Antiderivative (complex analysis)
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In complex analysis, a branch of mathematics, the antiderivative of a complex-valued function is a function whose complex derivative is the original function. As such, this concept is the analog of the antiderivative of a real-valued function, and these two notions have many similar properties as well as significant differences.
Formally, given an open set <math>U</math> in the complex plane and a function <math>g:U\to \mathbb C,</math> the antiderivative of <math>g</math> is a function <math>f:U\to \mathbb C,</math> whose complex derivative is <math>g</math>, <math>f'=g.</math>
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[edit] Uniqueness of antiderivative
The derivative of a constant function is zero. Therefore, any constant is an antiderivative of the zero function. If <math>U</math> is a connected set, then the constants are the only antiderivatives of the zero function. Otherwise, a function is an antiderivative of the zero function if and only if it is constant on each connected component of <math>U</math> (those constants need not be equal).
This observation can be used to establish that if a function <math>g:U\to \mathbb C</math> has an antiderivative, then that antiderivative is unique up to addition of a function which is constant on each connected component of <math>U</math>.
[edit] Existence of antiderivative
[edit] Necessary conditions
If <math>f</math> is an antiderivative of <math>g,</math> then <math>f</math> is differentiable, that is, holomorphic. Then <math>g</math> is also holomorphic. As such, for a function to admit an antiderivative it must be holomorphic, that is, locally expandable into a power series. This is in stark contrast to functions of a real variable, where continuity or even weaker assumptions on a function can guarantee that it has an antiderivative.
If <math>f</math> is an antiderivative of <math>g</math> on <math>U,</math> then given any piecewise C1 path <math>\gamma:[a, b]\to U</math> one can express the path integral of <math>g</math> over <math>\gamma</math> as
- <math>\int_\gamma\! g(z)\,dz=\int_a^b \!g\left(\gamma(t)\right)\gamma'(t)\, dt=\int_a^b \!f'\left(\gamma(t)\right)\gamma'(t)\,dt.</math>
By the chain rule and the fundamental theorem of calculus one then has
- <math>\int_\gamma \!g(z)\,dz=\int_a^b\! \frac{d}{dt}f\left(\gamma(t)\right)\,dt=f\left(\gamma(b)\right)-f\left(\gamma(a)\right).</math>
As such, the integral of <math>g</math> over <math>\gamma</math> does not depend on the actual path <math>\gamma</math>, but only on its endpoints.
This observation shows that not every holomorphic function <math>g</math> admits an antiderivative. For example, consider the reciprocal function, <math>g(z)= z^{-1}</math> defined on <math>\mathbb C\backslash\{0\}.</math> Given a nonzero number <math>z_0</math>, one can find a circle going through <math>z_0</math> which contains the origin inside of it, and a circle going through <math>z_0</math> which does not contain the origin inside of it. By the residue theorem, the path integral over the first circle will be non-zero, and over the second one will be zero. Therefore, there exist two paths starting and ending at <math>z_0</math> on which integral of <math>g</math> gives different values, which cannot happen if this <math>g</math> admits an antiderivative.
[edit] Sufficient conditions
So far we outlined to necessary conditions for a function <math>g</math> to have an antiderivative, those being the holomorphicity of <math>g</math>, and that the integral of <math>g</math> over any path depend only on the endpoints of the path. It turns out that these two conditions are sufficient.
Thus, if for example, the domain <math>U</math> of <math>g</math> is a simply connected set (in particular, a convex or star-convex set), then any <math>g</math> defined on <math>U</math> admits an antiderivative.
To prove that the conditions outlined above are sufficient for the existence of an antiderivative of <math>g</math> we can assume that the domain <math>U</math> of <math>g</math> is connected, as otherwise one can prove the existence of an antiderivative on each connected component. With this assumption, consider a point <math>z_0</math> in <math>U</math>, and for any <math>z</math> in <math>U</math> define
- <math>f(z)=\int_{\gamma}\! g(\zeta)\, d\zeta</math>
where <math>\gamma</math> is any path joining <math>z_0</math> to <math>z.</math> Such a path exists since <math>U</math> is assumed to be an open connected set, and the obtained <math>f(z)</math> is well-defined as it does not depend on the choice of the path.
This <math>f</math> is an antiderivative of <math>g</math>. Indeed let <math>z</math> be a point in <math>U</math> and consider a path <math>\gamma</math> from <math>z_0</math> to <math>z.</math> For any <math>w</math> in <math>U</math> close enough to <math>z</math> one can create a path from <math>z_0</math> to <math>w</math> by joining the path <math>\gamma</math> with a segment <math>[z, w]</math> from <math>z</math> to <math>w</math>. Then,
- <math>f(w)=\int_{\gamma\cup [z, w]}\! g(\zeta)\, d\zeta</math>
- <math>=\int_{\gamma}\! g(\zeta)\, d\zeta+\int_{[z, w]}\! g(\zeta)\, d\zeta</math>
- <math>=f(z)+\int_{[z, w]}\! g(\zeta)\, d\zeta.</math>
The last integral is approximately <math>(w-z)g(z)</math> for <math>w</math> very close to <math>z</math> (here we use the holomorphicity of <math>g</math>, although simple continuity would suffice). By moving <math>f(z)</math> on the left-hand side of the above equation, dividing by <math>w-z</math>, and setting <math>w\to z</math>, one obtains <math>f'(z)=g(z)</math> which was to be proved.
[edit] References
- Ian Stewart, David Tall (Mar 10, 1983). Complex Analysis. Cambridge University Press. ISBN 0-521-28763-4.
- Alan D Solomon (Jan 1, 1994). The Essentials of Complex Variables I. Research & Education Assoc. ISBN 0-87891-661-X.
