Without residue theory, determine $$\oint \limits_{C} \frac{2}{z^3+z}dz$$ if $C: \big|~z~-~\frac{i}{2}~\big|=1$ is positively oriented.

We first find that our integrand has three distinct singular points $z=\{-i,~0,~i\}$

Now if we draw a sketch of $C$, we see that only two of these singular points are within $C$.

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We can thus rewrite our integral as $$\oint \limits_{C} \frac{2}{z^3+z}dz = 2 \oint \limits_C \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz$$

Now we may define two closed, piecewise smooth curves $C_1, C_2$ around each singular point in $C$ as follows

enter image description here

We then have that \begin{align}\oint \limits_{C} \frac{2}{z^3+z}dz &= 2 \oint \limits_C \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz \\ &= 2 \bigg[ \underbrace{\oint \limits_{C_1} \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz}_{\displaystyle I_1} + \underbrace{\oint \limits_{C_2} \bigg(\frac{1}{z}\cdot \frac{1}{(z+i)(z-i)}\bigg)dz}_{\displaystyle I_2} \bigg] \end{align}

Now let $f(z) = \frac{1}{z(z+i)}$ then, from Cauchy Integral Formula, we know that \begin{align}I_1 &=\oint \limits_{C_1} \frac{f(z)}{(z-i)}dz \\ &= 2\pi i ~f(i) \\ &=-\pi i\end{align}

Now let $g(z) = \frac{1}{(z+i)(z-i)}$, then from Cauchy Integral Formula, we know that \begin{align}I_2 &= \oint \limits_{C_2} \frac{g(z)}{z}dz \\ &= 2\pi i ~ g(0) \\ &= 2\pi i\end{align}

So finally we have that \begin{align}\oint \limits_C \frac{2}{z^3 + z}dz = 2\big( -\pi i + 2\pi i\big) = 2 \pi i\end{align}

Does this seem correct?

  • $\begingroup$ Looks good to me, and the answer checks out with the residue calculation. $\endgroup$ – icurays1 Nov 13 '15 at 18:41
  • $\begingroup$ @icurays1 . Thank you! :). I just wanted to check it rather, since this is the first example I've worked through that had more than one singular point within $C$. $\endgroup$ – DJS Nov 13 '15 at 18:42
  • $\begingroup$ I posted an answer just so the question shows as answered. Cheers! $\endgroup$ – icurays1 Nov 13 '15 at 18:47
  • $\begingroup$ Of course, you are reproving a particular instance of a residue theorem, I hope you realize. $\endgroup$ – paul garrett Apr 17 '16 at 16:29
  • $\begingroup$ @paulgarrett Yes, that turned out to be exactly what needed to be done. When we had this question, the lecturer specifically wanted us to do it "from scratch" so that we could really see why the residue theorem "works the way it does" . :) . $\endgroup$ – DJS Apr 17 '16 at 16:31

It's correct. Essentially what you've done is exactly how the proof of the residue theorem goes anyway - you surround each isolated pole with a small contour and apply Cauchy's theorem to each individual pole. The result is the sum of the residues, multiplied by $2\pi i$, which is the residue theorem. Now you know why the residue formula is what it is!

  • $\begingroup$ Thank you! It's amazing to see how, even without using the residue theorem directly, we still manage to get to the residue theorem. Complex Analysis is by far one of the most beautiful subjects I've had so far. Everything just comes together so beautifully and elegantly. :D $\endgroup$ – DJS Nov 13 '15 at 18:49
  • $\begingroup$ Complex analysis is a magical dark art, and you're just getting to the good stuff. Have fun! $\endgroup$ – icurays1 Nov 13 '15 at 20:28

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