Using the Cauchy integral formula to evaluate $\int_{\gamma=(a,a)} \frac{z}{z^4-1} dz$.

I'm trying to understand how to use the Cauchy integral formula, but a bit confused as to how to use it as I cant seem to get the right answer!

$$\int_{\gamma=(a,a)} \frac{z}{z^4-1} dz$$ where $a\in\mathbb{R}$ and $a>0$ and $a\not= 1/2$.

please note: this must be solved using cauchy integral formula!

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shouldn't $a=1$ make much more problems? what you mean by gammer=(a,a)? – Dominic Michaelis Mar 14 '13 at 19:04
what is gammer=(a,a)? – NECing Mar 14 '13 at 19:06
Not sure, i have just found a question that i dont understand, i didnt make it up! Next to the integral sign, there is a gammer(a,a) – sarah Mar 14 '13 at 19:09
i think it means w=a and r=a? – sarah Mar 14 '13 at 19:10
@sarah I assume that by gammer you mean $\gamma$? – Alexander Gruber Mar 14 '13 at 19:16

The function $f(z)=\dfrac{z}{z^4-1}$ has poles at $\pm 1$ and $\pm i$. There are two cases:
1. When $0<a<1/2$, the circle with center $a$ and radius $a$ does not enclose any poles of $f$. Hence $\int_{\gamma(a,a)}f(z)\,dz=0$.
2. When $a>1/2$, the circle with center $a$ and radius $a$ encloses two poles of $f$, namely $1$ and $i$. The quickest way to evaluate the integral would be the residue theorem. But if we are allowed only the Cauchy integral formula, the partial fraction decomposition can be used: $$\dfrac{z}{z^4-1} = \frac{z}{4}\left(\frac{1}{z-1}-\frac{1}{z+1} +\frac{i}{z-i}-\frac{i}{z+i}\right)$$ By the Cauchy integral formula,
$$\int_{\gamma(a,a)}\frac{z}{4} \frac{1}{z-1}\,dz=2\pi i \frac14$$ and $$\int_{\gamma(a,a)}\frac{z}{4} \frac{i}{z-i}\,dz=2\pi i \frac{i^2}4$$ while the other two integrals are zero.