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Can someone help me: Assuming that none of the points $0$, $1$ and $-1$ lies on $L$, calculate all possible values of the integral $$ \mathop{\int}\limits_{L}{\frac{dz}{{z}{\mathrm{(}}{z}^{2}\mathrm{{-}}{1}{\mathrm{)}}}}\text{,} $$

for various positions of $ L $ .

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    I am just thinking about cauchy's integral and how we can use it – 2017-01-12
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    What's $L{}{}$? – 2017-01-12
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    I think it's any close rectifable curve – 2017-01-12
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    Is there any hint – 2017-01-12

1 Answers 1

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We have $z(z^2-1) = z(z+1)(z-1)$.

Calculating the residues, we have:

\begin{align} \text{Res}\left(\frac{1}{z(z+1)(z-1)},0\right)&=\lim\limits_{z\rightarrow0}\frac{1}{(z+1)(z-1)}=-1\\ \text{Res}\left(\frac{1}{z(z+1)(z-1)},1\right)&=\lim\limits_{z\rightarrow1}\frac{1}{z(z+1)}=\frac{1}{2}\\ \text{Res}\left(\frac{1}{z(z+1)(z-1)},-1\right)&=\lim\limits_{z\rightarrow-1}\frac{1}{z(z-1)}=\frac{1}{2} \end{align}

Now, by the Residue Theorem, we have

$$\oint_L\frac{dz}{z(z^2-1)}=2\pi i\sum_k\text{Res}\left(\frac{1}{z(z+1)(z-1)},z_k\right)$$

where $z_k$ are the zeros of the denominator of the function that are enclosed by L.

For example, if $L$ is a circle about the origin with radius $R>1$, we have

$$\oint_L\frac{dz}{z(z^2-1)}=2\pi i\left(-1+\frac{1}{2}+\frac{1}{2}\right)=0$$

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    Why did we take the sum of the values in the final step ? – 2017-01-24
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    The value of a contour integral in the complex plane along a closed path is the sum of all the residues at singularities of the integrand that are enclosed within this contour, multiplied with $2\pi i$. If you have a circle about the origin with radius $R>1$, then all three singularities of your function are enclosed within it, thus we sum over all the residues. – 2017-01-24