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let f(z)=1/z. Two contours joining the points -1 and 1 such that $$ \int_{c1} f(z)\,dz \ne \int_{c2} f(z)\,dz $$

we can choose a unit circle as a one contour. any other idea for another one??

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    I assume that you have given some direction to your contour $C_{1}$ , then how about the same contour but with the direction reversed,i think it will give you a negative sign due which they will not be equal! – 2017-02-25
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    yes, i think its correct. but can't we find a another contour, i mean polygon or something – 2017-02-25
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    Then how about a straight line from $-1$ to $i$ and another straight line from $i$ to $1$ ? , although you have to check the integral value ? – 2017-02-25
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    @BAYMAX but f is not continuous in that region. So can we choose that?? – 2017-02-25
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    please elaborate ? – 2017-02-25
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    @BAYMAX if G be an open subset of C and let p in G. If f(z) is continuous and analytic inG\{p} then we can say $ \int f(z) \,dz = 0 $. Am i correct?? – 2017-02-25
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54329/discussion-between-baymax-and-miraj). – 2017-02-25

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I would choose to half-circles, one going from $-1$ to $1$ through $i$ and the other through $-i$.

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    Can you elaborate your reason for this choice? even a few words – 2017-02-25
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    We know that a circle around the simple pole 0 will give a non-zero integral. Hence, cut this circle into two parts, they won't be equal. – 2017-02-25
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    but that's exactly the unit circle mentioned by OP, who asks for "another one" – 2017-02-25
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    No, a unit circle doesn't go from -1 to 1, cuz it comes back. – 2017-02-25