Univalent function

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]

Examples[]

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

Basic properties[]

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

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

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

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

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

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

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

given by ƒ(x) = x³. This function is clearly injective, but its derivative is 0 at x = 0, and its inverse is not analytic, or even differentiable, on the whole interval (−1, 1). Consequently, if we enlarge the domain to an open subset G of the complex plane, it must fail to be injective; and this is the case, since (for example) f(εω) = f(ε) (where ω is a primitive cube root of unity and ε is a positive real number smaller than the radius of G as a neighbourhood of 0).

See also[]

• Schlicht function

References[]

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