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Evaluate the integral

$$\int_{|z+1|=2} \frac{z^2}{4-z^2}dz$$

Solution : So $|z+1|=2$ is the circle of radius 2 centered at -1. Now inside this circle $\frac{z^2}{4-z^2}$ is analytic except for a simple pole at $z=-2$. We can write $f(z)=\frac{\phi (z)}{z+2}$ and $\phi (z)=\frac{z^2}{z-2}$. Therefore residue at -2 is -1. So the integral is $-2 \pi i$.

But the answer on the book is $2 \pi i$ is there an error in my calculation? Help please!

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1 Answer 1

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seriously? :)

$$\frac{z^2}{4-z^2} = \frac{z^2}{(2-z)(2+z)} = -\frac{z^2}{(z-2)(z+2)}$$

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  • $\begingroup$ Hi can you please put me out of my misery who is right? $\endgroup$ Jul 15, 2015 at 9:22
  • $\begingroup$ Your mistake in your working occurs where you define $\phi(z)=\frac{z^2}{z-2}$, when it should be the negative of that function: $\phi(z) = \frac{z^2}{2-z}$. Both john and the answers are right. Sorry. $\endgroup$ Jul 15, 2015 at 9:59
  • $\begingroup$ The book. You forgot a minus as you can see above. $\endgroup$
    – john
    Jul 15, 2015 at 10:00

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