Evaluate $$\int_{\gamma}\frac{z^2+2z}{z^2+4}dz$$ where the contour $\gamma$ is

1.) the circle of radius $2$ centered at $2i$, traversed once anti-clockwise.

2.) the unit circle centered at the origin, traversed once anti-clickwise.

So here we would have to use partial fractions: $$1+ \frac{2z-4}{z^2+4}.$$

Then for part 1.), $\gamma(t)=2 e^{it}+2i$.

And for part 2.), $\gamma(t)= e^{it}$.

I'm not sure what to do next to evaluate the integral for part 1.) and 2.).

  • $\begingroup$ Do you know the definition of $\int _\gamma f$? $\endgroup$ – Git Gud Nov 17 '13 at 16:35
  • $\begingroup$ Would that involve the Cauchy formula? $\endgroup$ – Luis_G Nov 17 '13 at 16:37
  • $\begingroup$ Are you allowed to use residue calculus? $\endgroup$ – Spine Feast Nov 17 '13 at 16:37
  • $\begingroup$ No. Do you know that $\int _\gamma f=\int \limits_0^{2\pi}f(e^{-i\theta})\cdot (-i)e^{-i\theta}\mathrm d\theta$? $\endgroup$ – Git Gud Nov 17 '13 at 16:38
  • $\begingroup$ 'DepeHb': I think we're meant to use Cauchy's Integral Theorem. 'Git Gud': I can't say I am familiar with that definition. Is what I've done so far correct? And what would be the next logical step? $\endgroup$ – Luis_G Nov 17 '13 at 16:50

$(1)$ Apply the residue theorem. $\int_{\gamma}\frac{z^2+2z}{(z-2i)(z+2i)}dz=2i\pi(\sum res_{z=z_k})$. Define $z_0:=z+2i.$ Thus, $res_{z=z_0}f(z)=\frac{z_k^2+2z_k}{2z_k}$, for $k=0$.

$(2)$ Notice that none of your singluar points are in your contour $\Rightarrow $$\int_{|z|=1}\frac{z^2+2z}{(z-2i)(z+2i)}dz=0$, by Cauchy Integral Formula.


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