# Contour integral of $\int_\gamma \frac{dz}{(2z+1)(z+3)^2}$

Im a little confused by the following integral question

Let $\gamma$ be the unit circle in $\mathbb{C}$ traversed in the anti-clockwise direction.

$\displaystyle\int_\gamma \dfrac{1}{(2z+1)(z+3)^2}$

z=-3 is outside the circle... and z=-$\frac{1}{2}$ is inside.

Hence.. $\displaystyle\int_\gamma \dfrac{1}{(2z+1)(z+3)^2}$ = $\displaystyle\int_\gamma \dfrac{\dfrac{1}{(z+3)^2}}{(2z+1)}$

And hence $f(x)=\dfrac{1}{(x+3)^2}$ and $f_o=\frac{1}{z}$? This is where im getting a little confused....

• Cauchy's integral formula. Note that the curve need not be a cicle of center $-1/2$. That it goes around $-1/2$ once in the anti-clockwise direction suffices. Commented May 3, 2013 at 15:45
• @julien Im just a little confused as its not in the form (z-a) on the denominator Commented May 3, 2013 at 15:49
• $2z+1=2(z-(-1/2))$ Commented May 3, 2013 at 15:50
• So i just take $z_0=\frac{-1}{2}$ so= $2\pi i \cdot f(\frac{-1}{2})$ where $f(x)=\dfrac{1}{(x+3)^2}$? Commented May 3, 2013 at 15:55

Hint! Cauchy Residue theorem says that your integral is $2\pi i \sum R_i$ where $R_i$ are the residues inside the closed contour, in you case the only pole inside is $-1/2$, so just take the residue at this point and multiply with $2 \pi i$
• Im not really sure if im doing this right to calculate the residue... $\displaystyle\lim_{z\rightarrow-\frac{1}{2}} \dfrac{(z+\frac{1}{2})(1)}{2(z-(-\frac{1}{2})(z+3)^2} = \dfrac{1}{2(z+3)^2}= \dfrac{1}{2(-\frac{1}{2}+3)^2} = \dfrac{2}{25}$ Hence $Res(f(z),-\frac{1}{2})= 2\pi i \cdot \frac{2}{25} =\frac{4}{25}\pi i$? Commented May 3, 2013 at 18:23
• DId you ever figure out if that is the correct answer? I thought for order 2 zeros we would have to take the derivative of the residue $f(z)$ before plugging in $-1/2$...Am curious! Commented May 3, 2022 at 5:57