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Univalent function

From Wikipedia, the free encyclopedia
Mathematical concept
For other uses, see Univalent.

In mathematics, in the branch of complex analysis, a holomorphic function on an open subset of the complex plane is called univalent if it is injective.[1] [2]

Examples

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The function f : z 2 z + z 2 {\displaystyle f\colon z\mapsto 2z+z^{2}} {\displaystyle f\colon z\mapsto 2z+z^{2}} is univalent in the open unit disc, as f ( z ) = f ( w ) {\displaystyle f(z)=f(w)} {\displaystyle f(z)=f(w)} implies that f ( z ) f ( w ) = ( z w ) ( z + w + 2 ) = 0 {\displaystyle f(z)-f(w)=(z-w)(z+w+2)=0} {\displaystyle f(z)-f(w)=(z-w)(z+w+2)=0}. As the second factor is non-zero in the open unit disc, z = w {\displaystyle z=w} {\displaystyle z=w} so f {\displaystyle f} {\displaystyle f} is injective.

Basic properties

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One can prove that if G {\displaystyle G} {\displaystyle G} and Ω {\displaystyle \Omega } {\displaystyle \Omega } are two open connected sets in the complex plane, and

f : G Ω {\displaystyle f:G\to \Omega } {\displaystyle f:G\to \Omega }

is a univalent function such that f ( G ) = Ω {\displaystyle f(G)=\Omega } {\displaystyle f(G)=\Omega } (that is, f {\displaystyle f} {\displaystyle f} is surjective), then the derivative of f {\displaystyle f} {\displaystyle f} is never zero, f {\displaystyle f} {\displaystyle f} is invertible, and its inverse f 1 {\displaystyle f^{-1}} {\displaystyle f^{-1}} is also holomorphic. More, one has by the chain rule

( f 1 ) ( f ( z ) ) = 1 f ( z ) {\displaystyle (f^{-1})'(f(z))={\frac {1}{f'(z)}}} {\displaystyle (f^{-1})'(f(z))={\frac {1}{f'(z)}}}

for all z {\displaystyle z} {\displaystyle z} in G . {\displaystyle G.} {\displaystyle G.}

Comparison with real functions

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For real analytic functions, unlike for complex analytic (that is, holomorphic) functions, these statements fail to hold. For example, consider the function

f : ( 1 , 1 ) ( 1 , 1 ) {\displaystyle f:(-1,1)\to (-1,1),円} {\displaystyle f:(-1,1)\to (-1,1),円}

given by f ( x ) = x 3 {\displaystyle f(x)=x^{3}} {\displaystyle f(x)=x^{3}}. This function is clearly injective, but its derivative is 0 at x = 0 {\displaystyle x=0} {\displaystyle x=0}, and its inverse is not analytic, or even differentiable, on the whole interval ( 1 , 1 ) {\displaystyle (-1,1)} {\displaystyle (-1,1)}. Consequently, if we enlarge the domain to an open subset G {\displaystyle G} {\displaystyle G} of the complex plane, it must fail to be injective; and this is the case, since (for example) f ( ε ω ) = f ( ε ) {\displaystyle f(\varepsilon \omega )=f(\varepsilon )} {\displaystyle f(\varepsilon \omega )=f(\varepsilon )} (where ω {\displaystyle \omega } {\displaystyle \omega } is a primitive cube root of unity and ε {\displaystyle \varepsilon } {\displaystyle \varepsilon } is a positive real number smaller than the radius of G {\displaystyle G} {\displaystyle G} as a neighbourhood of 0 {\displaystyle 0} {\displaystyle 0}).

See also

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Note

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  1. ^ (Conway 1995, p. 32, chapter 14: Conformal equivalence for simply connected regions, Definition 1.12: "A function on an open set is univalent if it is analytic and one-to-one.")
  2. ^ (Nehari 1975)

References

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This article incorporates material from univalent analytic function on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

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