Let [itex]\alpha[/itex] be circle in the complex plane centered at z=1 with radius r=3/2. I proceed by partial fraction decomposition and then use Cauchy's Integral Formula.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_\alpha \frac{z^7 -1}{z^6 - z^2}dz = \int_\alpha zdz - \int_\alpha \frac{1}{z^2}dz +\frac{i}{2}\int_\alpha \frac{z-i}{z^2 - i}dz +\frac{1}{2}\int_\alpha \frac{1-i}{z^2 + i}dz[/tex]

Now the first integral is analytic everywhere so is equal to zero, the second integral is equal to zero as well by Cauchy's Integral Formula. The third integral has only one of its two singularities inside [itex]\alpha[/itex], thus applying Cauchy's Integral Formula gives:

[tex]\frac{i}{2}\int_\alpha \frac{z-i}{z^2 - i}dz = \frac{i}{2}\int_\alpha \frac{\frac{z-i}{z+(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})}}{z-(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})}dz = \frac{i}{2}[2\pi i(\frac{(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})-i}{2(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})})] = \frac{-\pi}{2}(1-\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})[/tex]

The Fourth Integral similarly has only one of its two singularities inside [itex]\alpha[/itex], and so again applying Cauchy's Integral Formula:

[tex]\frac{1}{2}\int_\alpha \frac{1-i}{z^2 + i}dz = \frac{1}{2}\int_\alpha \frac{\frac{1-i}{z+(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})}}{z-(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})}dz = \frac{1}{2}[2\pi i(\frac{1-i}{2(\frac{1}{\sqrt{2}}-\frac{i}{\sqrt{2}})})]= \frac{\pi i}{2}(i+1)(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}) = \frac{-\pi i}{\sqrt{2}}[/tex]

Thus adding up these four integrals gives:

[tex]\frac{-\pi i + \sqrt{2}\pi + \pi}{2\sqrt{2}}[/tex]

Now according to the book the answer should be [itex]\frac{\pi i}{\sqrt{2}}[/itex], thus I have made some error, although after checking and rechecking everything I cannot find it, maybe someone could help me, thanks.

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# Homework Help: Line Integral That's Not Working Out

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