A topological proof
going in circles and counting
Let us suppose that the polynomial
has no root, so that for every complex number z
On this assumption, if we now allow z to describe any closed curve in the
Proof of fundamental theorem of algebra. |
We may, therefore, define the order of the origin O with respect to the function f(z) for any closed curve C as the net number of complete revolutions made by an arrow joining O to a point on the curve Γ traced out by the point representing f(z) as z traces out the curve C. As the curve C we shall take a circle with O as center and with radius t, and we define the function φ(t) to be the order of O with respect to the function f(z) for the circle about O with radius t. Clearly
It remains only to show that φ(t) = n for large values of t. We observe that on a circle of radius
we have the inequality
|f(z) - z ^{n}| = |a_{n-1} z ^{n-1} + a_{n-2} z ^{n-2} + ... + a_{0}| ≤ |a_{n-1}||z|^{n-1} + |a_{n-2}||z|^{n-2} + ... + |a_{0}| = t^{n-1}[|a_{n-1}| + ... +|a_{0}| /t^{n-1}] ≤ t^{n-1}[|a_{n-1}| + ... +|a_{0}|] < t^{n} = |z|^{n}. |
Since the expression on the left is the distance between the two points z^{n} and f(z), while the last expression on the right is the distance of the point z^{n} from the origin, we see that the straight line segment joining the two points f(z) and z^{n} cannot pass through the origin so long as z is on the circle of radius t about the origin. This being so, we may continuously deform the curve traced out by f(z) into the curve traced out by z^{n} without ever passing through the origin, simply by pushing each point f(z) along the segment joining it to z^{n}. Since the order of the origin will vary continuously and can assume only integral values during this deformation, it must be the same for both curves. Since the order for z^{n} is n, the order for f(z) must also be n. This completes the proof.
Reference
- R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, 1966.
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