Contour Integral Evaluation Explained

This problem asks us to evaluate a contour integral in the complex plane. The integral is \(\mathop \oint \nolimits_c \frac{1}{{{z^2} + 4}}dz\), and the contour C is a circle defined by \(|z - 2i| = 1\). To solve this, we will use Cauchy's Integral Formula.

Poles Identification and Location

First, we need to find the singularities (also known as poles) of the integrand \(f(z) = \frac{1}{{{z^2} + 4}}\). These are the points where the denominator becomes zero.

• Set the denominator to zero: \(z^2 + 4 = 0\)
• Solve for \(z\): \(z^2 = -4\)
• This gives us two poles: \(z = \sqrt{-4} = \pm 2i\)

So, the poles are at \(z_1 = 2i\) and \(z_2 = -2i\).

Next, we need to determine which of these poles lie inside the given contour C. The contour C is defined by \(|z - 2i| = 1\). This represents a circle in the complex plane with its center at \(2i\) and a radius of 1.

• For the pole \(z_1 = 2i\): Calculate its distance from the center of the contour: \(|2i - 2i| = |0| = 0\). Since \(0 < 1\) (the radius of the circle), the pole \(z_1 = 2i\) lies inside the contour C.
• For the pole \(z_2 = -2i\): Calculate its distance from the center of the contour: \(|-2i - 2i| = |-4i|\). The modulus of \(-4i\) is \(\sqrt{0^2 + (-4)^2} = \sqrt{16} = 4\). Since \(4 > 1\) (the radius of the circle), the pole \(z_2 = -2i\) lies outside the contour C.

Therefore, only the pole \(z = 2i\) contributes to the contour integral.

Cauchy's Integral Formula Application

Since only one simple pole lies inside the contour, we can apply Cauchy's Integral Formula. The formula states that for a simple closed contour C and a function \(f(z)\) that is analytic inside and on C, and a point \(a\) inside C, the integral is given by:

\[\mathop \oint \nolimits_C \frac{{f(z)}}{{z - a}}dz = 2\pi i f(a)\]

To use this formula, we need to rewrite our integrand \(\frac{1}{{{z^2} + 4}}\) in the form \(\frac{{f(z)}}{{z - a}}\).

• We know \(z^2 + 4 = (z - 2i)(z + 2i)\).
• The pole inside our contour is \(a = 2i\), so we want the denominator to be \((z - 2i)\).
• Therefore, we can set \(f(z) = \frac{1}{{z + 2i}}\).

Now, substitute these into Cauchy's Integral Formula:

\[\mathop \oint \nolimits_c \frac{1}{{{z^2} + 4}}dz = \mathop \oint \nolimits_c \frac{{\frac{1}{{z + 2i}}}}{{z - 2i}}dz\]

Here, \(f(z) = \frac{1}{{z + 2i}}\) and \(a = 2i\).

Next, evaluate \(f(a)\) at \(a = 2i\):

\[f(2i) = \frac{1}{{2i + 2i}} = \frac{1}{{4i}}\]

Finally, apply the formula:

\[\mathop \oint \nolimits_c \frac{1}{{{z^2} + 4}}dz = 2\pi i \cdot f(2i) = 2\pi i \cdot \frac{1}{{4i}}\]

Simplify the expression:

\[ = \frac{{2\pi i}}{{4i}} = \frac{{2\pi}}{{4}} = \frac{\pi}{2}\]

Thus, the value of the contour integral is \(\frac{\pi}{2}\).