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Where $C$ is the circle $|z|=\frac{3}{2}$, evaluate the following integral using the Cauchy integral formula.

$$\int_{C}\frac{e^{z}}{(z^2+1)(z^{2}-4)}dz$$ Clearly the simple poles at $z=\pm i$ are the ones that are inside the circle $C$, and the simple poles at $z=\pm 2$ are outside the circle.

I use partial fractions to simplify $$\frac{e^{z}}{z^{2}-4}$$ to get it equal to $$\frac{e^{z}}{4(z-2)}-\frac{e^{z}}{4(z+2)},$$ then define $$f(z)=\frac{e^{z}}{4},$$ which gives us $$\frac{f(z)}{z-2}-\frac{f(z)}{z+2},$$ then by the C.I.F I get that $$\int_{C}\frac{e^{z}}{(z^2+1)(z^{2}-4)}dz=\frac{\pi i}{2}[e^{i}-e^{-i}].$$

Could someone tell me if this is correct, thanks!

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  • $\begingroup$ Please refer to your notes: decomposing the function is quite unnecessary to compute its residues... $\endgroup$ – Did Jan 24 '18 at 17:05
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Since the only singularities enclosed by $|z|=3/2$ are at $\pm i$, the function $f(z)=\frac{e^z}{z^2-4}$ is analytic in and on $|z|=3/2$. Now, using partial fraction expansion, we have

$$\frac{1}{z^2+1}=\frac{1/i2}{z-i}-\frac{1/i2}{z+i}$$

Hence, from Cauchy's Integral Theorem we find

$$\begin{align} \oint_{|z|=3/2}\frac{e^z}{(z^2+1)(z^2-4)}\,dz&=\frac1{i2}\oint_{|z|=3/2}\frac{f(z)}{z-i}\,dz-\frac1{i2}\oint_{|z|=3/2}\frac{f(z)}{z+i}\,d\\\\ &z=\pi (f(i)-f(-i)\\\\ &=\pi \frac{e^{i}}{-5}-\pi \frac{e^{-i}}{-5}\\\\ &=\frac\pi5(e^{-i}-e^i)\\\\ &=-\frac{i2\pi\sin(1)}5 \end{align}$$

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  • $\begingroup$ @donantonio Hello my friend. Here are a few more references to satiate you. Enjoy. "Visual Complex Analysis" by Tristan Needham, "Introductory Complex Analysis" by Richard Silverman, "Lectures on Complex Integration" by Alexander Gogolin, "Applied Mathematics" by Charles Chui, First order hold ,a FT Table, FT property table, $\endgroup$ – Mark Viola Jan 18 '18 at 14:57
  • $\begingroup$ Would the cowardly down voter care to comment? $\endgroup$ – Mark Viola Jan 24 '18 at 18:47
  • $\begingroup$ Thanks for your solution I made a simple but fatal error in my first attempt as I did the partial fraction method on the wrong term , thank you very clear $\endgroup$ – Gibberish Jan 24 '18 at 20:52
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jan 24 '18 at 21:26
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Write the integral as a sum of integrals around $\{i,-i\}$. These integrals are easily evaluated using the Cauchy integral formula: $$ \frac{1}{2\pi i}\oint_{|z-i|=1/2}\frac{1}{(z-i)}\frac{e^z}{(z+i)(z^2-4)}dz = \frac{e^i}{(i+i)(i^2-4)} \\ \frac{1}{2\pi i}\oint_{|z+i|=1/2}\frac{1}{(z+i)}\frac{e^z}{(z-i)(z^2-4)}dz=\frac{e^{-i}}{(-i-i)((-i)^2-4)} \\ % \frac{1}{2\pi i}\oint_{|z-2|=1/2}\frac{1}{(z-2)}\frac{e^z}{(z+2)(z^2+1)}dz=\frac{e^2}{(2+2)(2^2+1)} \\ % \frac{1}{2\pi i}\oint_{|z+2|=1/2}\frac{1}{(z+2)}\frac{e^z}{(z-2)(z^2+1)}dz=\frac{e^{-2}}{(-2-2)((-2)^2+1)} $$ The answer is the sum of these two integrals.

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  • $\begingroup$ The integrals do not reflect the original integral, the path of integration of which does not enclose the singularities $\;\pm2\;$ , only $\;\pm i\;$ . Thus, the original integral cannot be equal to the sum of the above integrals. $\endgroup$ – DonAntonio Jan 16 '18 at 8:14
  • $\begingroup$ @DonAntonio : Thank you. I hadn't read carefully enough, and have corrected that now. $\endgroup$ – DisintegratingByParts Jan 16 '18 at 10:38

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