Hello, I am having difficulty matching one term in my Laurent series to that which mathematica tells me is the correct answer. For the function(adsbygoogle = window.adsbygoogle || []).push({});

[itex]f(z)=log\frac{1+z}{1-z}[/itex]

we know that there exists a k such that

[itex]Log|1+z|-Log|1-z|+i2\pi k[/itex]

Now, we know that the Taylor series of f is as below

[itex] f(z)=log\frac{1+z}{1-z}=\int \frac{1}{1+z}dz+\int \frac{1}{1-z} dz

=\int \sum_{n=0}^{\infty} (-z)^n dz+\int \sum_{n=0}^{\infty} dz=2 \sum_{n=0}^{\infty}\frac{z^(2n+1)}{2n+1} [/itex]

Now we would like to find the Laurent series. I used a similar approach resulting in

[itex] f(z)=log\frac{1+z}{1-z}=\int \frac{1}{z}\frac{1}{1+1/z}dz-\int \frac{1}{z}\frac{1}{1-1/z} dz =2 \sum_{n=1}^{\infty} \frac{z^{-(2n+1)}}{2n+1}[/itex]

Notice that since I used integration to find the coefficients, I was unable to determine the coefficient of the power z^-1 which according to mathematica should give me the following term in the series

[itex] \frac{2}{z}+i\pi [/itex]

I know that we can use the following definition to find this coefficient

[itex] b_{-1}=\frac{1}{2 \pi i} \oint_c f(z) dz [/itex]

However I do not know how to integrate the closed contour with a branch cut.

Also, it seems that the branch cut we choose is important to make our Laurent

series expansion consistent with the function, however what tells me what the

correct branch cut is?

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# Laurent series of a logarithm

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