# Graphical representation of complex roots to equations

1. Jul 3, 2010

### RK1992

I've never properly studied complex numbers but I will soon (in September). Basically:

We get taught from a young age that:
the real root of f(x)=x²-4 is where the graph of y=f(x) cuts the x axis

But is there a graphical representation of a complex root?

What's so special about the value x= +/- 2i if g(x)=x²+4 ? Is there a 3D graphical representation of this root?

2. Jul 3, 2010

### Hurkyl

Staff Emeritus
Well, remember that "real root of f(x)" really means a number a such that f(a)=0. It doesn't need any geometric interpretation to make sense.

Some people like to think in terms of geometry rather than algebra; so the particular correspondence between them you invoked says that the roots of f(x) correspond to the intersection of the parabola defined by y=f(x) and the line y=0, just like you described.

The similar geometric interpretation for complex-valued functions of complex numbers requires 4 dimensions to draw. We can often get away with projecting onto a three-dimensional image, though. (Of course, we have to project that onto a two-dimensional image so that we can draw it, and things get messy)

Another common way is to instead draw a before and after picture of the complex plane. e.g. put a grid on the "before" picture, and in the "after" picture we see a picture of how f transformed the grid.

For this function, a good "before" picture is to make the grid out of rays emanating from the origin and circles whose center is the origin.

The after picture consists of lines passing through the point -4 + 0i, and circles centered on that point. The spacing between the circles is unchanged. However, the spacing between the lines has doubled, and the grid overlaps itself -- e.g. the two rays emanating from the origin at angles x° and (x+180)° both map to the same ray emanating from -4 + 0i at an angle of (2x)°.

In this picture, we can estimate the roots of f by looking at 0 + 0i in the "after" picture, identifying the grid points that lie there, and then finding where they came from in the "before" picture.

3. Jul 3, 2010

4. Jul 3, 2010

### HallsofIvy

Staff Emeritus
A simpler (or more simple minded) way of looking at it:

The equation $x^2- 2ax+ a^2+ b^2= (x-a)^2+ b^2= 0$ has roots $x= a\pm bi$. It has complex roots, of course, because its graph does not cross the x-axis.

The vertex of the graph is where x= a so that itex]y= (x-a)^2+ b^2= b^2[/itex]. That is, the graph goes down to $(a, b^2)$ and then back up again. In general, if the graph of $y= x^2- ax+ b$ lies entirely above the x-axis, and its vertex is at $(x_0, y_0)$, then its roots are $x_0\pm i\sqrt{y_0}$.