4.2: Complex Line Integrals
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Line integrals are also called path or contour integrals. Given the ingredients we define the complex lineintegral \(\int_{\gamma} f(z)\ dz\) by
\[\int_{\gamma} f(z)\ dz := \int_{a}^{b} f(\gamma (t)) \gamma ' (t)\ dt. \label{4.2.1} \]
You should note that this notation looks just like integrals of a real variable. We don’t need the vectors and dot products of line integrals in \(R^2\). Also, make sure you understand that the product \(f(\gamma (t)) \gamma '(t)\) is just a product of complex numbers.
An alternative notation uses \(dz = dx + idy\) to write
\[\int_{\gamma} f(z)\ dz = \int_{\gamma} (u + iv) (dx + idy) \label{4.2.2} \]
Let’s check that Equations \ref{4.2.1} and \ref{4.2.2} are the same. Equation \ref{4.2.2} is really a multivariable calculus expression, so thinking of \(\gamma (t)\) as \((x(t), y(t))\) it becomes
\[\int_{\gamma} f(z) \ dz = \int_a^b [u(x(t), y(t)) + iv (x(t), y(t)] (x'(t) + iy'(t))dt \nonumber \]
but
\[u(x(t), y(t)) + iv (x(t), y(t)) = f(\gamma (t)) \nonumber \]
and
\[x'(t) + iy'(t) = \gamma '(t) \nonumber \]
so the right hand side of Equation \ref{4.2.2} is
\[\int_{a}^{b} f(\gamma (t)) \gamma '(t)\ dt. \nonumber \]
That is, it is exactly the same as the expression in Equation \ref{4.2.1}
Compute \(\int_{\gamma} z^2 \ dz\) along the straight line from 0 to \(1 + i\).
Solution
We parametrize the curve as \(\gamma (t) = t(1 + i)\) with \(0 \le t \le 1\). So \(\gamma '(t) = 1 + i\). The line integral is
\[\int z^2 \ dz = \int_{0}^{1} t^2 (1 + i)^2 (1 + i)\ dt = \dfrac{2i(1 + i)}{3}. \nonumber \]
Compute \(\int_{\gamma} \overline{z} \ dz\) along the straight line from 0 to \(1 + i\).
Solution
We can use the same parametrization as in the previous example. So,
\[\int_{\gamma} \overline{z} \ dz = \int_{0}^{1} t(1 - i) (1 + i)\ dt = 1. \nonumber \]
Compute \(\int_{\gamma} z^2\ dz\) along the unit circle.
Solution
We parametrize the unit circle by \(\gamma (\theta) = e^{i \theta}\), where \(0 \le \theta \le 2\pi\). We have \(\gamma '(\theta) = ie^{i\theta}\). So, the integral becomes
\[\int_{\gamma} z^2 \ dz = \int_{0}^{2\pi} e^{2i \theta} i e^{i \theta} \ d \theta = \int_{0}^{2\pi} ie^{3i\theta}\ d \theta = \dfrac{e^{3i\theta}}{3} \vert_{0}^{2\pi} = 0. \nonumber \]
Compute \(\int \overline{z}\ dz\) along the unit circle.
Solution
Parametrize \(C\): \(\gamma (t) = e^{it}\), with \(0 \le t \le 2\pi\). So, \(\gamma '(t) = ie^{it}\). Putting this into the integral gives
\[\int_{C} \overline{z}\ dz = \int_{0}^{2\pi} \overline{e^{it}} i e^{it} \ dt = \int_{0}^{2\pi} i \ dt = 2\pi i.\nonumber \]