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How to integrate $\int_{c}\frac{e^{2z}}{(z+1)^4}dz$ for $c:|z|=2$ and $c:|z-3|=1$?

My Attempt:

For $c:|z|=2$

$\int_{c}\frac{e^{2z}}{(z+1)^4}dz=\frac{2\pi i f^{(3)}(-1)}{3!}=\frac{2\pi i\times 2^{3} e^{-2}}{3!}=\frac{8\pi i}{3e^2}$

For $c:|z-3|=1$

$\int_{c}\frac{e^{2z}}{(z+1)^4}dz=0$ because $z=-1$ lies outside $|z-3|=1$ but I am not sure how to prove this.

Please help.

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    $\lvert -1 - 3 \rvert = 4 > 1$ – 2017-01-07
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    Your work looks correct. You're unsure how to prove that -1 lies outside the circle? – 2017-01-07
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    @Chappers, No I mean I am not sure how to prove the integral is zero. – 2017-01-07
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    It's zero because the function is analytic in the circle, but I'm curious, why did you think it's zero? :) – 2017-01-07
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    @benji, because $z=-1$ lies outside this circle. Is that the answer or do I have to prove that this means the integral is zero? – 2017-01-07
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    The only singularity is at $z=-1$, outside the contour, so the function is analytic on the interior of the contour...? – 2017-01-07
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    @Chappers, What I mean is the answer just to say that $z=−1$ lies outside the circle or do I have to prove the theorem? – 2017-01-07

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