# Kopie von Exploring the Fundamental Theorem of Algebra

What do we see? To the left is a circle centred at the origin with radius where . To the right there appears to be another circle, but this is not so. The closed path is the locus of where .
Check the following:
Rotating ), the locus of

*A*lies on the x-axis. The point,*z*, lies on this circle. We think of*z*as a complex number. Indeed,*z*can be**any**complex number. Here*p*is some polynomial with complex coefficients. Here*z*through one revolution about the circle causes*w*to rotate one revolution about its locus. As the radius of the circle diminishes (*w*approximates a circle of radius*r*centred at 1. As this radius increases, the locus of*w*behaves strangely! Try*A*= 0.4, 0.6, 0.8 and 1. Describe what you see.Now zoom out, so you can see the whole picture. Notice that the locus passes through the origin. Why is this? Now try with , the constant coefficient in . On the other hand, for sufficiently large , then the locus must, for some value of for some

*A*= 2, 3. Zoom out again. Describe what you see. Notice that as*z*rotates through a revolution,*w*does so twice. We say that the locus has**winding number**2 (about the origin). Try*A*= 5, 10. Zoom out and describe. What about 50? Now explore all of this for*n*= 3, 4, 5, ... (To do this, just right click on*w*and edit Object Properties.) Try any polynomial,*p*, of your choice, including ones with imaginary coefficients, and, in particular, with imaginary constant coefficients. No matter what polynomial,*p*, we choose, we see that for sufficiently small*r*, the locus of*w*is almost a (small) circle centred at*r*, the locus is almost a (large) circle (with winding number*n*) centred at the origin. (What is the radius of each of these circles?) If*r*, pass through the origin. Thus*z*, proving the**Fundamental Theorem of Algebra**. Of course, the word 'must' needs further scrutiny and relies on the fact that the locus of*w*varies**continuously**with*r*.## New Resources

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