Cauchy Integral Theorem with partial fractions

[ognik]

(Wish there was a solutions manual...). Please check my workings below

Show $ \int \frac{dz}{{z}^{2} + z} = 0 $ by separating integrand into partial fractions and applying Cauchy's Integral theorem for multiply connected regions. For 2 paths (i) |z| = R > 1 (ii) A square with corners $ \pm 2 \pm 2i $

They also say direct use of Cauchy's Integral theorem is 'illegal' - am I right, that is because the 2 singularities make it multiply connected, so instead we use Cauchy's Integral formula?

$ \int \frac{dz}{{z}^{2} + z} = \int \frac{dz}{z - 0} - \int\frac{dz}{z-(-1)} $
Both singularities - z=0 and z=-1 - are interior, for both contours. (Is it OK to have z-0 like I have? I think I saw something about ${z}_{0} \: must\ne 0$?

$ \therefore \int \frac{dz}{{z}^{2} + z} = 2\pi i - 2\pi i = 0$ for both paths. Is that correct?

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I wondered if I could do it without p/fractions, so for path(i) set $ z=Re^{i\theta}, dz=iRe^{i\theta} d\theta $

I simplified that to $ i\oint\frac{d\theta}{Re^{i\theta}+1} $ ... would appreciate a hint for the next step?
(I think I shouldn't have that R, just $\theta - 0 \le \theta \le 2\pi $ ?)

 

[Ackbach]

(Wish there was a solutions manual...). Please check my workings below

Show $ \int \frac{dz}{{z}^{2} + z} = 0 $ by separating integrand into partial fractions and applying Cauchy's Integral theorem for multiply connected regions. For 2 paths (i) |z| = R > 1 (ii) A square with corners $ \pm 2 \pm 2i $

They also say direct use of Cauchy's Integral theorem is 'illegal' - am I right, that is because the 2 singularities make it multiply connected, so instead we use Cauchy's Integral formula?

Right. However, you could use the theorem indirectly by modifying the given contours in such a way that you can retain the original values. Could you see how to do that?

$ \int \frac{dz}{{z}^{2} + z} = \int \frac{dz}{z - 0} - \int\frac{dz}{z-(-1)} $
Both singularities - z=0 and z=-1 - are interior, for both contours. (Is it OK to have z-0 like I have? I think I saw something about ${z}_{0} \: must\ne 0$?

That should be fine. I don't think $z_0\not=0$ is required. The statement of the theorem mentions nothing about the value of $z_0$.

$ \therefore \int \frac{dz}{{z}^{2} + z} = 2\pi i - 2\pi i = 0$ for both paths. Is that correct?

Looks good to me!

--------------------------------
I wondered if I could do it without p/fractions, so for path(i) set $ z=Re^{i\theta}, dz=iRe^{i\theta} d\theta $

I simplified that to $ i\oint\frac{d\theta}{Re^{i\theta}+1} $ ... would appreciate a hint for the next step?
(I think I shouldn't have that R, just $\theta - 0 \le \theta \le 2\pi $ ?)

Maybe. I'm not entirely sure where to go next.

 

[ognik]

Right. However, you could use the theorem indirectly by modifying the given contours in such a way that you can retain the original values. Could you see how to do that?

I would deform the R>1 disc with a thin intrusion from the left (looks a bit like a left-facing pacman :-)), that left both singularities exterior to the now simply connected region. This creates a new 'internal' path, from the beginning to the end of the break in C, same direction, I will call that C₂

Then, by Cauchy's Integral Theorem, $ \int_{C}^{} f(z)\,dz = \int_{C_2}^{} f(z)\,dz $,
IE $ \int_{C}^{} f(z)\,dz - \int_{C_2}^{} f(z)\,dz = 0$
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Can anyone suggest a next step with my earlier attempt at using $ Re^{i\theta} $? I considered using a power series, but that included R and theta, and didn't converge...