Give another proof of the fundamental theorem of algebra (see Problem 7.44) as follows. Let$\mathit{I}\mathbf{=}{\mathbf{\oint }}_{\left|z\right|\mathbf{=}\mathbf{R}}\frac{\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{z}\mathbf{\right)}}{\mathbf{f}\mathbf{\left(}\mathbf{z}\mathbf{\right)}}\mathit{d}\mathit{z}$about infinity, that is, in the negative direction around a very large circle C. Use the argument principle (7.8), and also evaluate I by finding the residue of f ‘(z)/f(z) at infinity; thus show that f(z) has n zeros inside C.

Polynomial are analytic functions in the whole complex plane, is an analytic function of z When$\left|z\right|\mathbf{\le }\mathit{R}$ so the argument principle applies and conclude that $\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }\mathbf{i}}$ must equal to the number Of zero of f counted with multiplicity.

Let f(z) be a polynomial of degree n .

Consider the rational function as follows:

$g\left(z\right)=\frac{f\text{'}\left(z\right)}{f\left(z\right)}$

As f(z) is polynomial of degree n as follows:

$g\left(z\right)=\frac{{\mathrm{nz}}^{\mathrm{n}}+\mathrm{lower}\mathrm{degree}\mathrm{terms}}{{\mathrm{z}}^{\mathrm{n}}+\mathrm{lower}\mathrm{degree}\mathrm{terms}}$

This makes it clear that,$\underset{\mathrm{z}\to \infty }{\lim }g\left(z\right)=n$.

Hence there exists a real constant R such that,$\left|g\left(z\right)-n\right|<\frac{1}{2}$

Consider the integral as follows:

$I={\oint }_{\left|z\right|=R}\frac{f\text{'}\left(z\right)}{f\left(z\right)}dz$

This can be written as:

$I={\oint }_{\left|z\right|=R}\frac{g\left(z\right)}{f\left(z\right)}dz$

Split the integrals as follows:

$$\begin{gathered} I_1={\oint }_{\left|z\right|=R}\frac{n}{z}dz{I}_2 \\ {I}_1={\oint }_{\left|z\right|=R}\frac{g\left(z\right)-n}{z}d\left|z\right| \end{gathered}$$

The integral $I_1$ is easy,$I_2=2\pi in$ .

As for,$I_2$ it would be bound for $I_2\left|g\left(z\right)-n\right|$.

$\left|I_2\right|\le {\oint }_{\left|z\right|=R}\frac{g\left(z\right)-n}{\left|z\right|}dz\le \frac{dz}{z}=\mathrm{\pi }$

Since polynomial is an analytic function in the whole complex plane, f is an analytic function of z when $\left|z\right|\le R$so the argument principle applies and concludes that $\frac{1}{2\mathrm{\pi i}}$ must equal to the number of zero of f Counted with multiplicity. Among other things, this means that $\frac{1}{2\mathrm{\pi i}}$must be an integer.

Hence $\left|\left(2\mathrm{\pi i}\right)-I_1\left(2\mathrm{\pi i}\right)\right|$ is an integer.

But,$\left|\frac{I}{\left(2\mathrm{\pi i}\right)}-\frac{I_1}{\left(2\mathrm{\pi i}\right)}\right|=\left|\frac{I_2}{\left(2\mathrm{\pi i}\right)}\right|\le \frac{1}{2}$.

Now, the only integer smaller than $\frac{1}{2}$ in absolute value is 0, so, $I=I_1$.

This implies that f has n zeroes when $\left|z\right|<R$.