kuniga.me > NP-Incompleteness > Complex Integration

05 Apr 2024

This post inaugurates a series with my notes on complex integration, corresponding to *Chapter 4* in Ahlfors’ Complex Analysis.

In this post we’ll cover different definitions of complex integrals - integrals of functions whose domain and image are complex numbers - and some basic properties.

To recap, in [2] we defined *complex derivatives* of a function $f: \Omega \rightarrow \mathbb{C}$, for $\Omega \in \mathbb{C}$, denoted by $f’(z)$, as:

It looks very similar to a real derivative except that we’re dealing with complex numbers, which makes the concept $h \rightarrow 0$ more nuanced.

How about the complex integral? Consider first the real-integral such as

\[(1) \qquad \int_a^b f(x) dx\]We can think of it as the sum of $f(x)$ for infinitesimal sub-intervals $dx$ over the a segment of the real-line. For complex numbers we can generalize it by letting it be any curve on the complex plane, in particular *parametric curves* as defined in [3] as we see next.

Consider the function $f(t): [a, b] \rightarrow \mathbb{C}$ where $t$ is from a real-number $a \le t \le b$. The line integral is defined as:

\[(2) \qquad \int_{a}^{b} f(t) dt\]This is very similar to the real integral $(1)$ with the exception that the image of $f(t)$ is complex (but not its domain). If we decompose it into its real and imaginary part, say, $f(t) = u(t) + iv(t)$ we can use the linearity of real integrals to obtain:

\[(3) \qquad \int_{a}^{b} f(t) dt = \int_{a}^{b} u(t) dt + i \int_{a}^{b} v(t) dt\]In other words we can define a line integral through the original real integral $(1)$. Using the linearity principle again, we can show that the complex line integral is also linear:

\[(4) \qquad \int_{a}^{b} cf(t) dt = c \int_{a}^{b} f(t) dt\]For a complex constant $c = \alpha + i \beta$.

Recall that the triangle inequality which states that $\abs{u + v} \le \abs{u} + \abs{v}$ for complex $u$ and $v$. Using (4) we can generalize this property to the complex line integral, as stated in *Theorem 1*.

**Theorem 1.**

Instead of defining over a real interval, we can define it over a curve or arc $\gamma$, as long as it is piecewise differentiable and that $f(z)$ is continuous over $z \in \gamma$:

\[(5) \quad \int_{\gamma} f(z) dz = \int_{a}^{b} f(g(t)) g'(t) dt\]Here we’re using variable substition $z = g(t)$ to define the line integral over a curve in terms of $(1.1)$.

We can also subdivide $\gamma$ into sub-curves $\gamma = \gamma_1 + \gamma_2 + \cdots + \gamma_n$ in which case the integral over it can be expressed as the sum of the integrals of its parts:

\[\int_{\gamma} f(z) dz = \int_{\gamma_1} f(z) dz + \int_{\gamma_2} f(z) dz + \cdots + \int_{\gamma_n} f(z) dz\]We can introduce the following notation:

\[(6) \quad \int_{\gamma} f(z) \overline{dz} = \overline{\int_{\gamma} \overline{f(z)} dz}\]Recall that we can think of a function of $C$ as one taking two real values, the real and imaginary part of $z$, so $f(z) = f(x, y)$. We can integrate over only $x$ and obtain a function of $y$. That is,

\[(7) \quad g(y) = \int_{\gamma} f(x, y) dx\]Analogously, we have:

\[(8) \quad h(x) = \int_{\gamma} f(x, y) dy\]With this notation, we can express $(7)$ and $(8)$ via $(5)$ and $(6)$:

\[(9) \quad \int_{\gamma} f(x, y) dx = \frac{1}{2}\paren{\int_{\gamma} f(z) dz + \int_{\gamma} f(z) \overline{dz}}\]and

\[(10) \quad \int_{\gamma} f(x, y) dy = \frac{1}{2i}\paren{\int_{\gamma} f(z) dz - \int_{\gamma} f(z) \overline{dz}}\]which follows from the identities: $x = (z + \overline{z})/2$ and $y = (z - \overline{z})/(2i)$. If we multiply $(10)$ by $i$ and add with $(9)$, we can express $\int_{\gamma} f(z) dz$ as a function of its “partial integrals”:

\[\int_{\gamma} f(z) dz = \int_{\gamma} f(x, y) dx + i \int_{\gamma} f(x, y) dy\]If we split the real and imaginary parts of the result of $f$, i.e. $f(z) = u(z) + iv(z)$, we can then obtain:

\[\int_{\gamma} f(z) dz = \int_{\gamma} (u(x, y) + iv(x, y))dx + i \int_{\gamma} (u(x, y) + iv(x, y))dy\]Grouping terms into real and imaginary parts:

\[\int_{\gamma} f(z) dz = \int_{\gamma} (u(x, y)dx - v(x, y)dy) + i \int_{\gamma} (u(x, y)dy + v(x, y)dx)\]Or using a more succinct notation (ommitting the parameters that can be inferred from context):

\[\int_{\gamma} f dz = \int_{\gamma} (udx - vdy) + i \int_{\gamma} (udy + vdx)\]We introduce yet another definition and notation, where we use the length of the differential and is defined as follows:

\[(11) \quad \int_{\gamma} f(z)\abs{dz} = \int_{a}^{b} f(g(t)) \abs{g'(t)} dt\]Which is similar to $(6)$ except that we multiply by the real $\abs{g’(t)}$ instead of the complex $g’(t)$. An analogous result of *Theorem 1* for arc-length integral is *Theorem 2*:

**Theorem 2.**

Which is nice because we can work with the contour integral notation instead of the parametric curve one.

**Curve length.** If $f(z) = 1$, then $(11)$ reduces to

and it corresponds to the length of $\gamma$. As an example, if $\gamma$ is a circle of radius $\rho$ centered at the origin, we can define the parametric curve $g(t) = \rho e^{it}$ for $0 \le t < 2 \pi$.

We have $g’(t) = g(t) = \rho e^{it}$ and that $\abs(\rho e^{it}) = \rho$ and that $f(g(t)) = 1$. Plugging this into $(11)$ gives us:

\[\int_{\gamma} \abs{dz} = \int_{a}^{b} \rho dt = 2\pi \rho\]