Evidently our candidate for a solution to the equation pz 0

Evidently our candidate for a solution to the equation p(z) = 0 should be the point z = a where |p(z)| attains a minimum. To confirm that this candidate works, we need to analyze the local behavior of the polynomial p near a. There exist a positive integer k, a nonzero complex number b, and a polynomial q such that p(z) = p(a) + b(z - a)^k + (z - a)^k+1q(z) for all z. Exercise 12.28. Why can the polynomial p be represented in this way?

Solution

p(z) = p(a) + b(z-a)^k + (z-a)^k+1q(z)

If we put z = a in the above equation, we get:

p(z) = p(a) + b(a-a)^k + (a-a)^k+1q(a)

This implies,

p(z) = p(a) which means z=a.

Hence proved.

We are also given that k is a positive integer which means that powers of (z-a) will always be a positive integer. Therefore, it will only give positive number. Hence, minimum value will always be p(a) as otherwise it will start adding values to it because of k being positive.

This explanation will help you in solving and understanding other questions as well.