so i've asked a lot of questions over the last couple of days regarding this topic, lets see if it's all come to fruition and i've made sense of this entire thing.
the question given is
Let $f(z) = \frac{e^{z}}{(z+1)(z+3)}$. Calculate the following integrals using cauchy formula
- $\int_{S_{4}^{+}(i)}f(z)dz$
- $\int_{S_{3}^{+}(1)}f(z)dz$
- $\int_{S_{3}^{+}(-1)}f(z)dz$
Where $S_{r}^{+}(a)$ is the positively orientated circle of radius r around the point a. So for question 1. $S_{4}^{+}(i)$ is the circle centered at i of radius 4....(sorry everyone)
Now, intuitively, The first and third questions have the same singularities within the contour, namely -1 and -3, so they should have the same answer...yes?
Next, i'm using cauchy's Formula ie $$g(z) = \frac{1}{2 \pi i}\int_{\gamma}\frac{g(w)}{w-z}dw$$ where $g(w)$ is holomorphic on the interior of $\gamma$
Further, following the advice of trancelocation, i can split the function up in order to help with the two singularities in my contour. so, let 1. $$f(z) = \int_{S_{4}^{+}(i)} \frac{e^{z}}{(z+1)(z+3)} dz = \frac{1}{2}\int_{S_{4}^{+}(i)}\left[ \frac{e^{z}}{z+1}-\frac{e^{z}}{z+3}\right] dz$$ $$=\frac{1}{2} \int_{S_{4}^{+}(i)}\frac{e^{z}}{z+1} dz - \frac{1}{2}\int_{S_{4}^{+}(i)}\frac{e^{z}}{z+3} dz $$ Now in the above two integrals i take $g(w) = e^{w}$ which is holomorphic on $S_{4}^{+}(i)$, and to co-incide with the formula i take $z = -1$ and $-3$ then $$f(z) = \frac{1}{2}\left[2\pi i ~g(-1) - 2 \pi i ~g(-3)\right] = \frac{2 \pi i }{2}\left[g(-1) - g(-3)\right] = \pi i \left[ e^{-1}-e^{-3}\right]$$
For the second one, $$g(z)=\frac{1}{2 \pi i}\int_{S_{3}^{+}(-1)}\frac{g(w)}{w-z}dw$$ i choose $$g(w) = \frac{e^w}{w+3}$$ as that makes $g(w)$ holomorphic on $S_{3}^{+}(1)$ then $z=-1$ this gives $$\int_{S_{3}^{+}(-1)}\frac{e^{z}}{(z+1)(z+3)}dz=2 \pi i~g(-1)=\int_{S_{3}^{+}(-1)}\frac{g(w)}{w+1}dw = 2 \pi i * \frac{1}{-2e} = \frac{-\pi i}{e}$$
either way, if you could check the maths to make sure that it's right i'd be greatly appreciative, i feel like im getting the hang of this i just want to make sure that i have before i move on.
Thanks for all the hard work.