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I am reading the Complex Analysis textbook by Ahlfors. In the complex integration chapter he defined complex integration over a curve $\gamma$ as $$\int_\gamma f(z)\,\mathrm{d}z = \int _a^b f(z(t))z'(t)\,\mathrm{d}t$$ where $z(t)$ is a parametrization of $\gamma$. He then define integrals with respect to $\mathrm{d}\bar z$ as $$\int_\gamma f(z)\,\mathrm{d}\bar z = \overline{\int_\gamma \overline{f(z)} \,\mathrm{d}z }$$

This seems to satisfies some nice properties - but I don't understand the intuition behind it. Can someone please give me some motivation or justification for this definition as to why this corresponds to integration with respect to $\bar z$? Is there a way to "derive" this from his original definition for integral?

PrincessEev
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Hanging Pawns
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2 Answers2

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Some Prequisites, Maybe:

One could (should?) pass the integral $\int_\gamma f(z) \, \mathrm{d}\overline{z}$ into the formal definition via a Riemann sum, which will look like $$ \int_\gamma f(z) \, \mathrm{d}z := \lim_{\|P\| \to 0} \sum_i f(p(\tau_i)) \Delta p_i $$ where:

  • $p(t)$ parameterizes the curve $\gamma$ for $t \in [a,b]$
  • $P := \{t_i\}_{i=0}^n$ is a partition $[a,b]$
  • $\|P\| := \max |t_i- t_{i-1}|$ (the mesh of the partition)
  • $\Delta p_i := p(t_i) - p(t_{i-1})$
  • $\tau_i \in [t_{i-1},t_i]$ (the tag points for our function: note that at the point $p(t)$ on the curve, $f$ has value $f(p(t))$)

There are probably some items I'm overlooking, but this is essentially the gist of it. Bear in mind, in particular, $p(t) \in \mathbb{C}$ for each $t \in [a,b]$.

It is only in sufficiently nice situations that we have $$ \int_\gamma f(z) \, \mathrm{d}z = \int_a^b f(p(t)) p'(t) \, \mathrm{d}t \tag{1} $$ so, naturally, one sees an association of $$ \mathrm{d}z \mapsto p'(t) \, \mathrm{d}t \newcommand{\dd}{\mathrm{d}} \newcommand{\z}{\overline{z}} \newcommand{\p}{\overline{p}} $$ This should make sense: in proving $(1)$, one uses the mean value theorem to see that small changes in $z$ along the curve take a form like the above (due to the rise of a $\Delta p_i$-like term that the mean value theorem handles).

A Rough & Dirty Derivation:

A proper, formal derivation via the Riemann sum is left to the reader as an exercise, but the spirit is mostly the same.

So, what might a small change $\dd \z$ look like? Well, if $z = p(t)$ for a particular $t$, then $\z = \p(t)$ for that same $t$. Then $$ \Delta \z = \p(t_i) - \p(t_{i-1}) = \overline{p(t_i) - p(t_{i-1})} $$ In the limit $\|P\| \to 0$, then, in the "sufficiently nice" situation, $$ \dd \z = \overline{p'(t) \, \dd t} $$ Of course, since $\dd t$ is a real quantity, $$ \dd \z = \overline{p'(t)} \, \dd t $$ and then $$\begin{align*} \int_\gamma f(z) \, \mathrm{d}\z &= \int_a^b f(p(t)) \, \overline{p'(t)} \, \dd t\\ &= \int_a^b \overline{\overline{f(p(t))} \, p'(t)} \, \dd t\\ &= \overline{\int_a^b \overline{f(p(t))} \, p'(t) \, \dd t}\\ &= \overline{\int_\gamma \overline{f(z)} \, \dd z}\\ \end{align*}$$ as desired.

PrincessEev
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(The first part of this answer is a tiny excerpt from a much longer answer to a broader question.)

Just as $\mathrm dz$ is essentially a shorthand for $\mathrm dx+i\mathrm dy$, similarly $\mathrm d\overline{z}$ is a shorthand for $\mathrm dx-i\mathrm dy=\overline{\mathrm dz}$.

If $f(z)=u(z)+iv(z)$, then $\mathrm d\overline{z}=\mathrm dx-i\mathrm dy$ yields: \begin{align*}\int_{C}f(z)\mathrm{d}\overline{z}&=\int_{C}\left(u\left(z\right)+iv\left(z\right)\right)\left(\mathrm{d}x\boxed{-}i\mathrm{d}y\right)\\&=\int_{C}\left(u\left(z\right)\mathrm{d}x+iv\left(z\right)\mathrm{d}x\boxed{-}iu\left(z\right)\mathrm{d}y\boxed{+}v\left(z\right)\mathrm{d}y\right)\end{align*}

For comparison,

\begin{align*}\overline{\int_{C}\overline{f(z)}\,\mathrm{d}z}&=\overline{\int_{C}\left(u\left(z\right)\boxed{-i}v\left(z\right)\right)\,\left(\mathrm{d}x+i\mathrm{d}y\right)}\\&=\overline{\int_{C}\left(u\left(z\right)\,\mathrm{d}x\boxed{-}iv\left(z\right)\,\mathrm{d}x+iu\left(z\right)\,\mathrm{d}y\boxed{+}v\left(z\right)\,\mathrm{d}y\right)}\\&=\int_{C}\left(u\left(z\right)\,\mathrm{d}x\boxed{+}iv\left(z\right)\,\mathrm{d}x\boxed{-}iu\left(z\right)\,\mathrm{d}y+v\left(z\right)\,\mathrm{d}y\right)\\&=\int_{C}f(z)\mathrm{d}\overline{z}\checkmark\end{align*}

From a comment:

how do we know that conjugation commutes with integration?

This is almost by definition. At the very beginning of the chapter in Ahlfors (or any other complex analysis text) on complex integration, the integral of a complex number is defined to split into the real and imaginary parts. So we have something like \begin{align*}\int \overline{a(t)+ib(t)}\,\mathrm dt&=\int a(t)+i\left(-b(t)\right) \mathrm dt\\&=\int a(t)\,\mathrm dt+i\int \left(-b(t)\right)\,\mathrm dt\text{ (by definition)}\\&=\int a(t)\,\mathrm dt-i\int b(t)\,\mathrm dt\\&=\overline{\int a(t)\,\mathrm dt+i\int b(t)\,\mathrm dt}\\&=\overline{\int a(t)+i b(t)\,\mathrm dt}\end{align*}

Mark S.
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