# how to calculate $\frac{1}{2\pi i} \int_{\gamma} \frac{2z}{(z-1)^ 4(z-3)}$

How to calculate $\frac{1}{2\pi i} \int_{\gamma} \frac{2z}{(z-1)^ 4(z-3)}dz$

When $\gamma = C_+(0,4)$ and where $\gamma = C_-(0,2)$.

I need to use the residuformula which states that is f is analytic in point a then:

$\frac{1}{2\pi i} \int_{C} \frac{f(z)}{(z-a)^{n+1}} dz = \frac{f^{(n)}}{n!}$.

or:

$\frac{1}{2\pi i} \int_{C} f(z) = \sum Res_{z=a_j} f(z)$.

Obviously on $\gamma = C_+(0,4)$ our residual points do not occur in the edges, so by the weak variant of cauchy's theorem this is $0$. The second one is a bit harder though, because both residual points, 0 and 3 occur in the edges. How do i solve such a problem?

Kees

• i see that in the point 0 is inside the circle, so i think i should only check z = 3. – Kees Til Mar 19 '15 at 19:53

It appears you are trying to determine the contour integral inside the annulus, correct? If that is the case, by Cauchy's theorem, we know if no poles are in the contour, then the integral is zero. If there are poles in the contour, the integral is equal to $2i\pi\sum\operatorname{Res}\{f(z);z_j\}$. Inside the annulus of radius $4$ and $2$, the only pole in this contour is when $z = 3$ which is a simple pole. Therefore, $$\frac{1}{2i\pi}\int_{C}\frac{2z}{(z-1)^4(z-3)}dz = \operatorname{Res}\{f(z);z=3\}$$ You say $z=0$ is a pole but $z=0$ is a zero of $f(z)$ not a pole.