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.
show that the function
Is univalent in the domain.
Let z1 and z2 be any two points in |z|<1. then
->z1-z2=0 and z1+z2+2?0 in |z|<1
This is clear from the fact
Thus f is one-one in |z|<1 and hence univalent in the domain.
One can prove that if and are two open connected sets in the complex plane, and
is a univalent function such that (that is, is surjective), then the derivative of is never zero, is invertible, and its inverse is also holomorphic. More, one has by the chain rule
for all in
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
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).
- ^ John B. Conway (1996) Functions of One Complex Variable II, chapter 14: Conformal equivalence for simply connected regions, page 32, Springer-Verlag, New York, ISBN 0-387-94460-5. Definition 1.12: "A function on an open set is univalent if it is analytic and one-to-one."
This article incorporates material from univalent analytic function on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.