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just in a bit of a situation and hoping someone can help i've been asked to find the following integral

$$\int_{S_{1}^{+}(-1)} \frac{dz}{z^2-1}$$ where $$S_{1}^{+}(-1)$$ is the positively orientated circle of radius 1 centred at -1

So far i've done the following steps

$$\int_{S_{1}^{+}(-1)} \frac{1}{z^2-1} = \frac{1}{2}\int_{S_{1}^{+}(-1)} \frac{dz}{z-1} - \frac{1}{2}\int_{S_{1}^{+}(-1)} \frac{dz}{z+1} $$

from here i note that $\frac{1}{2}\int_{S_{1}^{+}(-1)} \frac{dz}{z+1}$ is essentially a shift and rescaling of $\int_{S_{1}^{+}(0)} \frac{dz}{z}$ and so i equate that to $2 \pi i$ giving a final value of $- \frac{1}{2}\int_{S_{1}^{+}(-1)} \frac{dz}{z+1} = -\pi i$

but i'm unsure how to go about the other integral

i've yet to learn about residue theory or poles so i'm limited to essentially contour integration, deformation theorem and the fundamental theorem/cauchy integral theorems

any help would be greatly appreciated. cheers

Vaas
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    Do we get any hints about the meaning of the path, "$S_1^+(-1)$"? – Eric Towers Oct 23 '19 at 16:59
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    The function $\frac{1}{z-1}$ is holomorphic within the circle and a contour integral about the closed circle loop should get you zero. – edm Oct 23 '19 at 17:01
  • my apologies about the notation, i'll add that now, and Thanks you both. – Vaas Oct 23 '19 at 17:03

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