How can I show that the $ \lim_{R \rightarrow \infty}$ $\int_{|z|=R} \frac{dz}{(z)(z-3)^2}$ = 0? I have already proved that by ML - inequality, $\Big| \int_{|z|=R} \frac{dz}{(z)(z-3)^2} \Big|$ $\leq \frac{2\pi}{(R-3)^2}.$ Should I use the bound to find the limit of the integral?
2026-04-07 19:30:29.1775590229
Limit of Complex Integral is 0
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That's right. You simply use the fact that$$\lim_{R\to\infty}\frac{2\pi}{(R-3)^2}=0.$$