4.5: Examples

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Example \(\PageIndex{1}\)

Why can't we compute \(\int_{\gamma} \overline{z}\ dz\) using the fundamental theorem.

Solution

Because \(\overline{z}\) doesn’t have an antiderivative. We can also see this by noting that if \(\overline{z}\) had an antiderivative, then its integral around the unit circle would have to be 0. But, we saw in Example 4.2.4 that this is not the case.

Example \(\PageIndex{2}\)

Compute \(\int_{\gamma} \dfrac{1}{z}\ dz\) over each of the following contours

The line from 1 to \(1 + i\).
The circle of radius 1 around \(z = 3\).
The unit circle.

Solution

For parts (i) and (ii) there is no problem using the antiderivative \(\log (z)\) because these curves are contained in a simply connected region that doesn’t contain the origin.

(i)

\[\int_{\gamma} \dfrac{1}{z}\ dz = \log (1 + i) - \log (1) = \log (\sqrt{2}) + i \dfrac{\pi}{4}. \nonumber \]

(ii) Since the beginning and end points are the same, we get

\[\int_{\gamma} \dfrac{1}{z} \ dz = 0 \nonumber \]

(iii) We parametrize the unit circle by \(\gamma (\theta) = e^{i \theta}\) with \(0 \le \theta \le 2\pi\). We compute \(\gamma '(\theta) = ie^{i \theta}\). So the integral becomes

\[\int_{\gamma} \dfrac{1}{z} \ dz = \int_{0}^{2\pi} \dfrac{1}{e^{i \theta}} ie^{i \theta} \ dt = \int_{0}^{2\pi} i \ dt = 2\pi i. \nonumber \]

Notice that we could use \(\log (z)\) if we were careful to let the argument increase by \(2 \pi\) as it went around the origin once.

Example \(\PageIndex{3}\)

Compute \(\int_{\gamma} \dfrac{1}{z^2} \ dz\), where \(\gamma\) is the unit circle in two ways.

Using the fundamental theorem.
Directly from the definition.

Solution

(i) Let \(f(z) = -1/z\). Since \(f'(z) = 1/z^2\), the fundamental theorem says

\[\int_{\gamma} \dfrac{1}{z^2} \ dz = \int_{\gamma} f'(z) \ dz = f(\text{endpint}) - f(\text{start point}) = 0. \nonumber \]

It equals 0 because the start and endpoints are the same.

(ii) As usual, we parametrize the unit circle as \(\gamma (\theta = e^{i \theta}\) with \(0 \le \theta \le 2\pi\). So, \(\gamma '(\theta) = ie^{i \theta}\) and the integral becomes

\[\int_{\gamma} \dfrac{1}{z^2} \ dz = \int_{0}^{2 \pi} \dfrac{1}{e^{2i \theta}} ie^{i \theta}\ d \theta = \int_{0}^{2\pi} i e^{-i \theta}\ d \theta = -e^{-i \theta} \vert_{0}^{2\pi} = 0. \nonumber \]