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Evaluate $\int \frac{sin\pi z^2 + cos\pi z^2}{(z - 1)(z - 2)}dz$, where c is the circle $|z| = 3$

From taylor series

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The circle $|z| = 3$ has center at (0, 0) and radius 3. The point z = 1 and z = 2 lie inside the circle. Hence f(z) is analytic in C. Hence using partial fraction by removing

$f(z) = sin\pi z^2 + cos\pi z^2$

$\frac{1}{(z - 1)(z - 2)} = \frac{A}{(z - 1)} + \frac{B}{(z - 2)} \\ \frac{1}{(z - 1)(z - 2)} = \frac{(z - 2)A + (z - 1)B}{(z - 1)(z - 2)}$

Let z = 2 we get B = 1

And Let z = 1 we get A = 1

$\therefore \frac{1}{(z - 1)(z - 2)} = \frac{1}{(z - 2)} - \frac{1}{(z - 1)} \\ \therefore \int \frac{sin\pi z^2 + cos\pi z^2}{(z - 1) (z - 2)} \\ = \int \frac{f(z)dz}{z - 2} - \int \frac{f(z)dz}{z - 1} \\ = 2\pi i f(2) - 2\pi i f(1) \\ = 2\pi i (sin\pi 4 + cos4\pi) - 2\pi i(sin\pi + cos\pi) \\ = 2\pi i (1) - 2\pi i(-1) \\ = 4\pi i$

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