Question

Evaluate the integral 1/2$\pi$ i$\oint$dz /z²(z²+2z+3), where c is the circle |z|=3.

Answer 1

$\frac{1}{2\pi i}\oint\limits_{|z|=3} \frac{dz}{z^2(z^2+2z+3)}=?$

All the roots of the denominator are: $0, 0, -1+\sqrt{2}, -1-\sqrt{2}$. They all lie inside the contour $C_3=\{|z|=3\}$. Thus, if $R>3$, then there are no roots between the contours $C_3$ and $C_R=\{|z|=R\}$. Therefore, by the Cauchy integral theorem, for all $R>3$,

$\frac{1}{2\pi i}\oint\limits_{|z|=3} \frac{dz}{z^2(z^2+2z+3)}=\frac{1}{2\pi i}\oint\limits_{|z|=R} \frac{dz}{z^2(z^2+2z+3)}$

and therefore

$\frac{1}{2\pi i}\oint\limits_{|z|=3} \frac{dz}{z^2(z^2+2z+3)}=\lim\limits_{R\to+\infty}\frac{1}{2\pi i}\oint\limits_{|z|=R} \frac{dz}{z^2(z^2+2z+3)}$

But

$|\frac{1}{2\pi i}\oint\limits_{|z|=R} \frac{dz}{z^2(z^2+2z+3)}|\leq \frac{1}{2\pi}\oint\limits_{|z|=R} \frac{|dz|}{|z|^2|z^2+2z+3|}\leq$

$\leq \frac{1}{2\pi}\oint\limits_{|z|=R} \frac{|dz|}{|z|^2(|z|-3)^2}= \frac{1}{2\pi}\frac{2\pi R}{R^2(R-3)^2}=\frac{1}{R(R-3)^2}\to 0$

therefore, $\frac{1}{2\pi i}\oint\limits_{|z|=3} \frac{dz}{z^2(z^2+2z+3)}=\lim\limits_{R\to+\infty}\frac{1}{2\pi i}\oint\limits_{|z|=R} \frac{dz}{z^2(z^2+2z+3)}=0$.

Answer. $\frac{1}{2\pi i}\oint\limits_{|z|=3} \frac{dz}{z^2(z^2+2z+3)}=0$.