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In a section on the Cauchy Integration Formula in my complex analysis text, this problem is an exercise:

Evaluate$$\int_{\left|z\right|=4}{\frac{8\sin(z)}{(z-6)z^2}}\,dz$$

I'm failing to see how I can apply the formula to this problem. I've tried expanding the rational function $\frac{8}{(z-6)z^2}$ and applying partial fractions to little avail. I'm not too well versed in problems of this form to know what to attempt next. I'm failing to see how I can get this function into a form where my point of evaluation for the integration formula falls inside of the boundary $\left|z\right|=4$.

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One can write the integral as

$$\int_{|z| = 4} \frac{f(z)}{z^2} dz$$

where $f$ is the function

$$f(z) = \frac{8 \sin z}{z - 6}$$

The key point is that $f$ is holomorphic on and within the circle of radius $4$, since its only pole is at $z = 6$. Thus, the Cauchy integral formula implies that

$$\int_{|z| = 4} \frac{f(z)}{z^2} dz = 2 \pi i f'(0)$$

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  • $\begingroup$ According to my cursory reading of en.wikipedia.org/wiki/Cauchy's_integral_formula, shouldn't the RHS be $\frac{2\pi i}{2!}f''(0)$? $\endgroup$ – DanielV Mar 20 '14 at 3:23
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    $\begingroup$ @DanielV The general formulation is that $$f^{(n)}(a) = \frac{n!}{2\pi i} \oint_{\gamma} \frac{f(z)}{(z - a)^{n + 1}} dz$$ In this case, we choose $n = 1$, so only one derivative. $\endgroup$ – user61527 Mar 20 '14 at 3:23
  • $\begingroup$ Oh I missed the +1 in the denominator, thanks. Nicely explained. Is the formula correct even for $n=0$ ? $\endgroup$ – DanielV Mar 20 '14 at 3:24
  • $\begingroup$ Ah hah. This general form is what I needed! Thank you. I didn't quite grasp this from my reading, since my book uses a really nasty change of variable to introduce that form. $\endgroup$ – TamTamTam Mar 20 '14 at 3:36

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