Question

Let p(z) be a polynomial of degree n and let R>0 be sufficiently large so that p(z) is never zero in the domain ∣z∣>R. Let the contour C be the positively oriented circle centered at zero with radius R. Show that

1/2πi ∫ p′(z)/p(z) =n

Answer 1

The integral of p'(z)/p(z) along a sufficiently large positively oriented circle represents the number of zeros of the polynomial within the circle, which equals its degree n.

Step-by-step explanation:

The integral in question is an application of the argument principle from complex analysis, which relates the number of zeros and poles of a function inside a closed contour to a contour integral. We consider the positively oriented circle C centered at zero with radius R for a polynomial p(z) of degree n. Since p(z) is non-zero for |z| > R, it has all its zeros inside the circle. The integral evaluates to 2πi times the number of zeros inside C, counting multiplicities. The polynomial of degree n has exactly n zeros, so the integral evaluates to 2πi times n, and after dividing by 2πi, we obtain the result n.