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**ThePerfectHacker** That proof is **analytic**! You are using a fact about analytic functions on the complex plane which is proven using the knowledge of contour integration in the complex plane. This is not in any way an algebraic proof, it is 100% analytic.

**Theorem**. Let $\displaystyle P_{n}(z) = a_{z}z^{n}+ a_{n-1}z^{n-1} + \ldots + a_{1}z + a_0 $. Then $\displaystyle P_{n}(z) $ has at least one zero.

**Proof**. Assume to the contrary. Then $\displaystyle f(z) = \frac{1}{P_{n}(z)} $ is a bounded entire function, hence constant by Liouville's Theorem. Contradiction. We have:

$\displaystyle P_{n}(z) = a_{n}z^{n}+ a_{n-1}z^{n-1} + \cdots + a_{1}z + a_0 $

$\displaystyle \therefore |P_{n}(z)| = |z^n| \cdot \left|a_{n}+ \frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2} + \cdots + \frac{a_{1}}{z^{n-1}} + \frac{a_0}{z^n} \right| $

$\displaystyle \geq |z^n| \left( |a_n| - \left|\frac{a_{n-1}}{z} + \frac{a_{n-2}}{z^2} + \cdots + \frac{a_{1}}{z^{n-1}} + \frac{a_0}{z^n} \right| \right) $

$\displaystyle \geq |z^n| \left( |a_n|- \left(\underbrace{\frac{|a_{n-1}|}{|z|} + \frac{|a_{n-2}|}{|z^2|} + \cdots + \frac{|a_{1}|}{|z^{n-1}|} + \frac{|a_0|}{|z^n|}}_{Q} \right) \right) $

Consider two regions in the complex plane. The region inside and on the circle $\displaystyle |z| = R $ and the region outside the circle $\displaystyle |z| = R $. We choose $\displaystyle R $ so large such that $\displaystyle Q \leq \frac{|a_n|}{2} $ for all $\displaystyle z $.

$\displaystyle \geq |z^n| \left(|a_n|- \frac{|a_n|}{2} \right) = |z^n| \frac{|a_2|}{2} $

$\displaystyle \therefore |f(z)| = \frac{1}{P_{n}(z)|} \leq \frac{1}{|z^n| \frac{|a_n|}{2}} \leq \frac{1}{R^{n} \frac{|a_n|}{2}} = M $

So $\displaystyle f(z) $ is a bounded entire function and constant.

Why don't Bak and Newman do this?