Univalent function: Difference between revisions

Revision as of 04:48, 9 January 2014

Revision as of 12:46, 5 May 2014

Line 7:

One can prove that if <math>G</math> and <math>\Omega</math> are two open [[connected space|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_functionSurjective function|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>

Line 17:

== Comparison with real functions ==

For [[real number|real]] [[analytic function]]s, 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 ''&fnof;''(''x'')&nbsp;=&nbsp;''x''³. This function is clearly injective, but its derivative is 0 at ''x''&nbsp;=&nbsp;0, and its inverse is not analytic, or even differentiable, on the whole interval&nbsp;(&minus;1,&nbsp;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''(&epsilon;&omega;)&nbsp;= ''f''(&epsilon;) (where &omega; is a [[primitive root of unity|primitive cube root of unity]] and &epsilon; is a positive real number smaller than the radius of ''G'' as a neighbourhood of 0).