I want to compute the following :
1. $$\int_{\partial D_{2}(0)} \frac{e^{z}dz}{(z+1)(z-3)^{2}}dz,$$ 2. $$\int_{\partial D_{2}(-2i)}\frac{dz}{z^{2}+1} $$ 3. $$\int_{\partial D_{2}(0)} \frac{\sin z}{z+i} dz $$ 4. $$\int_{\partial D_{1}(0)} \frac{e^zdz}{(z-2)^3} dz,$$ where $D_{r}(c)$ denotes a disc with radius $r$ and center $c$.

(I will not write the curve in the integral )

  1. $$\int \frac{\frac{e^{z}}{z+1}}{(z-3)^{2}}=2\pi i f'''(3) = 2\pi i \frac{e^{3}(3-1)}{(3+1)^{3}}= \frac{e^{3}i}{16}.$$
  2. $i$ doesn't lie in the disc, so $$\int \frac{\frac{1}{z-i}}{(z+i)}dz = 2\pi i (\frac{1}{-i-i}) = -\pi. $$
  3. $$\int \frac{\sin zdz}{z+i} = 2\pi i \frac{\sin(i)}{-i} = -\pi i(\frac{e^{2}-2}{2e}).$$
  4. $0$ because $2$ does not lie in the disc.

I will be very glad if somebody could skim my answers and tell me if they are legit. Thanks for your attention.

  • $\begingroup$ You are missing a factorial in 1. See Cauchy differentiation formula on wiki. Besides, you did not compute the derivative correctly. $\endgroup$ – Sasha Nov 22 '11 at 17:04
  • $\begingroup$ I assume the rest are correct. Thanks. $\endgroup$ – Tashi Nov 22 '11 at 17:15

$(1)\,$ The only pole of the function in $\,\{z\;:\;|z|<2\}\,$ is $\,z=-1\,$ , so: $$Res_{z=-1}(f)=\lim_{z\to -1}\frac{e^z}{(z-3)^2}=\frac{1}{16e}$$ and $$\int_{\partial D_2(0)}\frac{e^z}{(z+1)(z-3)^2}\,dz=2\pi i\frac{1}{16e}=\frac{\pi i}{8e}$$

$(2)\,$ is fine.

$(3)\,$ Why did you divide by $\,-i\,$? Other than that it is correct, and also $\,(4)$

  • $\begingroup$ Would you care to comment on the justification of #2? Does the OP use the theorem, or the formula? $\endgroup$ – The Chaz 2.0 Oct 22 '13 at 23:15
  • $\begingroup$ (I ask because I achieved the result using partial fractions, which doesn't seem like what OP did...) $\endgroup$ – The Chaz 2.0 Oct 22 '13 at 23:21
  • 1
    $\begingroup$ @TheChaz2.0 , in (2) the OP is using Cauchy Formula $$f(a)=\frac1{2\pi i}\int\limits_\gamma \frac{f(z)}{z-a}dz$$ with the function $\;\frac1{z-i}\;$ , which is valid since this function is analytic on the integration path and the domain enclosed by it. $\endgroup$ – DonAntonio Oct 22 '13 at 23:33
  • $\begingroup$ Ah yes. Thank you! I should be learning about that soon :) $\endgroup$ – The Chaz 2.0 Oct 22 '13 at 23:40

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