Editing Fundamental theorem of algebra (section)

===Complex-analytic proofs===
Find a closed [[disk (mathematics)|disk]] ''D'' of radius ''r'' centered at the origin such that |''p''(''z'')|&nbsp;>&nbsp;|''p''(0)| whenever |''z''|&nbsp;≥&nbsp;''r''. The minimum of |''p''(''z'')| on ''D'', which must exist since ''D'' is [[compact set|compact]], is therefore achieved at some point ''z''<sub>0</sub> in the interior of ''D'', but not at any point of its boundary. The [[maximum modulus principle]] applied to 1/''p''(''z'') implies that ''p''(''z''<sub>0</sub>)&nbsp;=&nbsp;0. In other words, ''z''<sub>0</sub> is a zero of ''p''(''z'').

A variation of this proof does not require the maximum modulus principle (in fact, a similar argument also gives a proof of the maximum modulus principle for holomorphic functions). Continuing from before the principle was invoked, if ''a'' := ''p''(''z''<sub>0</sub>) ≠ 0, then, expanding ''p''(''z'') in powers of ''z'' − ''z''<sub>0</sub>, we can write

:<math>p(z) = a + c_k (z-z_0)^k + c_{k+1} (z-z_0)^{k+1} + \cdots + c_n (z-z_0)^n.</math>

Here, the ''c<sub>j</sub>'' are simply the coefficients of the polynomial ''z'' → ''p''(''z'' + ''z''<sub>0</sub>) after expansion, and ''k'' is the index of the first non-zero coefficient following the constant term. For ''z'' sufficiently close to ''z''<sub>0</sub> this function has behavior asymptotically similar to the simpler polynomial <math>q(z) = a+c_k (z-z_0)^k</math>. More precisely, the function

:<math>\left|\frac{p(z)-q(z)}{(z-z_0)^{k+1}}\right|\leq M</math>

for some positive constant ''M'' in some neighborhood of ''z''<sub>0</sub>. Therefore, if we define <math>\theta_0 = (\arg(a)+\pi-\arg(c_k)) /k</math> and let <math>z = z_0 + r e^{i \theta_0}</math> tracing a circle of radius ''r'' > 0 around ''z'', then for any sufficiently small ''r'' (so that the bound ''M'' holds), we see that

:<math>\begin{align}
|p(z)| &\le |q(z)| + r^{k+1} \left|\frac{p(z)-q(z)}{r^{k+1}}\right|\\[4pt]
&\le \left|a +(-1)c_k r^k e^{i(\arg(a)-\arg(c_k))}\right| + M r^{k+1} \\[4pt]
&= |a|-|c_k|r^k + M r^{k+1}
\end{align}</math>

When ''r'' is sufficiently close to 0 this upper bound for |''p''(''z'')| is strictly smaller than |''a''|, contradicting the definition of ''z''<sub>0</sub>. Geometrically, we have found an explicit direction θ<sub>0</sub> such that if one approaches ''z''<sub>0</sub> from that direction one can obtain values ''p''(''z'') smaller in absolute value than |''p''(''z''<sub>0</sub>)|.

Another analytic proof can be obtained along this line of thought observing that, since |''p''(''z'')|&nbsp;>&nbsp;|''p''(0)| outside ''D'', the minimum of |''p''(''z'')| on the whole complex plane is achieved at ''z''<sub>0</sub>. If |''p''(''z''<sub>0</sub>)|&nbsp;>&nbsp;0, then 1/''p'' is a bounded [[holomorphic function]] in the entire complex plane since, for each complex number ''z'', |1/''p''(''z'')|&nbsp;≤&nbsp;|1/''p''(''z''<sub>0</sub>)|. Applying [[Liouville's theorem (complex analysis)|Liouville's theorem]], which states that a bounded entire function must be constant, this would imply that 1/''p'' is constant and therefore that ''p'' is constant. This gives a contradiction, and hence ''p''(''z''<sub>0</sub>)&nbsp;=&nbsp;0.<ref>{{Cite book |last=Ahlfors |first=Lars |title=Complex Analysis |publisher=McGraw-Hill Book Company |edition=2nd |page=122}}</ref>

Yet another analytic proof uses the [[argument principle]]. Let ''R'' be a positive real number large enough so that every root of ''p''(''z'') has absolute value smaller than ''R''; such a number must exist because every non-constant polynomial function of degree ''n'' has at most ''n'' zeros. For each ''r''&nbsp;>&nbsp;''R'', consider the number

:<math>\frac{1}{2\pi i}\int_{c(r)}\frac{p'(z)}{p(z)}\,dz,</math>

where ''c''(''r'') is the circle centered at 0 with radius ''r'' oriented counterclockwise; then the [[argument principle]] says that this number is the number ''N'' of zeros of ''p''(''z'') in the open ball centered at 0 with radius ''r'', which, since ''r''&nbsp;>&nbsp;''R'', is the total number of zeros of ''p''(''z''). On the other hand, the integral of ''n''/''z'' along ''c''(''r'') divided by 2π''i'' is equal to ''n''. But the difference between the two numbers is

:<math>\frac{1}{2\pi i}\int_{c(r)}\left(\frac{p'(z)}{p(z)}-\frac{n}{z}\right)dz=\frac{1}{2\pi i}\int_{c(r)}\frac{zp'(z)-np(z)}{zp(z)}\,dz.</math>

The numerator of the rational expression being integrated has degree at most ''n''&nbsp;−&nbsp;1 and the degree of the denominator is ''n''&nbsp;+&nbsp;1. Therefore, the number above tends to 0 as ''r'' → +∞. But the number is also equal to ''N''&nbsp;−&nbsp;''n'' and so ''N''&nbsp;=&nbsp;''n''.

Another complex-analytic proof can be given by combining [[linear algebra]] with the [[Cauchy's integral theorem|Cauchy theorem]]. To establish that every complex polynomial of degree ''n''&nbsp;>&nbsp;0 has a zero, it suffices to show that every complex [[square matrix]] of size ''n''&nbsp;>&nbsp;0 has a (complex) [[eigenvalue]].<ref>A proof of the fact that this suffices can be seen [[Algebraically closed field#Every endomorphism of Fn has some eigenvector|here]].</ref> The proof of the latter statement is [[Proof by contradiction|by contradiction]].

Let ''A'' be a complex square matrix of size ''n''&nbsp;>&nbsp;0 and let ''I<sub>n</sub>'' be the unit matrix of the same size. Assume ''A'' has no eigenvalues. Consider the [[resolvent formalism|resolvent]] function

:<math> R(z)=(zI_n-A)^{-1},</math>

which is a [[meromorphic function]] on the complex plane with values in the vector space of matrices. The eigenvalues of ''A'' are precisely the poles of ''R''(''z''). Since, by assumption, ''A'' has no eigenvalues, the function ''R''(''z'') is an [[entire function]] and [[Cauchy's integral theorem|Cauchy theorem]] implies that

:<math> \int_{c(r)} R(z) \, dz =0.</math>

On the other hand, ''R''(''z'') expanded as a geometric series gives:

:<math>R(z)=z^{-1}(I_n-z^{-1}A)^{-1}=z^{-1}\sum_{k=0}^\infty \frac{1}{z^k}A^k\cdot</math>

This formula is valid outside the closed [[disc (mathematics)|disc]] of radius <math>\|A\|</math> (the [[operator norm]] of ''A''). Let <math>r>\|A\|.</math> Then

:<math>\int_{c(r)}R(z)dz=\sum_{k=0}^{\infty}\int_{c(r)}\frac{dz}{z^{k+1}}A^k=2\pi iI_n</math>

(in which only the summand ''k''&nbsp;=&nbsp;0 has a nonzero integral). This is a contradiction, and so ''A'' has an eigenvalue.

Finally, [[Rouché's theorem]] gives perhaps the shortest proof of the theorem.