Definition and Basic Properties

Section 4.1 Definition and Basic Properties

At first sight, complex integration is not really different from real integration. Let \(a, b \in \R\) and let \(g: [a,b] \to \C\) be continuous. Then we define

\begin{equation} \int_a^b g(t) \, \diff{t} \ := \ \int_a^b \Re g(t) \, \diff{t} \ + \ i \int_a^b \Im g(t) \, \diff{t} \, \text{.}\tag{4.1} \end{equation}

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This definition is analogous to that of integration of a parametric curve in \(\R^2\text{.}\) For a function that takes complex numbers as arguments, we typically integrate over a path \(\gg\) (in place of a real interval). If you meditate about the substitution rule for real integrals (Theorem A.0.6), the following definition, which is based on (4.1), should come as no surprise.

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Definition 4.1.1.

Suppose \(\gg\) is a smooth path parametrized by \(\gg (t) , \ a \leq t \leq b\text{,}\) and \(f\) is a complex function which is continuous on \(\gg\text{.}\) Then we define the integral of \(f\) on \(\gg\) as

\begin{equation*} \int_\gg f \ = \ \int_\gg f(z) \, \diff{z} \ := \ \int_a^b f(\gg(t)) \gg'(t) \, \diff{t} \, \text{.} \end{equation*}

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This definition immediately extends to paths that are piecewise smooth: Suppose \(\gg\) is parametrized by \(\gg (t) , \ a \leq t \leq b\text{,}\) which is smooth on the intervals \([a,c_1], [c_1,c_2], \dots, [c_{n-1},c_n], [c_n,b]\text{.}\)

Our footnote on Definition 1.4.11 about the subtlety of the definition of a smooth path applies also here, at the subdivision points \(c_i\text{.}\) Note that we do not require that the left and right derivatives match at these points.

Then, assuming again that \(f\) is continuous on \(\gg\text{,}\) we define

\begin{align*} \int_\gg f \ := \ \int_a^{c_1} f(\gg(t)) \gg'(t) \, \diff{t} \ + \amp \ \int_{c_1}^{c_2} f(\gg(t)) \gg'(t) \, \diff{t}\\ \ + \amp \ \cdots \ + \ \int_{c_n}^b f(\gg(t)) \gg'(t) \, \diff{t} \,\text{.} \end{align*}

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Example 4.1.2.

To see this definition in action, we compute the integral of the function \(f: \C \to \C\) given by \(f(z) = \zbar^2\) over several paths from 0 to \(1 + i\text{.}\)

Let \(\gg\) be the line segment from \(0\) to \(1 + i\text{.}\) A parametrization of this path is \(\gg(t) = t + i t , \ 0 \leq t \leq 1\text{.}\) Here \(\gg'(t) = 1+i\) and \(f(\gg(t)) = \left( t-it \right)^2\text{,}\) and so

\begin{equation*} \int_\gg f \ = \ \int_0^1 \left( t-it \right)^2 \left( 1+i \right) \diff{t} \ = \ 2(1-i) \int_0^1 t^2 \, \diff{t} \ = \ \frac{ 2 }{ 3 } (1-i) \, \text{.} \end{equation*}

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Let \(\gg\) be the arc of the parabola \(y=x^2\) from \(0\) to \(1 + i\text{.}\) A parametrization of this path is \(\gg(t) = t + i t^2 , \ 0 \leq t \leq 1\text{.}\) Now we have \(\gg'(t) = 1 + 2it\) and

\begin{equation*} f(\gg(t)) \ = \ \left( t - it^2 \right)^2 = t^2 - t^4 - 2it^3 \, \text{,} \end{equation*}

whence

\begin{align*} \int_\gg f \ \amp = \ \int_0^1 ( t^2 - t^4 - 2it^3 ) \left( 1 + 2it \right) \, \diff{t}\\ \amp = \ \int_0^1 ( t^2 + 3 t^4 - 2i t^5 ) \, \diff{t} \ = \ \frac{ 14 }{ 15 } - \frac i 3 \, \text{.} \end{align*}

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Let \(\gg\) be the union of the two line segments \(\gg_1\) from \(0\) to \(1\) and \(\gg_2\) from \(1\) to \(1+i\text{.}\) Parametrizations are \(\gg_1(t) = t , \ 0 \leq t \leq 1\text{,}\) and \(\gg_2(t) = 1 + it , \ 0 \leq t \leq 1\text{.}\) Hence

\begin{align*} \int_\gg f \ \amp = \ \int_{\gg_1} f + \int_{\gg_2} f\\ \amp = \ \int_0^1 t^2 \, \diff{t} + \int_0^1 (1-it)^2 \, i \, \diff{t}\\ \amp = \ \frac 1 3 + i \int_0^1 (1 - 2it - t^2) \, \diff{t}\\ \amp = \ \frac 1 3 + i \left( 1 - 2i \frac 1 2 - \frac 1 3 \right)\\ \amp = \ \frac 4 3 + \frac 2 3 i \,\text{.} \end{align*}

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It is apparent but nevertheless noteworthy that these integrals evaluate to different results; in particular—unlike in calculus—a complex integral does not simply depend on the endpoints of the path of integration.

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On the other hand, the complex integral has some standard properties, most of which follow from their real siblings in a straightforward way. Our first observation is that the actual choice of parametrization of \(\gamma\) does not matter. More precisely, if \(\gg (t) , \ a \leq t \leq b\) and \(\sigma (t) , \ c \leq t \leq d\) are parametrizations of a curve then we say that \(\sigma\) is a reparametrization of \(\gamma\) if there is an increasing piecewise smooth map of \([c,d]\) onto \([a,b]\) that takes \(\gg\) to \(\sigma\text{,}\) in the sense that \(\sigma = \gg \circ \tau\text{.}\)

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Proposition 4.1.3.

If \(\gg (t) , \ a \leq t \leq b\) is a piecewise smooth parametrization of a curve and \(\sigma (t) , \ c \leq t \leq d\) is a reparametrization of \(\gg\) then

\begin{equation*} \int_c^d f(\sigma(t)) \, \sigma'(t) \, \diff{t} \ = \ \int_a^b f(\gg(t)) \, \gg'(t) \, \diff{t} \, \text{.} \end{equation*}

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Example 4.1.4.

To appreciate this statement, consider the two parametrizations

\begin{equation*} \gamma(t) = e^{ i t } , \ 0 \le t \le 2 \pi \, , \qquad \text{ and } \qquad \sigma(t) = e^{ 2 \pi i \sin(t) } , \ 0 \le t \le \tfrac \pi 2 \, \text{,} \end{equation*}

of the unit circle. Then we could write \(\int_\gg f\) in the two ways:

\begin{equation*} \int_\gg f \ = \ i \int_0^{ 2 \pi } f \left( e^{ i t } \right) e^{ i t } \, \diff{t} \end{equation*}

and

\begin{equation*} \int_\gg f \ = \ 2 \pi i \int_0^{ \frac \pi 2 } f \left( e^{ 2 \pi i \sin(t) } \right) e^{ 2 \pi i \sin(t) } \cos(t) \, \diff{t} \, \text{.} \end{equation*}

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A quick substitution shows that the two integrals on the respective right-hand sides are indeed equal.

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Proposition 4.1.3 says that a similar equality will hold for any integral and any parametrization. Its proof is left as Exercise 4.5.9, which also shows that the following definition is unchanged under reparametrization.

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Definition 4.1.5.

The length of a smooth path \(\gg\) is

\begin{equation*} \length(\gg) \ := \ \int_a^b \left| \gg'(t) \right| \, \diff{t} \end{equation*}

for any parametrization \(\gg (t)\text{,}\) \(a \leq t \leq b\text{.}\) Naturally, the length of a piecewise smooth path is the sum of the lengths of its smooth components.

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Example 4.1.6.

Let \(\gamma\) be the line segment from \(0\) to \(1+i\text{,}\) which can be parametrized by \(\gamma(t) = t + it\) for \(0 \le t \le 1\text{.}\) Then \(\gamma'(t) = 1+i\) and so

\begin{equation*} \length(\gamma) \ = \ \int_0^1 | 1+i |\, \diff{t} \ = \ \int_0^1 \sqrt 2 \, \diff{t} \ = \ \sqrt 2 \, \text{.} \end{equation*}

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Example 4.1.7.

Let \(\gamma\) be the unit circle, which can be parametrized by \(\gamma(t) = e ^{ it }\) for \(0 \le t \le 2 \pi\text{.}\) Then \(\gamma'(t) = i \, e^{ it }\) and

\begin{equation*} \length(\gamma) \ = \ \int_0^{2 \pi} | i \, e^{ it } |\, \diff{t} \ = \ \int_0^{2 \pi} \diff{t} \ = \ 2 \pi \, \text{.} \end{equation*}

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Now we observe some basic facts about how the line integral behaves with respect to function addition, scalar multiplication, inverse parametrization, and path concatenation; we also give an upper bound for the absolute value of an integral, which we will make use of time and again.

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Proposition 4.1.8.

Suppose \(\gg\) is a piecewise smooth path, \(f\) and \(g\) are complex functions which are continuous on \(\gg\text{,}\) and \(c \in \C\text{.}\)

\(\ds \int_\gg (f + c \, g) \ = \ \int_\gg f \, + \, c \int_\gg g \, \text{.}\)
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If \(\gg\) is parametrized by \(\gg (t) , \ a \leq t \leq b\text{,}\) we define the path \(-\gg\) by \(-\gg(t) := \gg (a+b-t) , \ a \leq t \leq b\text{.}\) Then

\begin{equation*} \int_{-\gg} f \ = \ - \int_\gg f \,\text{.} \end{equation*}

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If \(\gg_1\) and \(\gg_2\) are piecewise smooth paths so that \(\gg_2\) starts where \(\gg_1\) ends, we define the path \(\gg_1\gg_2\) by following \(\gg_1\) to its end and then continuing on \(\gg_2\) to its end. Then

\begin{equation*} \int_{\gg_1\gg_2} f \ = \ \int_{\gg_1}f+\int_{\gg_2}f \,\text{.} \end{equation*}

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\(\ds \left| \int_\gg f \right| \ \leq \ \max_{z \in \gg} \left| f(z) \right| \cdot \length (\gg) \, \text{.}\)
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The path \(-\gg\) defined in (b) is the path that we obtain by traveling through \(\gg\) in the opposite direction.

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Proof.

(a) follows directly from the definition of the integral and Theorem A.0.4, the analogous theorem from calculus.

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(b) follows with the real change of variables \(s = a+b-t\text{:}\)

\begin{align*} \int_{-\gg} f \amp \ = \ \int_a^b f \left( \gg (a+b-t) \right) \left( \gg (a+b-t) \right)' \, \diff{t}\\ \amp \ = \ - \int_a^b f \left( \gg (a+b-t) \right) \gg' (a+b-t) \, \diff{t}\\ \amp \ = \ \int_b^a f \left( \gg (s) \right) \gg' (s) \, \diff{s}\\ \amp \ = \ - \int_a^b f \left( \gg (s) \right) \gg' (s) \, \diff{s}\\ \amp \ = \ - \int_\gg f \,\text{.} \end{align*}

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For (c), we need a suitable parametrization \(\gg(t)\) for \(\gamma_1\gamma_2\text{.}\) If \(\gamma_1\) has domain \([a_1,b_1]\) and \(\gamma_2\) has domain \([a_2,b_2]\) then we can use

\begin{equation*} \gg(t) := \begin{cases}\gg_1(t) \amp \text{ if } a_1\le t\le b_1 \, , \\ \gg_2(t-b_1+a_2) \amp \text{ if } b_1\le t\le b_1+b_2-a_2 \, , \end{cases} \end{equation*}

with domain \([a_1, b_1+b_2-a_2]\text{.}\) Now we break the integral over \(\gg_1\gg_2\) into two pieces and apply the change of variables \(s=t-b_1+a_2\text{:}\)

\begin{align*} \int_{\gg_1\gg_2} f \ \amp = \int_{a_1}^{b_1+b_2-a_2} f(\gg(t))\gg'(t)\,\diff{t}\\ \amp = \int_{a_1}^{b_1} f(\gg(t))\gg'(t)\,\diff{t} \ + \ \int_{b_1}^{b_1+b_2-a_2} f(\gg(t))\gg'(t)\,\diff{t}\\ \amp = \int_{\gg_1}f+\int_{\gg_2}f \, \text{.} \end{align*}

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The last step follows since \(\gg\) restricted to \([a_1,b_1]\) is \(\gg_1\) and \(\gg\) restricted to \([b_1,b_1+b_2-a_2]\) is a reparametrization of \(\gg_2\text{.}\)

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For (d), let \(\phi = \left( \Arg \int_\gg f \right)\text{.}\) Then \(\int_\gg f = \left| \int_\gg f \right| e^{ i \phi }\) and thus, since \(\left| \int_\gg f \right| \in \R\text{,}\)

\begin{align*} \left| \int_\gg f \right| \ \amp = \ e^{ -i\phi } \int_\gg f \ = \ \Re \left( e^{ -i \phi } \int_a^b f(\gg(t)) \gg'(t) \, \diff{t} \right)\\ \amp \ = \ \int_a^b \Re \left( f ( \gg (t) ) e^{ -i \phi } \gg' (t) \right) \diff{t} \,\\ \amp \leq \ \int_a^b \left| f ( \gg (t) ) e^{ -i \phi } \gg' (t) \right| \diff{t} \, \ = \ \int_a^b \left| f ( \gg (t) ) \right| \left| \gg' (t) \right| \diff{t}\\ \amp \leq \, \max_{ a \leq t \leq b } \left| f ( \gg (t) ) \right| \int_a^b \left| \gg' (t) \right| \diff{t} \, \ = \ \, \max_{z \in \gg} \left| f(z) \right| \cdot \length (\gg) \,\text{.} \end{align*}

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Here we have used Theorem A.0.5 for both inequalities.

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Example 4.1.9.

In Exercise 4.5.4, you are invited to show

\begin{equation*} \int_\gg \frac{\diff{z}}{z-w} \ = \ 2 \pi i \,\text{,} \end{equation*}

where \(\gg\) is any circle centered at \(w \in \C\text{,}\) oriented counter-clockwise. Thus Proposition 4.1.8(b) says that the analogous integral over a clockwise circle equals \(- 2 \pi i\text{.}\) Incidentally, the same example shows that the inequality in Proposition 4.1.8(d) is sharp: if \(\gg\) has radius \(r\text{,}\) then

\begin{equation*} 2 \pi \ = \ \left| \int_{ \gg } \frac{\diff{z}}{z-w} \right| \ \le \ \max_{ z \in \gg } \left| \frac{1}{z-w} \right| \length(\gg) \ = \ \frac 1 r \cdot 2 \pi r \, \text{.} \end{equation*}

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