## Homework Statement

count integral: $$\int\limits_C\frac{e^{2z}}{z^2-4}\mbox{d}z$$ where $$C$$ is closed curve containing $$z=\pm2$$

## The Attempt at a Solution

$$\ldots=\int\limits_C\frac{e^{2z}}{(z-2)(z+2)}\mbox{d}z$$
so function has two poles $$z=\pm2$$ and integral will be
$$\ldots=\int\limits_{C_1}\frac{e^{2z}}{(z-2)(z+2)}\mbox{d}z+\int\limits_{C_2}\frac{e^{2z}}{(z-2)(z+2)}\mbox{d}z=4\pi i$$
corrrect?

No. Isn't this the same integral you already posted except now the circle encloses both singularities? How come the exp term has completed disappeared?

yeah, its the same integral. yeah, that what i have done is nonsense, it should have been
$$\ldots=\int\limits_{C_1}\frac{\frac{e^{2z}}{z-2}}{z+2}\mbox{d}z+\int\limits_{C_2}\frac{\frac{e^{2z}}{z+2}}{z-2}\mbox{d}z$$ where $$C_1$$ is around $$z=-2$$ and $$C_2$$ is around $$z=2$$ and then $$\ldots=\frac{1}{-4}\cdot2\pi ie^{-4}+\frac14\cdot2\pi ie^4=\frac12\pi i\left(e^4-e^{-4}\right)$$
ok?

Yeah that works. But apparently you already know the residue theorem, so why are you using the Cauchy integral formula. If the function is meromorphic with simple poles, as is the case here, you can obtain the integral via the residue formula by inspection.