Univalent function

Template:Short description {{#invoke:other uses|otheruses}} 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.<ref>Template:Harv</ref><ref>Template:Harv</ref>

Examples

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

Basic properties

One can prove that if <math>G</math> and <math>\Omega</math> are two open connected sets in the complex plane, and

<math>f: G \to \Omega</math>

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

<math>(f^{-1})'(f(z)) = \frac{1}{f'(z)}</math>

for all <math>z</math> in <math>G.</math>

Comparison with real functions

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

<math>f: (-1, 1) \to (-1, 1) \, </math>

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

See also

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References

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