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 $U$ in the complex plane and a function $g:U\to \mathbb C,$ the antiderivative of $g$ is a function $f:U\to \mathbb C,$ whose complex derivative is $g$ , $f' = g .$

1 Uniqueness of antiderivative
2 Existence of antiderivative
- 2.1 Necessary conditions
- 2.2 Sufficient conditions
3 References
4 External links

[edit] Uniqueness of antiderivative

The derivative of a constant function is zero. Therefore, any constant is an antiderivative of the zero function. If $U$ 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 $U$ (those constants need not be equal).

This observation can be used to establish that if a function $g:U\to \mathbb C$ has an antiderivative, then that antiderivative is unique up to addition of a function which is constant on each connected component of $U$ .

[edit] Existence of antiderivative

[edit] Necessary conditions

If $f$ is an antiderivative of $g,$ then $f$ is differentiable, that is, holomorphic. Then $g$ 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 $f$ is an antiderivative of $g$ on $U,$ then given any piecewise C¹ path $\gamma:[a, b]\to U$ one can express the path integral of $g$ over $γ$ as

$\int_\gamma\! g(z)\,dz=\int_a^b \!g\left(\gamma(t)\right)\gamma'(t)\, dt=\int_a^b \!f'\left(\gamma(t)\right)\gamma'(t)\,dt.$

By the chain rule and the fundamental theorem of calculus one then has

$\int_\gamma \!g(z)\,dz=\int_a^b\! \frac{d}{dt}f\left(\gamma(t)\right)\,dt=f\left(\gamma(b)\right)-f\left(\gamma(a)\right).$

As such, the integral of $g$ over $γ$ does not depend on the actual path $γ$ , but only on its endpoints.

This observation shows that not every holomorphic function $g$ admits an antiderivative. For example, consider the reciprocal function, $g (z) = z - 1$ defined on $\mathbb C\backslash\{0\}.$ Given a nonzero number $z 0$ , one can find a circle going through $z 0$ which contains the origin inside of it, and a circle going through $z 0$ 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 $z 0$ on which integral of $g$ gives different values, which cannot happen if this $g$ admits an antiderivative.

[edit] Sufficient conditions

So far we outlined to necessary conditions for a function $g$ to have an antiderivative, those being the holomorphicity of $g$ , and that the integral of $g$ 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 $U$ of $g$ is a simply connected set (in particular, a convex or star-convex set), then any $g$ defined on $U$ admits an antiderivative.

To prove that the conditions outlined above are sufficient for the existence of an antiderivative of $g$ we can assume that the domain $U$ of $g$ is connected, as otherwise one can prove the existence of an antiderivative on each connected component. With this assumption, consider a point $z 0$ in $U$ , and for any $z$ in $U$ define

$f(z)=\int_{\gamma}\! g(\zeta)\, d\zeta$

where $γ$ is any path joining $z 0$ to $z .$ Such a path exists since $U$ is assumed to be an open connected set, and the obtained $f (z)$ is well-defined as it does not depend on the choice of the path.

This $f$ is an antiderivative of $g$ . Indeed, let $z$ be a point in $U$ and consider a path $γ$ from $z 0$ to $z .$ For any $w$ in $U$ close enough to $z$ one can create a path from $z 0$ to $w$ by joining the path $γ$ with a segment $[z, w]$ from $z$ to $w$ . Then,

$f(w)=\int_{\gamma\cup [z, w]}\! g(\zeta)\, d\zeta$

$=\int_{\gamma}\! g(\zeta)\, d\zeta+\int_{[z, w]}\! g(\zeta)\, d\zeta$

$=f(z)+\int_{[z, w]}\! g(\zeta)\, d\zeta.$

The last integral is approximately $(w - z) g (z)$ for $w$ very close to $z$ (here we use the holomorphicity of $g$ , although simple continuity would suffice). By moving $f (z)$ on the left-hand side of the above equation, dividing by $w - z$ , and setting $w\to z$ , one obtains $f'(z) = g (z)$ 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.