Contour Integral Problem Setup

We are asked to evaluate the contour integral \( \oint_C \frac{{{{\rm{z}}^{\rm{2}}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{4z}}}}\)dz.

The contour C is a circle centered at \(z = \frac{1}{4}\) with radius \( \frac{\pi}{4} \), traversed counter-clockwise.

The center of the circle is \(c = \frac{1}{4}\) and the radius is \(R = \frac{\pi}{4}\).

Singularities of the Integrand

The integrand is \(f(z) = \frac{{{{\rm{z}}^{\rm{2}}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{4z}}}}\). Singularities occur where the denominator is zero.

Set \({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{4z}} = 0\), which means \({\rm{sin(4z)}} = 0\).

This occurs when \(4z = n\pi\), where \(n\) is an integer (\(n \in \mathbb{Z}\)).

Thus, the singularities are located at \(z = \frac{n\pi}{4}\) for \(n = 0, \pm 1, \pm 2, \dots\).

Singularities Inside the Contour

The contour C is the circle \( \left|z - \frac{1}{4}\right| = \frac{\pi}{4} \).

A singularity \(z_n = \frac{n\pi}{4}\) is inside the contour if its distance from the center \(\frac{1}{4}\) is strictly less than the radius \(\frac{\pi}{4}\).

We check the distance \( \left|\frac{n\pi}{4} - \frac{1}{4}\right| \) against the radius \( \frac{\pi}{4} \).

• For \(n=0\): \(z_0 = 0\). \( \left|0 - \frac{1}{4}\right| = \frac{1}{4} \). Since \( \frac{1}{4} < \frac{\pi}{4} \) (approximately \(0.25 < 0.785\)), \(z=0\) is inside the contour.
• For \(n=1\): \(z_1 = \frac{\pi}{4}\). \( \left|\frac{\pi}{4} - \frac{1}{4}\right| = \frac{\pi - 1}{4} \). Since \( \frac{\pi - 1}{4} \approx 0.535 \), and \( 0.535 < 0.785 \), \(z = \frac{\pi}{4}\) is inside the contour.
• For \(n=-1\): \(z_{-1} = -\frac{\pi}{4}\). \( \left|-\frac{\pi}{4} - \frac{1}{4}\right| = \frac{\pi + 1}{4} \). Since \( \frac{\pi + 1}{4} \approx 1.035 \), and \( 1.035 > 0.785 \), \(z = -\frac{\pi}{4}\) is outside the contour.
• For \(n=2\): \(z_2 = \frac{2\pi}{4} = \frac{\pi}{2}\). \( \left|\frac{\pi}{2} - \frac{1}{4}\right| = \frac{2\pi - 1}{4} \). Since \( \frac{2\pi - 1}{4} \approx 1.32 \), and \( 1.32 > 0.785 \), \(z = \frac{\pi}{2}\) is outside the contour.

The only singularities inside the contour C are \(z = 0\) and \(z = \frac{\pi}{4}\).

Classifying Singularities

We need to determine the type and order of the singularities at \(z=0\) and \(z=\frac{\pi}{4}\).

• For \(z=0\): We evaluate the limit \( \lim_{z \to 0} f(z) \).
  \( \lim_{z \to 0} \frac{{{{\rm{z}}^{\rm{2}}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{4z}}}} \). Using the standard limit \( \lim_{x \to 0} \frac{\sin(ax)}{ax} = 1 \):
  \( \lim_{z \to 0} \frac{{{{\rm{z}}^{\rm{2}}}}}{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{4z}}}} = \lim_{z \to 0} \frac{z^2}{(4z)^2 \left(\frac{\sin(4z)}{4z}\right)^2} = \lim_{z \to 0} \frac{z^2}{16z^2 (1)^2} = \frac{1}{16} \).
  Since the limit exists and is finite, \(z = 0\) is a removable singularity.
• For \(z = \frac{\pi}{4}\): As \(z \to \frac{\pi}{4}\), \(\sin(4z) \to \sin(\pi) = 0\).
  Let \(z = \frac{\pi}{4} + w\). As \(z \to \frac{\pi}{4}\), \(w \to 0\).
  \( \sin(4z) = \sin(4(\frac{\pi}{4} + w)) = \sin(\pi + 4w) = -\sin(4w) \).
  Since \( \sin(4w) \approx 4w \) for small \(w\), \( \sin^2(4w) \approx (4w)^2 = 16w^2 \).
  The denominator behaves like \(16w^2\) near \(w=0\), which corresponds to \(16(z-\frac{\pi}{4})^2\) near \(z=\frac{\pi}{4}\). The numerator \(z^2\) approaches \((\frac{\pi}{4})^2 = \frac{\pi^2}{16} \ne 0\).
  The function has the form \(\frac{\text{non-zero}}{ (z-a)^2 \times (\text{analytic and non-zero at } a)} \). This indicates a pole of order 2 at \(z = \frac{\pi}{4}\).

Calculating Residues

According to the Residue Theorem, \( \oint_C f(z) dz = 2\pi i \sum (\text{Residues inside C}) \).

Residue at z = 0 (Removable Singularity)

The residue at a removable singularity is zero.

\(\text{Res}(f, 0) = 0\).

Residue at z = π/4 (Pole of Order 2)

For a pole of order \(m\) at \(a\), the residue can be found from the Laurent series \( f(z) = \sum_{n=-\infty}^{\infty} c_n (z-a)^n \), where \( \text{Res}(f, a) = c_{-1} \). For a pole of order 2, we look for the coefficient of \((z-a)^{-1}\).

Let \(z = \frac{\pi}{4} + w\). We expand \(f(\frac{\pi}{4} + w)\) around \(w=0\):

\( f(\frac{\pi}{4} + w) = \frac{(\frac{\pi}{4} + w)^2}{\sin^2(4(\frac{\pi}{4} + w))} = \frac{(\frac{\pi}{4} + w)^2}{\sin^2(\pi + 4w)} = \frac{(\frac{\pi}{4} + w)^2}{(-\sin(4w))^2} = \frac{(\frac{\pi}{4} + w)^2}{\sin^2(4w)} \).

We use the Taylor series for \(\sin(x)\) around \(x=0\): \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \).

\( \sin(4w) = 4w - \frac{(4w)^3}{6} + O(w^5) = 4w - \frac{32w^3}{3} + O(w^5) \).

\( \sin^2(4w) = \left(4w - \frac{32w^3}{3} + \dots\right)^2 = (4w)^2 - 2(4w)\left(\frac{32w^3}{3}\right) + \dots = 16w^2 - \frac{256w^4}{3} + \dots \).

Factor out \(16w^2\): \( \sin^2(4w) = 16w^2 \left(1 - \frac{16w^2}{3} + \dots\right) \).

Now write the full expression for \(f(\frac{\pi}{4} + w)\):

\( f(\frac{\pi}{4} + w) = \frac{(\frac{\pi}{4} + w)^2}{16w^2 \left(1 - \frac{16w^2}{3} + \dots\right)} = \frac{1}{16w^2} \left(\frac{\pi}{4} + w\right)^2 \left(1 - \frac{16w^2}{3} + \dots\right)^{-1} \).

Expand the terms using Taylor series around \(w=0\):

• \( \left(\frac{\pi}{4} + w\right)^2 = \frac{\pi^2}{16} + \frac{\pi}{2}w + w^2 \)
• \( \left(1 - \frac{16w^2}{3} + \dots\right)^{-1} = 1 + \frac{16w^2}{3} + O(w^4) \) (using \( (1-x)^{-1} = 1+x+\dots \))

Multiply the expansions inside the parenthesis:

\( \left(\frac{\pi^2}{16} + \frac{\pi}{2}w + w^2\right) \left(1 + \frac{16w^2}{3} + \dots\right) = \frac{\pi^2}{16}(1 + \dots) + \frac{\pi}{2}w(1 + \dots) + w^2(1 + \dots) \)

We are looking for the coefficient of \(w^{-1}\) in the final expansion of \(f(\frac{\pi}{4} + w)\). This means we need the coefficient of \(w^{1}\) in the product \( \left(\frac{\pi^2}{16} + \frac{\pi}{2}w + w^2\right) \left(1 + \frac{16w^2}{3} + \dots\right) \) before multiplying by \(\frac{1}{16w^2}\).

The term with \(w^1\) in the product is \( \frac{\pi}{2}w \cdot 1 = \frac{\pi}{2}w \).

So, the product expansion starts as \( \frac{\pi^2}{16} + \frac{\pi}{2}w + O(w^2) \).

Now, \( f(\frac{\pi}{4} + w) = \frac{1}{16w^2} \left( \frac{\pi^2}{16} + \frac{\pi}{2}w + O(w^2) \right) = \frac{\pi^2}{256w^2} + \frac{\pi}{32w} + O(1) \).

The coefficient of \(w^{-1}\) is \( \frac{\pi}{32} \).

Thus, \(\text{Res}(f, \frac{\pi}{4}) = \frac{\pi}{32}\).

Applying the Residue Theorem

The value of the contour integral is \( 2\pi i \) times the sum of the residues inside the contour C.

Sum of residues = \( \text{Res}(f, 0) + \text{Res}(f, \frac{\pi}{4}) = 0 + \frac{\pi}{32} = \frac{\pi}{32} \).

Value of the integral = \( 2\pi i \times \frac{\pi}{32} = \frac{2\pi^2 i}{32} = \frac{\pi^2 i}{16} \).

Final Result

The value of the contour integral is \( \frac{{{\rm{i}}{{\rm{\pi }}^2}}}{{16}} \).