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.

Title	proof of fundamental theorem of algebra (due to Cauchy)
Canonical name	ProofOfFundamentalTheoremOfAlgebradueToCauchy
Date of creation	2013-03-22 19:11:10
Last modified on	2013-03-22 19:11:10
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	8
Author	pahio (2872)
Entry type	Proof
Classification	msc 30A99
Classification	msc 12D99
Synonym	Cauchy proof of fundamental theorem of algebra