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Suppose we want to evaluate

$$\int_C f(z) \, d \overline{z}$$

(I'm intentionally being vague with how $f$ and $C$ are defined because my question doesn't focus on these details - see example below for such details). Suppose that the best way to go about this is to instead evaluate the integral with respect to $z$. How do we write $d\overline{z}$ in terms of $z$ (that is, what the relationship is between $d\overline{z}$ and $dz$)?

Take, for example, the following problem from Ahlfors's Complex Analysis:

If $P(z)$ is a polynomial and $C$ denotes the circle $|z-a| = R$, what is the value of $$\int_C P(z) \, d \overline{z} \, \, ?$$

After seeing a solution to the above problem, I suspected that the relationship was that $d\overline{z} = -dz$ based on this particular problem's solution. Is the relationship $d\overline{z} = -dz$ correct? Examples (as above) of this kind in the textbooks I've read are scarce and the explanations for how these integrals are evaluated are also scarce (if any exist?), hence I pose the question here.

EDIT:

Ahlfors tells us that

$$\int_C f(z) \, d \overline{z} = \overline{\int_C \overline{f(z)} \, dz}$$

The solution I found to the polynomial example mentioned above gives that

$$\int_C \overline{P(z)} \, dz = 2\pi i P'(a)R^2 $$

$$\implies \overline{\int_C \overline{P(z)} \, dz} = -2\pi i P'(a)R^2$$

EDIT Question: Is the above chain of results meant to imply that the conjugation of the integral changes the answer by a $(-)$ sign?

Aram Nazaryan
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    Ahlfors defines integrals with respect to $\bar z$ by “double conjugation.” In my edition that is on the third page of chapter “4 Complex Integration.” – Martin R Mar 02 '24 at 17:17
  • @MartinR Thank you. I re-examined the solution to the above example and the solution does indeed utilize that. In that solution, it seems that the final conjugation of the integral changed the underlying integral's answer by a minus sign; does this seem correct? – Aram Nazaryan Mar 02 '24 at 17:29
  • It is not clear to me what you mean. But $d\overline{z} = -dz$ is definitely not correct, since $\int_C P(z) dz = 0$ by Cauchy's integral theorem. – Martin R Mar 02 '24 at 17:34
  • @MartinR Ah I understand. I made an edit to my question to elucidate my concern about the $(-)$ sign – Aram Nazaryan Mar 02 '24 at 17:35
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    That solution (but not the result) is wrong. $\int_C \overline{P(z)} , dz = 2\pi i \overline{P'(a)}R^2$, and then conjugation gives the final result. The complex conjugate of $i$ is $-i$, but conjugation does not “change the sign of the integral.” – Martin R Mar 02 '24 at 17:40
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    $d\bar z$ means just that when one parametrize the curve by $z=z(t)$ - for example assuming here $a=0$ by a translation that doesn't change the d since d(constant)=$0$ we get $z=Re^{i\theta}$ so $dz=iRe^{\theta}d\theta$ hence $d\bar z=-iRe^{-i\theta}d\theta$ since $R,d\theta$ are real; then writing $P(z)=\sum a_kR^ke^{ik\theta}$ the only term that doesn't vanish by orthogonality is the $k=1$ which gives the final result $-2\pi ia_1R^2$ etc – Conrad Mar 02 '24 at 17:46

1 Answers1

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(As you found out in the meantime,) Ahlfors defines integration with respect to $\bar z$ by “double conjugation:” $$ \int_C f(z) \, d \overline{z} = \overline{\int_C \overline{f(z)} \, dz} \, . $$ If $\gamma:[0, 1]\to \Bbb C$ is a parametrization of $C$ then $$ \int_C f(z) \, d \overline{z} = \int_0^1 f(\gamma(t)) \overline{\gamma'(t)} \, dt \, . $$ There is no relationship $d \bar z = -dz$, that impression is perhaps caused by a wrong intermediate result in the solution of the exercise.

With respect to the exercise: For all integers $n$ is $$ \int_C (\bar z - \bar a)^n dz = i R^{n+1} \int_0^{2 \pi} e^{i(1-n)t} \, dt = \begin{cases} 2 \pi i R^2 & \text{ if } n = 1 \, ,\\ 0 & \text{ if } n \ne 1 \, . \end{cases} $$ It follows that for $P(z) = \sum_{k=0}^n a_k (z-a)^k$ $$ \int_C \overline{P(z)} \, dz = 2 \pi i R^2 \overline{a_1} = 2 \pi i R^2 \overline{P'(a)} \, . $$ (That intermediate result is wrong in your solution.) Conjugation then gives the result $$ \int_C P(z) \, d\bar z = -2\pi i R^2 P'(a). $$

That final conjugation changes $i$ to $-i$ and $\overline{P'(a)}$ to $P'(a)$, it does not “change the sign” of the integral.

Martin R
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  • Thank you so much for your detailed explanation. Somehow, I managed to completely overstep the fact that $2\pi i P'(a)R^2$ a product of complex numbers, some of which are strictly real – Aram Nazaryan Mar 02 '24 at 18:05