Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

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

1 Answer 1

$(1)\,$ 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 $\,(4)$

share|improve this answer
    
Would you care to comment on the justification of #2? Does the OP use the theorem, or the formula? –  The Chaz 2.0 Oct 22 '13 at 23:15
    
(I ask because I achieved the result using partial fractions, which doesn't seem like what OP did...) –  The Chaz 2.0 Oct 22 '13 at 23:21
1  
@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. –  DonAntonio Oct 22 '13 at 23:33
    
Ah yes. Thank you! I should be learning about that soon :) –  The Chaz 2.0 Oct 22 '13 at 23:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.