1. ## Complex Antiderivitive

Let $D = \mathbb{C} - \{0\}$.

Is there a function $F: D \rightarrow \mathbb{C}$

such that $d/dz(F(z)) = 1/z$ $\forall z \in D$?

Similarly, is there a function $F: \mathbb{C} \rightarrow \mathbb{C}$
such that such that $d/dz(F(z)) = \overline{z}$ $\forall z \in \mathbb{C}$?

Is this as simple as integrating? In class, we have only covered what it means when a complex function is integrated over a path. What such path should be used? Or is there such a thing as a "pathless" complex integral?

Thanks!

2. ## Re: Complex Antiderivitive

Surely if you've reached contour integration, you've proved that a function of a complex variable being differentiable is equivalent to being analytic which is in turn equivalent to being infinitely differentiable (these statements are NOT equivalent in general for functions of a real variable). Furthermore, assuming D is simply connected, you know that an analytic function on D possesses an anti-derivative on D (Cauchy Integral Formula). Finally, the Cauchy-Riemann equations give NECESSARY conditions for which a function is analytic (i.e., if they fail, the function is NOT analytic; if they hold, it proves nothing). Under the additional assumption of continuity of first partial derivatives of some function, then the converse is true (i.e. they are sufficient).

I'm sure you can reformulate the above into a decisive proof. I suggest reviewing an getting completely straight the hypotheses and implications of the basic theorems of complex analysis right away (Cauchy-Riemann, Cauchy Integral, Cauchy-Goursat, Morera, analyticity, etc.), otherwise you're doomed to be constantly confused.

BTW, every integral (in the 1-dimensional case) is a path integral; when you integrate on the real line, you are still integrating on a path. The various forms of the fundamental theorem of calculus (Stoke's Theorem) show that under certain conditions, an integral over a "closed" manifold can be replaced by another integral on the boundary of said manifold. As it applies to your course, a path integral can be replaced by integrating the anti-derivative on the boundary of the path (i.e. the end-points) which then reduces to the difference of the anti-derivative (or "potential") function. If the contour is closed, you can again apply the anti-derivative on the "boundary" and obtain zero since you will be taking the difference of the anti-derivative at the same point (this is the essence of Morera's Theorem; the converse to Cauchy's Theorem). Green's Theorem also applies for closed contours (again, with certain assumptions on the domain, f, etc.).

3. ## Re: Complex Antiderivitive

Hi, Thanks very much for the reply!

Actually, we haven't covered the Cauchy Integral Formula, where it appears that it has the statement:
Assuming D is simply connected, you know that an analytic function on D possesses an anti-derivative on D (Cauchy Integral Formula).
But to be sure I understand, is the following line of reasoning valid?

If, for a simply connected subset D of C,

a) the Cauchy-Riemann equations hold for F on D
and
b) the first partials for the functions u & v in the above equations are continuous on D,

then I can conclude F is analytic,

and thus by the Cauchy Integral Formula, I may conclude F has an antiderivative on D?

4. ## Re: Complex Antiderivitive

That is correct. Cauchy-Riemann equations + continuity of first partials => analytic f(z) => F(z) exists s.t. F'(z) = f(z) for all z in a simply connected domain D on which f(z) is analytic (btw, you should make sure you know the definition of a 'domain' D). The converses are also true (<=), so each statement is equivalent (<=>). It's very important as you go on to firmly establish the various equivalences of analyticity of complex functions (for instance, differentiability, infinite differentiability, power series representation, vanishing integrals on closed contours, etc. are things you can add to that list of equivalences, under appropriate hypotheses), since they will come up often (I assume you're taking a first course in complex variables).

Therefore, you should conclude f(z) = 1/z has an anti-derivative in some particular D (Be careful about which D you pick for this function! It is a function of very special consideration in elementary complex analysis, along with its anti-derivative Log(z), and has a surprising property where about it's NOT analytic), while conj(z) does not.

Btw, it is a good exercise to use the Cauchy-Riemann equations to derive the computational method (via partial differentiation of components) about which you can get the derivative of an analytic function (in two ways!). Of course, once you prove the results, you can just differentiate with respect to the complex variable z as you would a real variable x without difficulty.

And one last food for thought since I've already said too much...keep in mind (since a lot of students think otherwise I have noticed), the Cauchy-Riemann equations are really by themselves nothing special. They are just a criterion - by that I mean a necessary condition (i.e. they help you eliminate candidates for analytic functions). They are derived purely from that fact that if f(z) is differentiable, then the derivative (i.e. limit) should be path-independent (think back to third semester calculus when you did limits of functions of two variables). When you plug the conditions for approach along the horizontal and vertical directions into the definition of the derivative, out pops the Cauchy-Riemann equations (we could have chosen any other arbitrary two independent directions to obtain a different set of (probably more complicated, and possibly less useful) equations. Now, what makes them special is the big theorem that says they are sufficient for analyticity if the partials are continuous (follows from Taylor's theorem, but a bit harder to prove).