<s>[INST] How to calculate this line integral?(complex plane) 
There are counter-clockwise two curves, $C_1$ and $C_2$ on complex plane.
$C_1$ ; ${1 \over 3} \leq \vert z \vert \leq 2$ (on the 1st Quadrant.)
$C_2$ ; ${1 \over 3} \leq \vert z \vert \leq 2$ (on the 4th Quadrant)
Calculate $\int_{C_1} {z^3+1 \over z^4 + 4z -1} dz$ - $\int_{C_2} {z^3+1 \over z^4+4z-1} dz$

My attempt) By Rouche thm, $z^4 + 4z -1$ have 3 roots on the curve, $C$: ${1 \over 3} \leq \vert z \vert \leq 2$
Since,  $\int_{C_1} {z^3+1 \over z^4 + 4z -1} dz$ - $\int_{C_2} {z^3+1 \over z^4+4z-1} dz$ = ${1 \over 2} {\int_{C} {z^3+1 \over z^4 + 4z -1} dz} $
Therefore, by argumented principle 
${1 \over 2} {\int_{C} {z^3+1 \over z^4 + 4z -1} dz} $ = ${1 \over 8}  {\int_{C} {4(z^3+1) \over z^4 + 4z -1} dz} = {1 \over 8} \times 2\pi i \times (3-0) = {3 \over 4}\pi i  $
But the answer was $0$ 
What point do I have a mistake?
Any help would be thanksful.
 [/INST] The denominator has one root close to $\frac14$ outside the region and a triple of roots close to the roots of $z^3+4=0$, where the conjugate pair $\frac{1\pm i\sqrt3}2\sqrt[3]4$ has one root inside $C_1$ and $C_2$, respectively. 

Let $\zeta$ be the root inside $C_1$. Then the integral of $\frac{P(z)}{Q(z)}$ over the curve $C_1$ is equal to the residuum $2\pi i \frac{P(ζ)}{Q'(ζ)}=2\pi i \frac{ζ^3+1}{4ζ^3+4}=\frac{\pi i}2$.
Likewise the integral over $C_2$ is equal to the residual value $2\pi i \frac{P(\bar ζ)}{Q'(\bar ζ)}=2\pi i \frac{\bar ζ^3+1}{4\bar ζ^3+4}=\frac{\pi i}2$.
Their difference is clearly zero.

As $P(z)=\frac14Q'(z)$, the integral over any closed curve $C$ is clearly $\frac{\pi i}2$ times the winding number of the image under $Q$ of the curve around zero, which is the number of roots of $Q$ inside the curve. As $Q$ has real coefficients, the roots come in conjugate pairs, thus giving equal integrals over $C_1$ and $C_2$ independent of the number of roots contained within.
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