![In particular, we can choosez≥ R when |a
z
+
b
z
2
+….
q
z
n|≤
1
2 ,since it tends to 0.
So|P(z)
zn |≥1+
1
2
>
1
2
so|P(z)|≥
1
2
R
n
, whenever ≥ R . So | 1
P(z)|≤
1
2
, thus | 1
P(z)|is
bounded above in the domain [
−1
2
,
1
2
].
Then we take the help of a theorem which says, an entire function which is
bounded in the complex plane is a constant. This is Liouville’s theorem in
complex analysis.
Now if we assume that P(z)has no zeroes, or no roots, then | 1
P(z)|is never
infinity and hence bounded above. The function
1
P(z)
is bounded from below
and above, analytic and holomorphic ; so it is an entire function and hence a
constant by Liouville’s theorem it is a constant ! This implies that P(z) is a
constant. But P(z) is a function of z and changes as z changes, so P(z) not a
constant. Hence the contradiction. So our assumption that P(z) has no root is
false (| 1
P(z)| must blow up as →0 ) . Thus P(z)must have a root.
Last but not the least, if a polynomial of degree n has more than n roots, say
n+1 roots, The polynomial must be a constant or an identity, satisfied by every
value of the variable. Since the polynomial when divided by a factorz−a leaves
a remainder 0 and a quotient which is a polynomial of degree n – 1 . The
quotient becomes 0 after n number of steps, even if it contains repeated roots.
Nice! Would it seem a bit clumpy if we don’t prove Liouville’s theorem right
here? Maybe. So let us prove it below, of course a way more than that.
Liouville’s theorem: Every bounded entire function is a constant
This was already proved by Cauchy before it was named after Liouville. Entire
functions are Analytic everywhere in the complex plane.
First Cauchy Riemann equations for analytic functions. Let f (z)=u(x , y)+iv (x, y)
be an analytic function . Since these are functions of two variables, we can](https://image.slidesharecdn.com/proofoffundamentaltheoremofalgebraevenforclass11student-250113121308-5c39340e/85/Proof-of-Fundamental-theorem-of-algebra-even-for-class-11-student-docx-2-320.jpg)
Proof of Fundamental theorem of algebra even for class 11 student.docx
- 1. Fundamental Theorem of Algebra A polynomial of n-th degree has n roots. The theorem says that every polynomial of degree n with complex coefficients has n roots. It suffices to show that every polynomial of degree n with complex coefficients has a roots. For, if r is a root of the polynomial P (z)=zn +a zn−1 +bzn−2 +….q, it would be divisible by z−r by factor theorem, and the resulting polynomial is of degree less than n and it must have a root also. Continuing this process we are sure that the polynomial shall have maximum n roots; no matter if some of the roots may be repeated. If the coefficient ofz n is not 1 then we can divide the polynomial throughout by that coefficient and begin again. A function is said to be analytic at a point if its derivative exist at each point in a neighborhood of that point. It may be a complex function or even real function. A function is said to be differentiable if its derivative exist at each point of its domain. A holomorphic function is a complex analytic function which is infinitely derivable. An entire function is a complex valued function which is holomorphic in the entire complex plane. It is also called an integral function. We want to state that any polynomial is differentiable, analytic and holomorphic and entire easily verifiable. Now , at the first step of proving the fundamental theorem of algebra, let us divide the polynomial by ,zn and it gives P(z) zn =1+ a z + b z2 +…. q zn ,wheneverz≠0 . Now |P(z) zn |≥1+ |a z + b z2 +…. q zn|,wheneverz≠0. Or, |P(z) zn |≥1 when lim ¿z→∞ |a z + b z 2 +…. q z n|=0¿ .
- 2. In particular, we can choosez≥ R when |a z + b z 2 +…. q z n|≤ 1 2 ,since it tends to 0. So|P(z) zn |≥1+ 1 2 > 1 2 so|P(z)|≥ 1 2 R n , whenever ≥ R . So | 1 P(z)|≤ 1 2 , thus | 1 P(z)|is bounded above in the domain [ −1 2 , 1 2 ]. Then we take the help of a theorem which says, an entire function which is bounded in the complex plane is a constant. This is Liouville’s theorem in complex analysis. Now if we assume that P(z)has no zeroes, or no roots, then | 1 P(z)|is never infinity and hence bounded above. The function 1 P(z) is bounded from below and above, analytic and holomorphic ; so it is an entire function and hence a constant by Liouville’s theorem it is a constant ! This implies that P(z) is a constant. But P(z) is a function of z and changes as z changes, so P(z) not a constant. Hence the contradiction. So our assumption that P(z) has no root is false (| 1 P(z)| must blow up as →0 ) . Thus P(z)must have a root. Last but not the least, if a polynomial of degree n has more than n roots, say n+1 roots, The polynomial must be a constant or an identity, satisfied by every value of the variable. Since the polynomial when divided by a factorz−a leaves a remainder 0 and a quotient which is a polynomial of degree n – 1 . The quotient becomes 0 after n number of steps, even if it contains repeated roots. Nice! Would it seem a bit clumpy if we don’t prove Liouville’s theorem right here? Maybe. So let us prove it below, of course a way more than that. Liouville’s theorem: Every bounded entire function is a constant This was already proved by Cauchy before it was named after Liouville. Entire functions are Analytic everywhere in the complex plane. First Cauchy Riemann equations for analytic functions. Let f (z)=u(x , y)+iv (x, y) be an analytic function . Since these are functions of two variables, we can
- 3. differentiate is along any path at any pointz0 , particularly along x keeping y unchanged and then keeping along y keeping x unchanged, and that does not matter. So f '(z) = lim ∆ x →0 u(x+∆ x, y)−u(x , y) ∆x +i lim ∆x →0 v(x+∆ x , y)−v (x , y) ∆ x , Or, f '(z) = ∂u ∂ x +i ∂v ∂ x , keeping y fixed. Similarly f '(z) = ∂v ∂ y + ∂u i∂ y , keeping x fixed, i.e. f '(z) = ∂v ∂ y −i ∂u ∂ y . Thus the Cauchy Riemann equations f '(z) = ∂u ∂ x +i ∂v ∂ x = ∂v ∂ y −i ∂u ∂ y hold for any analytic function. Separating real and imaginary parts, ∂u ∂x = ∂v ∂ y , ∂v ∂x = −∂u ∂ y . These little creatures have vast and varied consequences , interesting, and important. In fact this is in the backbone of complex analysis, conformal theory, Vector Calculus, Riemann geometry, Potential theory, Harmonic analysis, differential equations of boundary value problems, wave equations, heat conduction and diffusion equations, Electromagnetics, Fluid dynamics et al. It is an easy set of steps of calculation to prove the reverse also, that Cauchy Riemann equations guarantee analyticity. Partially differentiating Cauchy Riemann equations by x and y again, we get, ∂2 u ∂x2 = ∂2 v ∂ x∂ y , ∂2 v ∂ y∂ x = −∂2 u ∂2 y .
- 4. Since the functions u and v possess continuous partial derivatives as assumed, ∂ 2 v ∂x ∂ y = ∂ 2 v ∂ y ∂ x , so we get, ∂ 2 u ∂x 2 + ∂ 2 u ∂ 2 y =0, and similarly ∂ 2 v ∂x 2 + ∂ 2 v ∂ 2 y =0. Functions satisfying this condition, such as u and v each, whether in complex analysis, or in vector analysis, in two dimensions or more, are called harmonic functions. The operator such as∇ 2 u= ∂2 u ∂ x2 + ∂2 u ∂2 y =( ∂2 ∂x2 + ∂2 ∂2 y )u Is called Laplacian operator. In vector analysis, in 3 dimensions Gradient operator ∇=( ∂ ∂ x , ∂ ∂ y , ∂ ∂ x )=( ∂ ∂ x i+ ∂ ∂ y j+ ∂ ∂x k) , treated as a vector operator where i , j,k are unit vectors in x, y and z directions, acts on a scalar point functionf (x , y, z) in 3 variables gives gradient of the function , a normal vector to the level surface of f (x , y, z) such as equipotential surfaces in electrostatic fieldE at the point (x , y , z), electrostatic force at a point per unit charge, when a test charge is placed at the point; E=∇ f =( ∂ ∂ x , ∂ ∂ y , ∂ ∂ x )f =(∂ f ∂x i+ ∂f ∂ y j+ ∂f ∂ x k) In the direction of maximum increase in potential, that is the direction of electric field at the point. Divergence of a vector field. Thus the region where there is electrostatic potential f (x , y, z)at each point , becomes a vector field , such as electrostatic field E=∇ f defined at each point . Divergence of the vector E=Ex i+ Ey j+¿ Ez k is given by ∇ . E=( ∂ ∂x i+ ∂ ∂ y j+ ∂ ∂ x k).(Ex i+Ey j+Ez k) Or, ∇ . E= ∂ Ex ∂ x i+ ∂ Ey ∂ y j+ ∂z ∂ x k .
- 5. Divergence comes in Gauss theorem in Electrostatics according to which the total flux of electric field passing out of a closed surface S is equal to total charge enclosed inside the surface, I.e., volume integral of charge density. i.e., ∬ S ❑ E.nds=∭ V ❑ (∇. E)dV =∫ V ❑ ρdV =Q . where V is total volume enclosed within the closed surface S, Q is total charge and is charge density, ρ is charge per unit volume at any point. This is same as total flux out of a unit sphere containing a charge at its center. If isolated charges are there inside the closed surface, total flux out of the surface is same as sum of the fluxes out of individual unit spheres containing isolated charges. A small elementds on the surface S, i.e. its vector areands , n being unit vector perpendicular to this small area ds is integrated over the entire surface. Just counting light sources inside a closed surface by total light passing through the closed surface. Equivalent of this theorem appears in gravitation, magneto statics, and continuity equation in fluid flow. Now Gauss’s theorem can be restated as ρ= ∬ S ❑ E.nds V = ∭ V ❑ (∇. E)dV V , if ρ is constant over the volume (isotropic). Thus for a small infinitesimal region dV bounded by surface dV , ρ= E.nds dV =∇ .E average value of E=∇ f over the enclosing surface. This is differential form of Gauss’ theorem as we are familiar in connection with electrostatics. All these excursions to interesting fields is just to show that ∇ . E=∇.∇f =¿ ( ∂ ∂ x i+ ∂ ∂ y j+ ∂ ∂ x k).(∂ f ∂x i+ ∂f ∂ y j+ ∂f ∂ x k) ¿ ∂ 2 f ∂x 2 + ∂ 2 f ∂ 2 y + ∂ 2 f ∂ x 2 =∇ 2 f ,∇2 being Laplacian operator. The volume average of charge enclosed inside closed surface S.
- 6. In a region devoid of any charges, ∇ 2 f =0, as ∇ 2 f =Q=0. This ∂2 f ∂x2 + ∂2 f ∂2 y is value of a function at a point is average value of the flux over the boundary of the enclosing curve, for any harmonic function f . So to call a function harmonic or analytic is same thing. It is just another version of Cauchy’s integration formula f (a)= 1 2πi ∮ C ❑ f (z) z−a at the root of complex analysis. No need of separately proving this formula, it directly springs out of Gauss’ theorem! The theorems such as Gauss’ theorem, Stoke’s theorem, and Green’s theorem etc. transform line integrals to surface integrals and surface integrals to volume integrals and are of great use in vector calculus. Moreover, they have forms independent of any particular coordinate systems. Now we prove Liouville’s theorem . Since the analytic functionf is bounded, and it is an harmonic function just proved, let if ha ve two supposedly different values f 1, f 2at two different points . The values are volume averages over any sphere around the corresponding points. If we take two large spheres of same volume one around each point, the volumes two spheres would have an arbitrarily negligible proportion of their volume common to both and since f is bounded, the two values being volume averages , shall differ by an arbitrarily small amount. In effect f 1=f 2, i.e. f is constant. Now all the gaps in proof of fundamental theorem of algebra have been plugged, easily accessible for any class 11 student!