Understanding the Concept:
According to Cauchy's Integral Formula for higher-order derivatives, if a function \( f(z) \) is analytic inside and along a simple closed contour \( C \), and \( a \) is a point enclosed within \( C \), then:
\[
\oint_C \frac{f(z)}{(z-a)^{n+1}} \, dz = \frac{2\pi i}{n!} f^{(n)}(a)
\]
where \( f^{(n)}(a) \) is the \(n\)-th derivative of \( f(z) \) evaluated at the point \( a \).
Step 1: Identify Singularities and Contour Boundaries
The integrand expression is:
\[
\frac{e^z}{z^3}
\]
The singularity occurs where the denominator equals zero:
\[
z^3 = 0 \implies z = 0
\]
This represents a pole of order \( 3 \) located at the origin \( z = 0 \).
The given integration contour is the circle \( C: |z| = 1 \), which is centered at the origin with a radius of 1. Since the pole at \( z = 0 \) lies inside this unit circle, it contributes to the contour integral.
Step 2: Map parameters to Cauchy's Formula
Comparing the given integral with the generalized formula:
\[
\oint_C \frac{e^z}{(z-0)^3} \, dz
\]
We identify:
• \( f(z) = e^z \) (which is an entire function and analytic everywhere)
• Singular point \( a = 0 \)
• Power exponent \( n + 1 = 3 \implies n = 2 \)
Step 3: Differentiate and Evaluate
Find the second derivative (\( n = 2 \)) of the function \( f(z) \):
\[
f'(z) = \frac{d}{dz}(e^z) = e^z
\]
\[
f''(z) = \frac{d^2}{dz^2}(e^z) = e^z
\]
Evaluate this derivative at the singular point \( a = 0 \):
\[
f''(0) = e^0 = 1
\]
Substitute these values into Cauchy's derivative theorem formula:
\[
\oint_C \frac{e^z}{z^3} \, dz = \frac{2\pi i}{2!} \cdot f''(0) = \frac{2\pi i}{2} \cdot (1) = \pi i
\]
The integral evaluates to \( \pi i \), matching Option (C).