# proof of fundamental theorem of algebra (due to Cauchy)

We will prove that any equation

 $f(z)\;:=\;z^{n}\!+\!a_{1}z^{n-1}\!+\!a_{2}z^{n-2}\!+\ldots+\!a_{n-1}z\!+\!a_{n% }\;=\;0,$

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

Proof.  We can suppose that  $a_{n}\neq 0$.  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 $\mathbb{R}^{2}$.  Let  $c:=\sum_{j=1}^{n}|a_{j}|$; it is positive.  Using the triangle inequality   we make the estimation

 $\displaystyle|f(z)|$ $\displaystyle\;=\;|z|^{n}\left|1+\frac{a_{1}}{z}+\frac{a_{2}}{z^{2}}+\ldots+% \frac{a_{n}}{z^{n}}\right|$ $\displaystyle\;\geqq\;\left(1-\frac{|a_{1}|}{|z|}-\frac{|a_{2}|}{|z|^{2}}+% \ldots-\frac{|a_{n}|}{|z|^{n}}\right)$ $\displaystyle\;\geqq\;\left(1-\frac{|a_{1}|}{|z|}-\frac{|a_{2}|}{|z|}+\ldots-% \frac{|a_{n}|}{|z|}\right)$ $\displaystyle\;=\;|z|^{n}\left(1-\frac{c}{|z|}\right)\;\geqq\;\frac{1}{2}|z|^{% n},$

being true for  $|z|>\max\{1,\,2c\}$.  Denote  $r:=\max\{1,\,2c,\,\sqrt[n]{2|a_{n}|}\}$.  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

 $g(x,\,y)\;=\;|f(z)|\;\geqq\;\frac{1}{2}|z|^{n}\;>\;\frac{1}{2}r^{n}\;\geqq\;% \frac{1}{2}\left(\sqrt[n]{2|a_{n}|}\right)^{n}\;=\;|a_{n}|\;>\;0.$

Thus

 $g(x_{0},\,y_{0})\;\leqq\;g(0,\,0)\;=\;|a_{n}|\;<\;|f(z)|\quad\mbox{for}\;\;|z|% \;>\;r.$

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  $z_{0}:=x_{0}\!+\!iy_{0}$,   $z:=z_{0}\!+\!u$  and

 $f(z)\;=\;f(z_{0}\!+\!u)\;:=\;b_{n}\!+\!b_{n-1}u\!+\!b_{n-2}u^{2}\!+\ldots+b_{1% }u^{n-1}\!+\!u^{n}.$

Then  $b_{n}=f(z_{0})\neq 0$  by the antithesis.  Moreover, denote

 $c_{j}\;:=\;\frac{b_{j}}{b_{n}}\quad(j\;=\;1,\,2,\,\ldots,\,n),\;\;c_{0}\;:=\;% \frac{1}{b_{n}}.$

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

 $f(z)\;=\;b_{n}(1\!+\!c_{n-k}u^{k}\!+\!c_{n-k-1}u^{k+1}\!+\ldots\!+\!c_{0}u^{n}).$

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

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

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

 $|\underbrace{c_{n-k-1}u^{k+1}\!+\ldots+\!c_{0}u^{n}}_{h(u)}|\;\leqq\;|c_{n-k-1% }|\varrho^{k+1}\!+\ldots+\!|c_{0}|\varrho^{n}\;\leqq\;(|c_{n-k-1}|\!+\ldots+\!% |c_{0}|)\varrho^{k+1}\;:=\;R\varrho^{k+1}$

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

 $\displaystyle|f(z)|$ $\displaystyle\;=\;|b_{n}|\!\cdot\!|1\!-\!p\varrho^{k}+h(u)|$ $\displaystyle\;\leqq\;|b_{n}|\left[|1\!-\!p\varrho^{k}|+|h(u)|\right]$ $\displaystyle\;\leqq\;|b_{n}|\left[1\!-\!p\varrho^{k}\!+\!R\varrho^{k+1}\right]$ $\displaystyle\;\leqq\;|b_{n}|\left[1\!-\!\varrho^{k}(p\!-\!R\varrho)\right]$ $\displaystyle\;\leqq\;|b_{n}|\left[1\!-\!\varrho^{k}(p\!-\!R\cdot\frac{p}{2R})\right]$ $\displaystyle\;\leqq\;|b_{n}|\left[1\!-\!\frac{1}{p}\!\cdot\!\frac{p}{2}\right]$ $\displaystyle\;\leqq\;\frac{|b_{n}|}{2}\;<\;|b_{n}|\;=\;|f(z_{0})|,$

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) ProofOfFundamentalTheoremOfAlgebradueToCauchy 2013-03-22 19:11:10 2013-03-22 19:11:10 pahio (2872) pahio (2872) 8 pahio (2872) Proof msc 30A99 msc 12D99 Cauchy proof of fundamental theorem of algebra