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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.

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  • $\begingroup$ 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 $\frac{3}{2}$. $\endgroup$ – Daniel Fischer Nov 2 '14 at 20:31
  • $\begingroup$ Ha, ok. I guess i need to refresh on circles, I havnt done them in awhile. Thanks $\endgroup$ – cele Nov 2 '14 at 20:40
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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 $C$.

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  • $\begingroup$ do you know where I could find a good example of how to work the Cauchy Integral? $\endgroup$ – cele Nov 2 '14 at 21:02
  • $\begingroup$ 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. $\endgroup$ – Git Gud Nov 2 '14 at 21:12
  • $\begingroup$ for b)how can the integral exist if z=1 and z=2 are not in the circle?? $\endgroup$ – cele Nov 2 '14 at 21:13
  • $\begingroup$ @cele With 'exists' I mean it makes sense. I'm not saying it's not zero. $\endgroup$ – Git Gud Nov 2 '14 at 21:14
  • $\begingroup$ 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? $\endgroup$ – cele Nov 2 '14 at 21:26

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