Example 6.5. Consider the following diagram

Evaluate \(\displaystyle {I = \int z^2 dz}\) on a smooth curve from point \(O\) to point \(B\).

Solution
If we evaluate \(I\) along the straight line \(OB\) with equation. We call this \(\displaystyle {I_1 = \int \limits _{OB}z^2dz}\) \(z = x + iy\) in terms of \(y\). \begin {align*} z^2 & = (x + iy )^2 = (x^2 - y^2) + i 2xy\quad 0\leq y\leq 1\\ & = (4y^2 - y^2) + i (2(2y)y)\\ & = 3y^2 + i4y^2 \end {align*}

\begin {align*} I_1 & = \int ^1_0(3y^2 + 4y^2)(2+ i)dy\\ & = (3 + 4i)(2 + i) \int ^1_0 y^2 dy\\ & = \frac {2}{3} + \frac {11}{3}i\\\\ \end {align*}

OR, we may evaluate \(I\) via the path \(OAB\), call it \(I_2\) \[I_2 = \int \limits _{OA}z^2dz + \int \limits _{AB}z^2dz\]

Now, on \(OA,\, z(x) = x + 0i = x,\quad 0 \leq x \leq 2\)

On \(AB,\, z(y) = 2 + yi,\quad 0\leq y \leq 1\) \begin {align*} I_2 & = \int ^2_0x^2dx + \int ^1_0(2 + iy)^2dy\\ & = \frac {8}{3} + i\Bigg [\int ^1_0(4 - y^2)dy + 4i\int ^1_0ydy\Bigg ]\\ & = \frac {8}{3} + i \Big (4 - \frac {1}{3}\Big ) - 2\\ & = \frac {2}{3} + \frac {11}{3}i\\ \end {align*}

Note that \(I_1 = I_2\), thus the value of \(I\) is independent of path.

So if the question was to evaluate \(\displaystyle {\int z^2 dz}\) around the closed path

We have \(I_1 - I_2 = 0\)

Example 6.6. Evaluate \(\displaystyle {\int _C \frac {dz}{z - z_0}}\) where \(C\) is the contour around the region \(\,R_1 < \left |z - z_0\right | < R_2\).

Solution

Clearly, \(C\) denotes the inner and outer boundaries of the region \begin {align*} \int _C \frac {dz}{z - z_0} & = \int _{C_1} \frac {dz}{z - z_0} + \int _{C_2} \frac {dz}{z - z_0}\\\\ & = \int ^{2\pi }_0\frac {iR_1e^{i\theta }d\theta }{R_1e^{i\theta }} + \int ^{2\pi }_0 \frac {iR_2e^{i\theta }d\theta }{R_2e^{i\theta }}\\\\ & = i\int ^{2\pi }_0d\theta + i \int ^{2\pi }_0 d\theta \\\\ & = 2\pi i - 2\pi i \\ & = 0\\\\ \end {align*}

Example 6.7. Evaluate \(\displaystyle {\int \big ( z + 2\overline {z}\big )dz}\) where \(C\) is the contour given by

Here, contour \(C\) is \(OABC\).

Example 6.8. Evaluate \(\displaystyle {\int \frac {dz}{z^{1/4}}}\) where \(C\) is the circle \(\left |z\right | = 1\)

1.

where the branch used is the axe \(dz\). Where \(1^{1/4} = -i\),

2.

and the integration is to start from \(z = i\).

Working
We make \(z^{1/4}\) single valued if \(z = re^{i\theta }\), we see that for branches of \(z^{1/4} = w\) are \[ w_k = r^{1/4}\exp \Bigg [i\frac {\theta _0 + 2k\pi )}{4} \Bigg ],\quad k = 0, 1, 2, 3\] Setting \(z= 1\), so that \(\left |z\right | = 1,\quad Arg(z) = 0\)

\begin {align*} \text {We get}\quad w_k(1) & = (1)^{1/4}\exp \Bigg (\frac {k\pi i}{2}\Bigg )\, , \quad k = 0, 1, 2, 3\\\\ w_0(1) & = 1\\ w_1(1) & = i\\ w_2(1) & = -1\\ w_3(1) & = -i \end {align*}

Now, since the centre involves \(\left |z\right | = 1\), or \(C\), we have \(z = e^{i\theta }\) and \(dz = ie^{i\theta }d\theta \)

1.

The branch to be selected is \begin {align*} z^{1/4} & = w_3(1) = r^{1/4} \exp \Bigg (i\Bigg (\frac {\theta + 2(3)\pi }{4}\Bigg )\Bigg )\\ & = r^{1/4} \exp \Big (\frac {i\theta _0}{4}\Big ) \exp \Big (\frac {i6\pi }{4}\Big )\\ & = \exp \Big (\frac {i\theta _0}{4}\Big )(-i)\\ & = (-i)\exp \Big (\frac {i\theta _0}{4}\Big )\\ \end {align*}

\begin {align*} \text {Thus}\quad \int _C \frac {1}{z^{1/4}}dz & = \int _{\pi /2}^{5\pi /2}\frac {i e^{i\theta }}{-ie^{i\theta /4}}d\theta \\\\ & = - \int _{\pi /2}^{5\pi /2} e^{i3\theta /4}d\theta \\ \end {align*}

2.

Start the integration from \(z = 1\) \begin {align*} \int _C \frac {1}{z^{1/4}}dz = \int ^{2\pi }_0 \frac {i e^{i\theta }}{-ie^{i\theta /4}}d\theta \\ \end {align*}

Example 6.9.

1.

Evaluate \(\displaystyle {\int _C \ln \overline {z}\,dz}\) where \(C\) is the unit circle \(\left |z\right | = 1\). Use the branch \(\ln 1 = 4\pi i\) and integration starts from \( z = 1\).

2.

Show that \(\displaystyle {\left |\displaystyle {\int _C \frac {e^{iz}}{z}}\right |\leq 2\pi e^R}\) where \(C\) is \(\left |z\right | = R\)