Understanding the Definition of $\int_\gamma f\ \overline{dz}$

Solution 1:

If we go back to the definition, for a continuously differentiable path $\gamma \colon [a,b]\to \mathbb{C}$, and a function $f$ continuous on the trace of $\gamma$, we have

$$\int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\cdot \gamma'(t)\,dt.$$

If we write $f = g + ih$ with real-valued functions $g$ and $h$, and similarly $\gamma = \alpha + i\beta$, we obtain

$$\begin{align} \int_\gamma f(z)\,dz &= \int_a^b f(\gamma(t))\cdot\gamma'(t)\,dt\\ &= \int_a^b \left(g(\gamma(t)) + ih(\gamma(t))\right)\cdot\left(\alpha'(t) + i\beta'(t)\right)\,dt\\ &= \int_a^b \left(g(\gamma(t))\alpha'(t) - h(\gamma(t))\beta'(t)\right) + i\left(g(\gamma(t))\beta'(t) + h(\gamma(t))\alpha'(t)\right)\,dt\\ &= \int_a^b g(\gamma(t))\alpha'(t) - h(\gamma(t))\beta'(t)\,dt + i \int_a^b g(\gamma(t))\beta'(t) + h(\gamma(t))\alpha'(t)\,dt. \end{align}$$

By the analogous expansion, we obtain

$$\int_\gamma \overline{f(z)}\,dz = \int_a^b g(\gamma(t))\alpha'(t) + h(\gamma(t))\beta'(t)\,dt + i\int_a^b g(\gamma(t))\beta'(t) - h(\gamma(t))\alpha'(t)\,dt.$$

So the answer to the first question is in general no, we only have

$$\int_\gamma \overline{f(z)}\,dz = \overline{\int_\gamma f(z)\,dz}$$

if $\int_a^b h(\gamma(t)) \alpha'(t)\,dt = \int_a^b h(\gamma(t))\beta'(t)\,dt = 0$.

Note that the function $f$ as well as the differential $dz$ are complex-valued, and to get the conjugate of a product, you must conjugate both factors, so

$$\overline{\int_\gamma f(z)\,dz} = \int_\gamma \overline{f(z)}\;\overline{dz}.$$

An alternative and maybe easier to digest definition for $\int_\gamma f(z)\,\overline{dz}$ is

$$\int_\gamma f(z)\,\overline{dz} := \int_a^b f(\gamma(t))\cdot \overline{\gamma'(t)}\,dt.$$

Informally,

$$dz = \gamma'(t)\cdot dt;\qquad \overline{dz} = \overline{\gamma'(t)}\cdot dt.$$

Solution 2:

Let $f=u+iv$ for real valued functions $u$ and $v$, then \begin{align} \int f\,\mathrm d\overline z &= \int (u+iv)(\mathrm dx-i\,\mathrm dy) = \int \left((u\,\mathrm dx + v\,\mathrm dy)+i\,(v\,\mathrm dx-u\,\mathrm dy)\right) \\ &= \int (u\,\mathrm dx + v\,\mathrm dy)+ i\int \,(v\,\mathrm dx-u\,\mathrm dy) \\ &= \overline{\int (u\,\mathrm dx + v\,\mathrm dy)- i\int \,(v\,\mathrm dx-u\,\mathrm dy)} \\ &= \overline{\int \left((u\,\mathrm dx + v\,\mathrm dy)- i \,(v\,\mathrm dx-u\,\mathrm dy)\right)} \\ &= \overline{\int (u-iv)(\mathrm dx+i\,\mathrm dy)} = \overline{\int \overline f\,\mathrm dz}. \end{align}