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Ok, so I have the following problem that I am working on. It says to evaluate $$\int \frac{z}{(z-1)(z-2)}dz$$ where C are given by \begin{align} a)& \ \ C:\lvert z \rvert=\frac12\\ b)& \ \ C:\vert z+1 \rvert=1\\ c)& \ \ C:\lvert z-1 \rvert=\frac12\\ d)& \ \ C:\lvert z \rvert=4 \end{align}

So, my first thought was to use the Cauchy Integral Formula but after graphing

$a)$, which has zeroes at $1$ and $2$, which both lie outside the graph of the circle given by $a)$ so I believe then that $a)$ would be $0$

For $b)$ I have a circle centered at (-1,0) with radius of 1, which means that I would need to use Cauchy's Formula.

For $c)$ it is a circle centered at (1,0) with radius of $\frac12$

And for $d)$ I have a circle centered at the origin with radis of 4 so both zeroes would work.

Is my thinking correct?? If not, can someone help with understanding the problem.

Micah
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cele
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  • a) is right. For th next, $\lvert z - a\rvert = r$ defines a circle with centre $a$ and radius $r$, so $\lvert z-1\rvert = \frac{1}{2}$ is a circle with centre $1$ and radius $\frac{1}{2}$, not a circle centered at the origin with radius $\frac{3}{2}$. – Daniel Fischer Nov 02 '14 at 20:31
  • Ha, ok. I guess i need to refresh on circles, I havnt done them in awhile. Thanks – cele Nov 02 '14 at 20:40

1 Answers1

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So, my first thought was to use the Cauchy Integral Formula but after graphing $a)$, which has zeroes at $1$ and $2$, which both lie outside the graph of the circle given by $a)$ so I believe then that $a)$ would be $0$

Correct.

For $b)$ then, I would think that it does not exist since $C:\vert z+1 \rvert=1$n is a point at 0

I don't understand what you're saying, but the integrand function has no singularities in the given circle, the integral exists.

For $c)$ it is a circle centered at the origin, with radius $\frac32$ so I could use $z=1$ but not $z=2$ since it lies outside the circle.

No, it's a circle centered at $1$ with radius $\frac 1 2$. Use Cauchy's integral formula for an appropriate integrand.

And for $d)$ I have a circle centered at the origin with radis of 4 so both zeroes would work.

Again, I don't understand what you mean with 'both zeroes would work', but note that you can't use Cauchy's integral formula (not directly anyway). Try partial fractions and then use Cauchy's integral formula.

Note that for all $z\in \mathbb C\setminus \{1,2\}$ one has $\dfrac z{(z-1)(z-2)}=\dfrac 2{z-2}-\dfrac 1 {z-1}$.

Thus $\displaystyle \int _C\dfrac z{(z-1)(z-2)}=\int _C\dfrac 2{z-2}-\dfrac 1 {z-1}=2\cdot 2\pi i-2\pi i=2\pi i$.

The second equality is a consequence of Cauchy's integral formula and the fact that $z\mapsto 2$ and $z\mapsto 1$ are holomorphic functions inside $C$ (they are so everywhere, really) and $1$ and $2$ are inside $C$.

Git Gud
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  • do you know where I could find a good example of how to work the Cauchy Integral? – cele Nov 02 '14 at 21:02
  • MSE is a good place. Here are a couple of answers of mine in which I used Cauchy's integral formula: 1, 2. Feel free to ask anything you don't understand in any of my answers. – Git Gud Nov 02 '14 at 21:12
  • for b)how can the integral exist if z=1 and z=2 are not in the circle?? – cele Nov 02 '14 at 21:13
  • @cele With 'exists' I mean it makes sense. I'm not saying it's not zero. – Git Gud Nov 02 '14 at 21:14
  • i would claim then that a and b are both 0. For c, z+1 is the only point interior to C, but when I plug into f(z) for z=1 using Cauchy's Formula, I also get 0. Is this right? – cele Nov 02 '14 at 21:26
  • The first two are correct, c) however is wrong. What are you using as $f$ on Cauchy's integral formula? Note the second answer that I linked above, in the second bullet I address a mistake the OP made. I suspect you did the same. – Git Gud Nov 02 '14 at 21:30
  • ok, so then would $-2\pi i$ be correct? That is using $\frac {z}{(z-2)}$ as $f(z)$ – cele Nov 02 '14 at 21:36
  • @cele Yes, that's correct. – Git Gud Nov 02 '14 at 21:40
  • How do I understand what to use as my f(z)? Ive heard the term holomorphic used, and that means that it is a complex function, but Im not sure how to tell that it is holomorphic – cele Nov 02 '14 at 21:45
  • @cele Well, in this kind of problem it's easy to choose. The function must be holomorphic in the circle, so you choose according to that. Do you know the definition of holomorphicity? – Git Gud Nov 02 '14 at 22:11
  • Im not so sure. Ive looked up the definition but honestly, after looking at definitions all day, it does not seem clear to me. – cele Nov 02 '14 at 23:06
  • I'm assuming that f(z) is not just the function itself, because if it were, I would just get 0 for the answer. – cele Nov 02 '14 at 23:12
  • @cele It can't be the integrand. To use Cauchy's integral formula you need a ratio that looks like $\dfrac {g(z)}{z-z_0}$. If you take the integrand, there is no ratio. You need to adapt what you're given to the formula. Holomorphicity is just a different word for differentiability, which I'm guessing you know what it is. – Git Gud Nov 02 '14 at 23:24
  • well, the only other way i thought to do it was to take $2\pi if(1) + 2\pi i f(2)$ where $\frac {z}{z-1}$ is used for the first and $\frac {z}{z-2}$ i used for the second, therefore giving me $4\pi i$ as the result – cele Nov 02 '14 at 23:33
  • @cele That's what I told you to do, but you did it wrong. I'll help you out with this one. – Git Gud Nov 03 '14 at 00:04
  • @cele See the edit to the answer. – Git Gud Nov 03 '14 at 00:38
  • Gut, although I see that you mentioned it before, I believe that I completely looked over the partial fractions step. – cele Nov 03 '14 at 00:49