Editing Fundamental theorem of algebra (section)

==Proofs==
All proofs below involve some [[mathematical analysis]], or at least the [[Topology|topological]] concept of [[continuous function|continuity]] of real or complex functions. Some also use [[Derivative|differentiable]] or even [[Analytic function|analytic]] functions. This requirement has led to the remark that the Fundamental Theorem of Algebra is neither fundamental, nor a theorem of algebra.<ref>{{Cite book|last1=Aigner|first1=Martin|url=http://worldcat.org/oclc/1033531310|title=Proofs from the book|last2=Ziegler|first2=Günter|publisher=Springer|year=2018|isbn=978-3-662-57264-1|pages=151|oclc=1033531310}}</ref>

Some proofs of the theorem only prove that any non-constant polynomial with ''real'' coefficients has some complex root. This lemma is enough to establish the general case because, given a non-constant polynomial {{math|''p''}} with complex coefficients, the polynomial
:<math>q=p\overline{p},</math>
has only real coefficients, and, if {{math|''z''}} is a root of {{math|''q''}}, then either {{math|''z''}} or its conjugate is a root of {{math|''p''}}. Here, <math>\overline{p}</math> is the polynomial obtained by replacing each coefficient of {{mvar|p}} with its [[complex conjugate]]; the roots of <math>\overline{p}</math> are exactly the complex conjugates of the roots of {{mvar|p}}.

Many non-algebraic proofs of the theorem use the fact (sometimes called the "growth lemma") that a polynomial function ''p''(''z'') of degree ''n'' whose dominant coefficient is 1 behaves like ''z<sup>n</sup>'' when |''z''| is large enough. More precisely, there is some positive real number ''R'' such that

:<math>\tfrac{1}{2}|z^n|<|p(z)|<\tfrac{3}{2}|z^n|</math>

when |''z''|&nbsp;>&nbsp;''R''.

===Real-analytic proofs===
Even without using complex numbers, it is possible to show that a real-valued polynomial ''p''(''x''): ''p''(0) ≠ 0 of degree ''n'' > 2 can always be divided by some quadratic polynomial with real coefficients.<ref>{{Cite journal |last=Basu |first=Soham |date=October 2021|title=Strictly real fundamental theorem of algebra using polynomial interlacing|journal=[[Australian Mathematical Society|Bulletin of the Australian Mathematical Society]] |language=en |volume=104 |issue=2 |pages=249–255 |doi=10.1017/S0004972720001434 |mr=4308140|doi-access=free }}</ref> In other words, for some real-valued ''a'' and ''b'', the coefficients of the linear remainder on dividing ''p''(''x'') by ''x''<sup>2</sup> − ''ax'' − ''b'' simultaneously become zero.

: <math>p(x) = (x^2 - ax - b) q(x) + x\,R_{p(x)}(a, b) + S_{p(x)}(a, b),</math>

where ''q''(''x'') is a polynomial of degree ''n'' − 2. The coefficients ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') are independent of ''x'' and completely defined by the coefficients of ''p''(''x''). In terms of representation, ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') are bivariate polynomials in ''a'' and ''b''. In the flavor of Gauss's first (incomplete) proof of this theorem from 1799, the key is to show that for any sufficiently large negative value of ''b'', all the roots of both ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') in the variable ''a'' are real-valued and alternating each other (interlacing property). Utilizing a [[Sturm's theorem|Sturm-like]] chain that contain ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') as consecutive terms, interlacing in the variable ''a'' can be shown for all consecutive pairs in the chain whenever ''b'' has sufficiently large negative value. As ''S''<sub>''p''</sub>(''a'', ''b'' = 0) = ''p''(0) has no roots, interlacing of ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') in the variable ''a'' fails at ''b'' = 0. Topological arguments can be applied on the interlacing property to show that the locus of the roots of ''R''<sub>''p''(''x'')</sub>(''a'', ''b'') and ''S''<sub>''p''(''x'')</sub>(''a'', ''b'') must intersect for some real-valued ''a'' and ''b'' < 0.

===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.


===Topological proofs===
[[File:Koreny.gif|thumb|right|Animation illustrating the proof on the polynomial <math>x^5-x-1</math>|150x150px]]

Suppose the minimum of |''p''(''z'')| on the whole complex plane is achieved at ''z''<sub>0</sub>; it was seen at the proof which uses Liouville's theorem that such a number must exist. We can write ''p''(''z'') as a polynomial in ''z''&nbsp;−&nbsp;''z''<sub>0</sub>: there is some natural number ''k'' and there are some complex numbers ''c<sub>k</sub>'', ''c''<sub>''k''&nbsp;+&nbsp;1</sub>, ..., ''c<sub>n</sub>'' such that ''c<sub>k</sub>''&nbsp;≠&nbsp;0 and:

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

If ''p''(''z''<sub>0</sub>) is nonzero, it follows that if ''a'' is a ''k''<sup>th</sup> root of −''p''(''z''<sub>0</sub>)/''c<sub>k</sub>'' and if ''t'' is positive and sufficiently small, then |''p''(''z''<sub>0</sub>&nbsp;+&nbsp;''ta'')|&nbsp;<&nbsp;|''p''(''z''<sub>0</sub>)|, which is impossible, since |''p''(''z''<sub>0</sub>)| is the minimum of |''p''| on ''D''.

For another topological proof by contradiction, suppose that the polynomial ''p''(''z'') has no roots, and consequently is never equal to 0. Think of the polynomial as a map from the complex plane into the complex plane. It maps any circle |''z''|&nbsp;=&nbsp;''R'' into a closed loop, a curve ''P''(''R''). We will consider what happens to the [[winding number]] of ''P''(''R'') at the extremes when ''R'' is very large and when ''R'' = 0. When ''R'' is a sufficiently large number, then the leading term ''z<sup>n</sup>'' of ''p''(''z'') dominates all other terms combined; in other words,

:<math>\left | z^n \right | > \left | a_{n-1} z^{n-1} + \cdots + a_0 \right |.</math>

When ''z'' traverses the circle <math>Re^{i\theta}</math> once counter-clockwise <math>(0\leq \theta \leq 2\pi),</math> then <math>z^n=R^ne^{in\theta}</math> winds ''n'' times counter-clockwise <math>(0\leq \theta \leq 2\pi n)</math> around the origin (0,0), and ''P''(''R'') likewise. At the other extreme, with |''z''|&nbsp;=&nbsp;0, the curve ''P''(0) is merely the single point ''p''(0), which must be nonzero because ''p''(''z'') is never zero. Thus ''p''(0) must be distinct from the origin (0,0), which denotes 0 in the complex plane. The winding number of ''P''(0) around the origin (0,0) is thus 0. Now changing ''R'' continuously will [[homotopy|deform the loop continuously]]. At some ''R'' the winding number must change. But that can only happen if the curve ''P''(''R'') includes the origin (0,0) for some ''R''. But then for some ''z'' on that circle |''z''|&nbsp;=&nbsp;''R'' we have ''p''(''z'') = 0, contradicting our original assumption. Therefore, ''p''(''z'') has at least one zero.

===Algebraic proofs===
These proofs of the Fundamental Theorem of Algebra must make use of the following two facts about real numbers that are not algebraic but require only a small amount of analysis (more precisely, the [[intermediate value theorem]] in both cases):
* every polynomial with an odd degree and real coefficients has some real root;
* every non-negative real number has a square root.

The second fact, together with the [[quadratic formula]], implies the theorem for real quadratic polynomials. In other words, algebraic proofs of the fundamental theorem actually show that if ''R'' is any [[real-closed field]], then its extension ''C'' = ''R''({{radic|−1}}) is algebraically closed.

====By induction====
As mentioned above, it suffices to check the statement "every non-constant polynomial ''p''(''z'') with real coefficients has a complex root". This statement can be proved by induction on the greatest non-negative integer ''k'' such that 2<sup>''k''</sup> divides the degree ''n'' of ''p''(''z''). Let ''a'' be the coefficient of ''z<sup>n</sup>'' in ''p''(''z'') and let ''F'' be a [[splitting field]] of ''p''(''z'') over ''C''; in other words, the field ''F'' contains ''C'' and there are elements ''z''<sub>1</sub>, ''z''<sub>2</sub>, ..., ''z<sub>n</sub>'' in ''F'' such that

:<math>p(z)=a(z-z_1)(z-z_2) \cdots (z-z_n).</math>

If ''k''&nbsp;=&nbsp;0, then ''n'' is odd, and therefore ''p''(''z'') has a real root. Now, suppose that ''n''&nbsp;=&nbsp;2''<sup>k</sup>m'' (with ''m'' odd and ''k''&nbsp;>&nbsp;0) and that the theorem is already proved when the degree of the polynomial has the form 2<sup>''k''&nbsp;−&nbsp;1</sup>''m''′ with ''m''′ odd. For a real number ''t'', define:

:<math>q_t(z)=\prod_{1\le i<j\le n}\left(z-z_i-z_j-tz_iz_j\right).</math>

Then the coefficients of ''q<sub>t</sub>''(''z'') are [[symmetric polynomial]]s in the ''z<sub>i</sub>'' with real coefficients. Therefore, they can be expressed as polynomials with real coefficients in the [[elementary symmetric polynomial]]s, that is, in −''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., (−1)''<sup>n</sup>a<sub>n</sub>''. So ''q<sub>t</sub>''(''z'') has in fact ''real'' coefficients. Furthermore, the degree of ''q<sub>t</sub>''(''z'') is ''n''(''n''&nbsp;−&nbsp;1)/2&nbsp;=&nbsp;2<sup>''k''−1</sup>''m''(''n''&nbsp;−&nbsp;1), and ''m''(''n''&nbsp;−&nbsp;1) is an odd number. So, using the induction hypothesis, ''q<sub>t</sub>'' has at least one complex root; in other words, ''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>''&nbsp;+&nbsp;''tz<sub>i</sub>z<sub>j</sub>'' is complex for two distinct elements ''i'' and ''j'' from {1, ..., ''n''}. Since there are more real numbers than pairs (''i'', ''j''), one can find distinct real numbers ''t'' and ''s'' such that ''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>''&nbsp;+&nbsp;''tz<sub>i</sub>z<sub>j</sub>'' and ''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>''&nbsp;+&nbsp;''sz<sub>i</sub>z<sub>j</sub>'' are complex (for the same ''i'' and ''j''). So, both ''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>'' and ''z<sub>i</sub>z<sub>j</sub>'' are complex numbers. It is easy to check that every complex number has a complex square root, thus every complex polynomial of degree 2 has a complex root by the quadratic formula. It follows that ''z<sub>i</sub>'' and ''z<sub>j</sub>'' are complex numbers, since they are roots of the quadratic polynomial ''z''<sup>2</sup>&nbsp;−&nbsp; (''z<sub>i</sub>''&nbsp;+&nbsp;''z<sub>j</sub>'')''z''&nbsp;+&nbsp;''z<sub>i</sub>z<sub>j</sub>''.

Joseph Shipman showed in 2007 that the assumption that odd degree polynomials have roots is stronger than necessary; any field in which polynomials of prime degree have roots is algebraically closed (so "odd" can be replaced by "odd prime" and this holds for fields of all characteristics).<ref>Shipman, J. [http://www.jon-arny.com/httpdocs/Gauss/Shipman%20Intellig%20Mod%20p%20FTA.pdf Improving the Fundamental Theorem of Algebra].  ''The Mathematical Intelligencer'', volume&nbsp;29 (2007), number&nbsp;4, pp.&nbsp;9–14.</ref> For axiomatization of algebraically closed fields, this is the best possible, as there are counterexamples if a single prime is excluded. However, these counterexamples rely on −1 having a square root. If we take a field where −1 has no square root, and every polynomial of degree ''n''&nbsp;∈&nbsp;''I'' has a root, where ''I'' is any fixed infinite set of odd numbers, then every polynomial ''f''(''x'') of odd degree has a root (since {{nowrap|(''x''<sup>2</sup> + 1)<sup>''k''</sup>''f''(''x'')}} has a root, where ''k'' is chosen so that {{nowrap|deg(''f'') + 2''k'' ∈ ''I''}}).

====From Galois theory====
Another algebraic proof of the fundamental theorem can be given using [[Galois theory]]. It suffices to show that '''C''' has no proper finite [[field extension]].<ref>A proof of the fact that this suffices can be seen [[Algebraically closed field#The field has no proper finite extension|here]].</ref> Let ''K''/'''C''' be a finite extension. Since the [[Normal extension#Normal closure|normal closure]] of ''K'' over '''R''' still has a finite degree over '''C''' (or '''R'''), we may assume [[without loss of generality]] that ''K'' is a [[normal extension]] of '''R''' (hence it is a [[Galois extension]], as every algebraic extension of a field of [[characteristic (algebra)|characteristic]] 0 is [[separable extension|separable]]). Let ''G'' be the [[Galois group]] of this extension, and let ''H'' be a [[Sylow theorems|Sylow]] 2-subgroup of ''G'', so that the [[order (group theory)|order]] of ''H'' is a power of 2, and the [[index of a subgroup|index]] of ''H'' in ''G'' is odd. By the [[fundamental theorem of Galois theory]], there exists a subextension ''L'' of ''K''/'''R''' such that Gal(''K''/''L'')&nbsp;=&nbsp;''H''. As [''L'':'''R''']&nbsp;=&nbsp;[''G'':''H''] is odd, and there are no nonlinear irreducible real polynomials of odd degree, we must have ''L''&nbsp;= '''R''', thus [''K'':'''R'''] and [''K'':'''C'''] are powers of 2. Assuming by way of contradiction that [''K'':'''C''']&nbsp;>&nbsp;1, we conclude that the [[p-group|2-group]] Gal(''K''/'''C''') contains a subgroup of index 2, so there exists a subextension ''M'' of '''C''' of degree&nbsp;2. However, '''C''' has no extension of degree&nbsp;2, because every quadratic complex polynomial has a complex root, as mentioned above. This shows that [''K'':'''C'''] = 1, and therefore ''K'' = '''C''', which completes the proof.

===Geometric proofs===
There exists still another way to approach the fundamental theorem of algebra, due to J. M. Almira and A. Romero: by [[Riemannian geometry|Riemannian geometric]] arguments. The main idea here is to prove that the existence of a non-constant polynomial ''p''(''z'') without zeros implies the existence of a [[Flat manifold|flat Riemannian metric]] over the sphere '''S'''<sup>2</sup>. This leads to a contradiction since the sphere is not flat.

A Riemannian surface (''M'', ''g'') is said to be flat if its [[Gaussian curvature]], which we denote by ''K<sub>g</sub>'', is identically null. Now, the [[Gauss–Bonnet theorem]], when applied to the sphere '''S'''<sup>2</sup>, claims that

:<math>\int_{\mathbf{S}^2}K_g=4\pi,</math>

which proves that the sphere is not flat.

Let us now assume that ''n'' > 0 and

:<math>p(z) = a_0 + a_1 z + \cdots + a_n z^n \neq 0</math>

for each complex number ''z''. Let us define

:<math>p^*(z) = z^n p \left ( \tfrac{1}{z} \right ) = a_0 z^n + a_1 z^{n-1} + \cdots + a_n.</math>

Obviously, ''p*''(''z'')&nbsp;≠ 0 for all ''z'' in '''C'''. Consider the polynomial ''f''(''z'')&nbsp;=&nbsp;''p''(''z'')''p*''(''z''). Then ''f''(''z'')&nbsp;≠ 0 for each ''z'' in '''C'''. Furthermore,

:<math>f(\tfrac{1}{w}) = p \left (\tfrac{1}{w} \right )p^* \left (\tfrac{1}{w} \right ) = w^{-2n}p^*(w)p(w) = w^{-2n}f(w).</math>

We can use this functional equation to prove that ''g'', given by

:<math>g=\frac{1}{|f(w)|^{\frac{2}{n}}}\,|dw|^2 </math>

for ''w'' in '''C''', and

:<math>g=\frac{1}{\left |f\left (\tfrac{1}{w} \right ) \right |^{\frac{2}{n}}}\left |d\left (\tfrac{1}{w} \right ) \right |^2 </math>

for ''w''&nbsp;∈&nbsp;'''S'''<sup>2</sup>\{0}, is a well defined Riemannian metric over the sphere '''S'''<sup>2</sup> (which we identify with the extended complex plane '''C'''&nbsp;∪&nbsp;{∞}).

Now, a simple computation shows that

:<math>\forall w\in\mathbf{C}: \qquad  \frac{1}{|f(w)|^{\frac{1}{n}}} K_g=\frac{1}{n}\Delta \log|f(w)|=\frac{1}{n}\Delta \text{Re}(\log f(w))=0,</math>

since the real part of an analytic function is harmonic. This proves that ''K<sub>g</sub>''&nbsp;=&nbsp;0.