu/manila/webfiles/calcNSF/JavaCode/…, but there are others... $\endgroup$ May 27, 2013 at 7:08
  • $\begingroup$ +1 for the same reason as OP commented above! I'm just leaving this comment for me here, to clarify my confusion. In short, what the graphic is saying is that IF we only allow the outputs of the function to be real values, then $x$ can take on ANY real value, or only values on a single line in the complex plane (more specifically, if the equation is in the form $a(x-b)+x$, then $x$ is only allowed to take on complex values of the form $x=(b+ki)$, since otherwise the output of the function would be complex). $\endgroup$ Jul 13, 2019 at 16:19
  • $\begingroup$ When we allow $x$ to take on those values, we get an "upside-down" parabola, but growing into the complex direction. The reason its upside down is because $i^2$ will always make the coefficient on the $a$ the negative of what it was before. All the author of this article wanted to show was that the parabola allowed to extend into the complex domain will have the same roots as an upside down version of our original parabola, only that in the complex direction instead of the real direction! $\endgroup$ Jul 13, 2019 at 16:21

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