The value of the contour integral $\frac{1}{2\pi i} \oint_C \frac{\cos(\pi z)}{(z - 1)(z - 2)^2} \, dz$ evaluated counter-clockwise along the circle $C: |z| = 1.5$ is:

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### 1. Strategy & Theoretical Principles By Cauchy's Residue Theorem: $$\oint_C f(z) \, dz = 2\pi i \sum \text{Res}(f, z_k)$$ where $z_k$ are the singularities lying strictly inside the closed contour $C$. ### 2. Analytical Formulation The integrand $f(z) = \frac{\cos(\pi z)}{(z - 1)(z - 2)^2}$ has: - A simple pole at $z = 1$ - A pole of order 2 at $z = 2$ The contour is the circle $|z| = 1.5$. - The pole $z = 1$ lies inside $C$ since $|1| = 1 < 1.5$. - The pole $z = 2$ lies outside $C$ since $|2| = 2 > 1.5$. ### 3. Step-by-Step Calculation Therefore, only the residue at $z = 1$ is enclosed: $$\text{Res}(f, 1) = \lim_{z \to 1} (z - 1) f(z) = \lim_{z \to 1} \frac{\cos(\pi z)}{(z - 2)^2} = \frac{\cos(\pi)}{(1 - 2)^2} = \frac{-1}{(-1)^2} = -1$$ Applying Cauchy's theorem: $$\oint_C f(z) \, dz = 2\pi i \times (-1) = -2\pi i$$ If the question asks for the imaginary coefficient or real value of $\frac{1}{i} \oint$, let's formulate question as $\frac{1}{i} \oint_C f(z) dz = -2\pi \approx -6.283$. Let's ensure the question specifies $\frac{1}{2\pi i} \oint_C$ so the answer is an integer $-1.0$. Let's rephrase to: "The value of the contour integral $\frac{1}{2\pi i} \oint_C \frac{\cos(\pi z)}{(z - 1)(z - 2)^2} \, dz$ along $C: |z| = 1.5$ is:" Then the exact answer is $-1.0$ (tolerance: -1.0 to -1.0)! ### 4. Distractor Forensics - Candidate enclosing both poles will attempt to calculate the second order residue at $z=2$, wasting time or introducing extraneous terms. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Always check which poles are interior to the given contour $|z| = 1.5$. - **Standard Reference:** Churchill, R.V. & Brown, J.W., *Complex Variables and Applications*, 9th Edition, McGraw-Hill, Chapter 6.

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