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Question 5.19: Use Rouche´’s theorem (Problem 5.18) to prove that every pol...

Use Rouche´’s theorem (Problem 5.18) to prove that every polynomial of degree n has exactly n zeros (fundamental theorem of algebra).

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Suppose the polynomial to be a_0+a_1 z+a_2 z^2+\cdots+a_n z^n, where a_n \neq 0. Choose f(z)=a_n z^n and g(z)=a_0+a_1 z+a_2 z^2+\cdots+a_{n-1} z^{n-1}.

If C is a circle having center at the origin and radius r > 1, then on C we have

\begin{aligned}\left|\frac{g(z)}{f(z)}\right|=& \frac{\left|a_0+a_1 z+a_2 z^2+\cdots+a_{n-1} z^{n-1}\right|}{\left|a_n z^n\right|} \leq \frac{\left|a_0\right|+\left|a_1\right| r+\left|a_2\right| r^2+\cdots+\left|a_{n-1}\right| r^{n-1}}{\left|a_n\right| r^n} \\& \leq \frac{\left|a_0\right| r^{n-1}+\left|a_1\right| r^{n-1}+\left|a_2\right| r^{n-1}+\cdots+\left|a_{n-1}\right| r^{n-1}}{\left|a_n\right| r^n}=\frac{\left|a_0\right|+\left|a_1\right|+\left|a_2\right|+\cdots+\left|a_{n-1}\right|}{\left|a_n\right| r}\end{aligned}

Then, by choosing r large enough, we can make |g(z)/f(z)| < 1, i.e., |g(z)| < |f(z)|. Hence, by Rouche´’s theorem, the given polynomial f(z) + g(z) has the same number of zeros as f(z) =a_nz^n. But, since this last function has n zeros all located at z = 0, f(z) + g(z) also has n zeros and the proof is complete.

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