Suppose that p(z) is a polynomial in the complex variable z and that p(z) is never zero. Then 1/p(z) is an entire (analytic everywhere) function. By Cauchy's theorem,

I(r) := ∫ (1/zp(z)) dz = 2π i/p(0),

where the integral I(r) is taken over a circle of radius r and centered at the origin. But

|I(r)| ≤ ∫ (1/|z| |p(z)| ) |dz| ≤ (1/r m(r)) ∫ |dz|,

where m(r) is the smallest value of |p(z)| when z ranges on the circle, while the latter integral in the display above is, clearly, the circumference of the circle, i.e. 2 π r. Combining the above we see that

m(r) ≤ p(0).

But this can't be true for all r because, as r → ∞, we have m(r) → ∞. QED

## No comments:

## Post a Comment