# Cauchy Integral Formula, mistake in solution?

I found some complex analysis problems and solutions online but I think one of the solutions is wrong. I got $-\sqrt{3}\pi i$ as the answer to (ii) here. So there is a mistake in this solution right?

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Generally it helps to provide the work you've done so far on a problem if you're asking for help. | Also, $z^2-2z-3=(z-3)(z+2)$? That's a typo. The solution to (ii) looks correct. – anon Apr 9 '12 at 8:29

$$2\pi i \left. \frac{\sin \pi z/3}{(z-3)} \right|_{z=-1} =2\pi i\frac{\sin (-\pi /3)}{(-1)-3}=(2\pi i) \frac{-\sqrt{3}/2}{-4}=\frac{\sqrt{3}\,\pi}{4}i.$$
Your answer is missing the $(z-3)$'s evaluation in the denominator it looks like.