Cauchy Integral Evaluation

This solution evaluates the contour integral $\oint \frac{Z^3-4}{3Z-i} dZ$ using Cauchy's Integral Formula. The integration is assumed to be counter-clockwise and enclosing the singularity.

Singularity Location

The integrand is $ \frac{Z^3-4}{3Z-i} $. A singularity occurs where the denominator equals zero:

$3Z - i = 0 \implies Z = \frac{i}{3}$

Let $a = \frac{i}{3}$. This point is assumed to lie inside the contour of integration.

Cauchy Formula Application

Cauchy's Integral Formula states that for an analytic function $g(z)$ inside and on a simple closed contour $C$, with $a$ being a point inside $C$:

$\oint_C \frac{g(z)}{z-a} dz = 2\pi i g(a)$

The given integral is rewritten to fit this form:

$\oint \frac{Z^3-4}{3Z-i} dZ = \oint \frac{1}{3} \frac{Z^3-4}{Z - i/3} dZ$

Here, $g(Z) = Z^3 - 4$, which is analytic everywhere, and $a = \frac{i}{3}$.

Function $g(a)$ Evaluation

Evaluate the analytic function $g(Z)$ at the singularity $a = \frac{i}{3}$:

$g\left(\frac{i}{3}\right) = \left(\frac{i}{3}\right)^3 - 4$ $g\left(\frac{i}{3}\right) = \frac{i^3}{27} - 4$ $g\left(\frac{i}{3}\right) = \frac{-i}{27} - 4 \quad (\text{since } i^3 = -i)$

Integral Value Computation

Apply Cauchy's Integral Formula result, incorporating the factor of $\frac{1}{3}$:

$ \text{Value} = \frac{1}{3} \times \left( 2\pi i \times g\left(\frac{i}{3}\right) \right) $ $ \text{Value} = \frac{1}{3} \times 2\pi i \times \left( \frac{-i}{27} - 4 \right) $ $ \text{Value} = \frac{2\pi i}{3} \left( \frac{-i}{27} - 4 \right) $ $ \text{Value} = \left( \frac{2\pi i}{3} \times \frac{-i}{27} \right) - \left( \frac{2\pi i}{3} \times 4 \right) $ $ \text{Value} = \frac{-2\pi i^2}{81} - \frac{8\pi i}{3} $ $ \text{Value} = \frac{-2\pi (-1)}{81} - \frac{8\pi i}{3} \quad (\text{since } i^2 = -1) $ $ \text{Value} = \frac{2\pi}{81} - \frac{8\pi i}{3} $

The calculated value of the integral is $\frac{2\pi}{81} - \frac{8\pi i}{3}$.