proof of fundamental theorem of algebra (due to Cauchy)

We will prove that any equation

$$f(z):=z^n+a_1{z}^{n-1}+a_2{z}^{n-2}+\mathrm{\dots }+a_{n-1}z+a_n=\mathrm{\hspace{0.33em}0},$$

where the coefficients $a_j$ are complex numbers and  $n\geqq 1$,  has at least one root (http://planetmath.org/Equation) in$ℂ$.

Proof.  We can suppose that  $a_n\ne 0$.  Denote  $z:=x+iy$  where  $x,y$ are real.  Then the function

$$g(x,y):=|f(z)|=|f(x+iy)|$$

is defined and continuous in the whole ${ℝ}^2$.  Let  $c:={\sum }_{j=1}^n|a_j|$; it is positive.  Using the triangle inequality we make the estimation

$|f(z)|$$\mathrm{\hspace{0.33em}}={|z|}^n\left|1+{\displaystyle \frac{a_1}{z}}+{\displaystyle \frac{a_2}{z^2}}+\mathrm{\dots }+{\displaystyle \frac{a_n}{z^n}}\right|$

$\mathrm{\hspace{0.33em}}\geqq \left(1-{\displaystyle \frac{|a_1|}{|z|}}-{\displaystyle \frac{|a_2|}{{|z|}^2}}+\mathrm{\dots }-{\displaystyle \frac{|a_n|}{{|z|}^n}}\right)$

$\mathrm{\hspace{0.33em}}\geqq \left(1-{\displaystyle \frac{|a_1|}{|z|}}-{\displaystyle \frac{|a_2|}{|z|}}+\mathrm{\dots }-{\displaystyle \frac{|a_n|}{|z|}}\right)$

$\mathrm{\hspace{0.33em}}={|z|}^n\left(1-{\displaystyle \frac{c}{|z|}}\right)\geqq {\displaystyle \frac{1}{2}}{|z|}^n,$

being true for  $|z|>\max \{1,\mathrm{\hspace{0.17em}2}c\}$.  Denote  $r:=\max \{1,\mathrm{\hspace{0.17em}2}c,\sqrt[n]{2|a_n|}\}$.  Consider the disk $x^2+y^2\leqq r^2$.  Because it is compact, the function  $g(x,y)$  attains at a point  $(x_0,y_0)$  of the disk its absolute minimum value (infimum) in the disk.  If  $|z|>r$,  we have

$$g(x,y)=|f(z)|\geqq \frac{1}{2}{|z|}^n>\frac{1}{2}{r}^n\geqq \frac{1}{2}{\left(\sqrt[n]{2|a_n|}\right)}^n=|a_n|>\mathrm{\hspace{0.33em}0}.$$

Thus

Hence $g(x_0,y_0)$ is the absolute minimum of $g(x,y)$ in the whole complex plane.  We show that $g(x_0,y_0)=0$.  Therefore we make the antithesis that  $g(x_0,y_0)>0$.

Denote  $z_0:=x_0+i{y}_0$,   $z:=z_0+u$  and

$$f(z)=f(z_0+u):=b_n+b_{n-1}u+b_{n-2}{u}^2+\mathrm{\dots }+b_1{u}^{n-1}+u^n.$$

Then  $b_n=f(z_0)\ne 0$  by the antithesis.  Moreover, denote

$$c_j:=\frac{b_j}{b_n}\mathit{\hspace{1em}}(j=\mathrm{\hspace{0.33em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots },n),c_0:=\frac{1}{b_n}.$$

and assume that  $c_{n-1}=c_{n-2}=\mathrm{\dots }=c_{n-k+1}=0$  but  $c_{n-k}\ne 0.$  Thus we may write

$$f(z)=b_n(1+c_{n-k}{u}^k+c_{n-k-1}{u}^{k+1}+\mathrm{\dots }+c_0{u}^n).$$

If  $c_{n-k}=p(\cos \alpha +i\sin \alpha )$  and  $u=\varrho (\cos \phi +i\sin \phi )$, then

$$c_{n-k}{u}^k=p{\varrho }^k[\cos (\alpha +k\phi )+i\sin (\alpha +k\phi )]$$

by de Moivre identity.  Choosing  $\varrho \leqq 1$  and  $\phi =\frac{\pi -\alpha }{k}$  we get $c_{n-k}{u}^k=-p{\varrho }^k$  and can make the estimation

$$|\underset{h(u)}{\underbrace{c_{n-k-1}{u}^{k+1}+\mathrm{\dots }+c_0{u}^n}}|\leqq |c_{n-k-1}|{\varrho }^{k+1}+\mathrm{\dots }+|c_0|{\varrho }^n\leqq (|c_{n-k-1}|+\mathrm{\dots }+|c_0|){\varrho }^{k+1}:=R{\varrho }^{k+1}$$

where $R$ is a constant.  Let now  $\varrho =\min \{1,\sqrt[k]{\frac{1}{p}},\frac{p}{2R}\}$.  We obtain

$|f(z)|$$\mathrm{\hspace{0.33em}}=|b_n|\cdot |1-p{\varrho }^k+h(u)|$

$\mathrm{\hspace{0.33em}}\leqq |b_n|\left[|1-p{\varrho }^k|+|h(u)|\right]$

$\mathrm{\hspace{0.33em}}\leqq |b_n|\left[1-p{\varrho }^k+R{\varrho }^{k+1}\right]$

$\mathrm{\hspace{0.33em}}\leqq |b_n|\left[1-{\varrho }^k(p-R\varrho )\right]$

$\mathrm{\hspace{0.33em}}\leqq |b_n|\left[1-{\varrho }^k(p-R\cdot {\displaystyle \frac{p}{2R}})\right]$

$\mathrm{\hspace{0.33em}}\leqq |b_n|\left[1-{\displaystyle \frac{1}{p}}\cdot {\displaystyle \frac{p}{2}}\right]$

which result is impossible since $|f(z_0|$ was the absolute minimum.  Consequently, the antithesis is wrong, and the proof is settled.