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Let $\mathcal{C}$ be the circle $|z + 2| = 3$ described in the anti-clockwise (i.e. positive) sense in the complex plane. Evaluate:

$\int_{\mathcal{C}}\frac{1}{z^3(z+4)}dz$

I tried to find the partial fraction of

$\frac{1}{z^3(z+4)}=\frac{A}{z}+\frac{B}{z^2}+\frac{C}{z^3}+\frac{D}{(z+4)}$.

I got the expression

$\int_{\mathcal{C}}\frac{1}{z^3(z+4)}dz=\int_{\mathcal{C}}\frac{\frac{1}{64}}{z}+\frac{-\frac{1}{16}}{z^2}+\frac{\frac{1}{4}}{z^3}+\frac{-\frac{1}{64}}{(z+4)}dz=2\pi i\{\frac{1}{64}-\frac{1}{64}\}=0$.

Without the help of partial fraction can I able to solve the problem in less time?. Please suggest me alternative methods.

Here singularities inside the circle. How to evaluate the integral, if the singularities lies on the sircle? Please provide examples

2 Answers2

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Here we have a simple pole $z=-4$ and a pole $z=0$ of order $3$ which lie inside the circle $|z+2|=3$, then we see with residue theorem $$\operatorname{Res}_{z=-4}\frac{1}{z^3(z+4)}=\lim_{z\to-4}(z+4)\frac{1}{z^3(z+4)}=\dfrac{-1}{64}$$ $$\operatorname{Res}_{z=0}\frac{1}{z^3(z+4)}=\lim_{z\to0}\frac{1}{2!}\dfrac{d^2}{dz^2}\left(z^3\frac{1}{z^3(z+4)}\right)=\dfrac{1}{64}$$ $$\int_{\mathcal{C}}\frac{1}{z^3(z+4)}dz=2\pi i\operatorname{Res}_{z=-4,0}\frac{1}{z^3(z+4)}=2\pi i\left(\dfrac{-1}{64}+\dfrac{1}{64}\right)=\color{blue}{0}$$

yu qian
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Nosrati
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There is no singularity outside the circle $|z+2|=3$. So the Integral equals to $\int_{{|z|=R},{R\to\infty}} \frac{1}{z^3(z+1)}dz $ $\leqslant$$\int_{{|z|=R},{R\to\infty}} |\frac{1}{z^3(z+1)}||dz|$ $\leqslant$$lim_{{R\to\infty}} \frac{1}{R^3(R-1)}2\pi R$ =0

yu qian
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  • I don't understand, which theorem you are using. –  Nov 04 '17 at 09:49
  • Cauchy integral theorem. let U be an open subset of C which is simply connected, let f : U → C be a holomorphic function, and so the integral is zero over a rectifiable path in U whose start point is equal to its end point. Cauchy integral theorem.You can use the real axis to cut the area {z| |z+2|>3，|z|<R},so you get two simply connected domains (one above the real axis, one below). and sum up the integral over the two boundaries, you can find that the integral over the path(-infinity,-5)and (1,infinity) cancel. So the integral over |z+2|=3 equals the integral over |z|=R – yu qian Nov 04 '17 at 18:20