$ \lim_{R \rightarrow \infty } \oint_ {|z|=R} \frac{ 2z - 1 }{ z^{3} - 2z^{2} + 3z} = 0$
I'm struggling with this problem. Could someone please help me figure out?
I have decomposed integrand into $-\frac{1}{3z} + \frac{z+4}{3(z^{2}-2z+3)}$ and I know that first term is always $-\frac{2 \pi i}{3}$ regardless of $R$ (or opposite sign when computed clockwise) but I don't know how to compute the next term.
If $\lvert z\rvert=R$ and $R$ is large enough, you have$$\left\lvert\frac{2z-1}{z^3-2z^2+3z}\right\rvert\leqslant\frac{2R+1}{R^3-R^2-3R}$$and therefore$$\left\lvert\oint_{\lvert z\rvert=R}\frac{2z-1}{z^3-2z^2+3z}\,\mathrm dz\right\rvert\leqslant2\pi R\frac{2R+1}{R^3-2R^2-3R}.$$Now, use the fact that$$\lim_{R\to\infty}2\pi R\frac{2R+1}{R^3-2R^2-3R}=0.$$