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?
(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.