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$$\int_c \frac{z}{z^2 + 4z + 3}dz$$

where C is a circle centered at -1 with radius 2.

$$\int_c \frac{z}{z^2 + 4z + 3}dz = \int_c \frac{z}{(z+3)(z+1)}dz$$

So both singularities z = -1 and z = -3 but z = -3 is on the circle. So can we use Cauchy's formula later on?

Using partial fractions:

$$\int_c \frac{z}{(z+3)(z+1)}dz = \frac{A}{z+3} + \frac{B}{z+1} : z = A(z+1) + B(z+3)$$ and when z = -1, -1 = 2B => $B = \frac{-1}{2} $ and when z = -3 : -2A = -3 so A = $\frac{2}{3}$

so we get:

$$\frac{3}{2} \frac{1}{z+3}dz + \frac{-1}{2} \frac{1}{z+1}dz$$

so if we apply Cauchy's integral formula to both:

$$\frac{3}{2} 2 \pi i (1) + \frac{-1}{2} 2 \pi i (1) = 3 \pi i + - \pi i = 2 \pi i$$

But is that right?

Jwan622
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    To my understanding, Cauchy's formula or the residue theorem can't be used directly when the contour itself contains a singularity of the integrand. Depending on context, you usually appeal to the Cauchy principal value, or just tweak the contour somehow such that, in a limit, the tweaked contour approaches the original contour. Might be worth looking into. – PrincessEev Dec 11 '20 at 03:26
  • As a Lebesgue integral your integral does not exist. – Kavi Rama Murthy Dec 11 '20 at 05:14
  • So... what is the integral? Or is the question invalid? – Jwan622 Dec 12 '20 at 17:59

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