A classic proof for the [url=https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra]fundamental theorem of algebra[/url] (that is, each non-constant polynomial has a root among the complex numbers) uses topology.

It is to prove that [math]f(v)[/math] has a root in the complex plane, that is, a [math]v[/math] is required (on the left) for which [math]f(v)[/math] (on the right) goes through the origin ([color=#ff0000]X[/color]).[br]The circles on the left are always transformed to a closed curve on the right which wind around the origin, if [math]r[/math] is sufficiently large. On the other hand, if [math]r=0[/math], the point on the left is transformed to a point on the right. Between these two extremal situations the right curve continuously changes. This results in a position when the right curve obviously crosses the origin.[br]You can try some activities out:[br][list=1][*]Change the radius of the circle to find an image of the circle which crosses the origin on the right.[/*][*]Trace the circles to remember which radiuses were already checked.[/*][*]Show the image of the circle to immediately know that a radius will be fine or not.[/*][*]Trace the images of the circles to have a general view about the map [math]f(v)[/math].[/*][*]Change [math]f(v)[/math] to different polynomials. Learn how the higher order polynomials behave.[/*][*]You can also try non-polynomial functions. For example, [math]f(v)=\sin(x)[/math] is an interesting option.[/*][/list]