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Evaluate: $$\int_C\left[\frac{6}{(z-i)^2}+\frac{2}{z-1}+1-3(z-i)^2\right]dt$$

where $C$ is the circle $|z-i|=4$ traversed once counterclockwise.

$C$ in this question is a circle so I must parameterize it, so a suitable parametrization for $C$ is $z(t)=z_0 +re^{it}$, $0 \leq t \leq\pi$, correct? I can split up the integrate into four parts. If someone can help me get started in the first part corresponding to: $$\int_C\frac{6}{(z-i)^2}dt$$

then I am sure that I will be able to do the rest. Thanks!

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I assume this is homework. If so, I suspect that you are not meant to explicitly evaluate the integral as the value can be read off directly from the expansion you have. –  copper.hat Feb 12 '13 at 6:58
@copper.hat No, it is not homework. It is practice problems I am doing. The answer is $0$ which I dont understand why? –  Q.matin Feb 12 '13 at 7:02
There are certain theorems which make this problem rather trivial. Are there any restrictions to what you can use? –  EuYu Feb 12 '13 at 7:04
@EuYu One theorem that I know I can use is $\int^b_af(z(t))z'(t)dt$ –  Q.matin Feb 12 '13 at 7:08
Try the residue theorem. (I suspect the answer should be $4 \pi i $.) –  copper.hat Feb 12 '13 at 7:14

1 Answer 1

up vote 2 down vote accepted

I am assuming you do not have access to Cauchy's integral formula. I will solve the first part of the integral directly, I will leave the rest to you as you requested.

Parameterizing $C$ as $4e^{it} + i$, we have $$\oint_C\frac{6}{(z-i)^2}\ dz = \int_0^{2\pi} \frac{24ie^{it}}{16e^{2it}}\ dt = \int_0^{2\pi}\frac{6i}{4}e^{-it}\ dt$$ The function $e^{-it}$ is holomorphic, so by the fundamental theorem of calculus, we have $$ \int_0^{2\pi}\frac{6i}{4}e^{-it}\ dt = -\frac{6}{4}e^{-it}\bigg|^{t=2\pi}_{t=0} = 0$$

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A couple of questions Euyu, why is there an $i$ in $24ie^{it}$ and how did you get that answer directly? Also, you used the $\int^b_af(z(t))z'(t)dt$ formula to get the denominator corerct? –  Q.matin Feb 12 '13 at 7:33
@Q.matin The $i$ comes from the derivative of $i + 4e^{iz}$. Remember that we not only substitute the value of the parameterization, but also multiply by it's derivative (the $z'(t)$ term) when taking the integral. The transition from the contour integral into $$\int_a^bf(z(t))z'(t)\ dt$$ is the first equality. –  EuYu Feb 12 '13 at 7:34
Ah, now I see what I did wrong. When I used the formula I ignorantly only used it once for the denominator and did not use it for the numerator as well. Thanks a lot!! –  Q.matin Feb 12 '13 at 7:38
@Q.matin You're very welcome. Try your hand with the rest of the question, if you have any problems then I will be happy to help later. –  EuYu Feb 12 '13 at 7:39
@Q.matin The $i$ disappeared at the end because we integrated $e^{-ix}$, which introduced a term of $\frac{-1}{i}$. –  EuYu Feb 12 '13 at 14:27

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