Graphically solving for complex roots -- how to visualize? 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?

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?
 A: 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.  
A: 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:

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.
