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I have found a link determining the Laurent series expansion in a punctured disk of unit radius.Here's the link https://math.stackexchange.com/a/231228/251057. Here I have understood everything except the substitution "With substitution $z=\frac{1}{u}$ we get $b_k=\frac{1}{2 \pi i} \oint_\gamma\frac{g(\frac{1}{u})}{u^{-k-1}u^2}du=\frac{1}{2 \pi i} \oint_\gamma\frac{f(u)}{u^{-k+1}}du=a_{-k}.$" when we substitute $z$ by $\frac {1} {u}$ then how does the contour change as in the case of Riemann integration when variable of integration is substituted then accordingly we had to change the limit of the integration. Now here in the case of contour integration if the above substitution were taken then $dz=-\frac {1} {u^2}\ du$. But how the limits of the integration i.e. in this case the contour of the integration changes due to this substitution.

I am in a fix. Please help me in undestanding this concept.

Thank you in advance.

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Actually substitution will also affect the path. However in this case we can assume that the path is circular and since $|z|$ is constant then so is $|u|=1/|z|$. However the argument of $z$ and $u$ is opposite which means that the circular path will be taken in opposite direction.

Now it's not given that the path is circular, but as the result is independent of the path given that the winding number is unchanged we can assume that it's circular. In effect this means that the new path must be a path with opposite winding number.

Note that this has already been taken care of in the formula as you're also missing the minus sign from the substitution - remember that reversion of the path is also negating the result. The path in the formulae is the same.

skyking
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  • I think we also have to use Cauchy's deformation theorem here. Isn't it? –  Nov 01 '17 at 14:26
  • @A.Chattopadhyay Yes and no. In the post you referred to $\gamma$ was specified to be the unit circle, but if $\gamma$ would have been any curve you would use it or the residue theorem... – skyking Nov 01 '17 at 14:30