Lioville's Theorem In Complex Analysis | Mathquery
Lioville's Theorem:-
Statement :
If a function f(z) is analytic for all limit values of z and is bounded then f(z) is constant .
Proof :
Let z₁ , z₂ be any two point of the z-plane the contour C to be a large circle of radius R centred at origin and containing the point z₁ , z₂ . Therefore |r₁|<R and |z₂|<R . Also as f(z) is bounded therefore ∃ a positive M such that f|(z)|≤M for all z.
By Cauchy's Integral formula we have
f(z₁) = 1/2πi ∫ f(z) dz /z-z₁
c
f(z₂) = 1/2πi ∫ f(z) dz /z-z₂
c
∴ f(z₁) - f(z₂) = 1/2πi ∫ f(z) dz /z-z₁
c
- 1/2πi ∫ f(z) dz /z-z₂
c
= 1/2πi ∫ (z₁-z₂) f(z) dz /(z-z₁)(z-z₂)
c
|f(z₁)-f(z₂)|
= |1/2πi ∫(z₁-z₂)f(z) dz /(z-z₁)(z-z₂)|
c
≤ |1/2πi|∫|z₁-z₂||f(z)||dz|/|z-z₁||z-z₂|
c
≤ 1/2π |z₁-z₂|M ∫ |dz|/(|z|-|z₁|)(|z|-|z₂|)
c
as |f(z)|<M
= 1/2π |z₁-z₂|M ∫dz /(R-|z₁|)(R-|z₂|)
c
as |z|=R
Again |z|=R ⇒z= R exp(iθ)
∴ dz = R exp(iθ). i .dθ
∴ |dz| = R dθ
2π
= 1/2π |z₁-z₂|M .R ∫ dθ
0
/ R² (1-|z₁|/R)(1-|z₂|/R)
= 1/2π |z₁-z₂| M .2π
--->0 as R-->∞
/ R(1-|z₁|/R)(1-|z₂|/R)
∴ f(z₁) - f(z₂) = 0 or f(z₁) = f(z₂)
As z₁ and z₂ are arbitrary it follows that f(z) is constant .
= |1/2πi ∫(z₁-z₂)f(z) dz /(z-z₁)(z-z₂)|
c
≤ |1/2πi|∫|z₁-z₂||f(z)||dz|/|z-z₁||z-z₂|
c
≤ 1/2π |z₁-z₂|M ∫ |dz|/(|z|-|z₁|)(|z|-|z₂|)
c
as |f(z)|<M
= 1/2π |z₁-z₂|M ∫dz /(R-|z₁|)(R-|z₂|)
c
as |z|=R
Again |z|=R ⇒z= R exp(iθ)
∴ dz = R exp(iθ). i .dθ
∴ |dz| = R dθ
2π
= 1/2π |z₁-z₂|M .R ∫ dθ
0
/ R² (1-|z₁|/R)(1-|z₂|/R)
= 1/2π |z₁-z₂| M .2π
--->0 as R-->∞
/ R(1-|z₁|/R)(1-|z₂|/R)
∴ f(z₁) - f(z₂) = 0 or f(z₁) = f(z₂)
As z₁ and z₂ are arbitrary it follows that f(z) is constant .
About The Scientist :-
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Liouville's Theorem
Joseph Liouville (24 March 1809 – 8 September 1882)was a French mathematician.For more.
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