# Line Integral That's Not Working Out

1. Dec 9, 2011

### Poopsilon

Let $\alpha$ 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.

$$\int_\alpha \frac{z^7 -1}{z^6 - z^2}dz = \int_\alpha zdz - \int_\alpha \frac{1}{z^2}dz +\frac{i}{2}\int_\alpha \frac{z-i}{z^2 - i}dz +\frac{1}{2}\int_\alpha \frac{1-i}{z^2 + i}dz$$

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 $\alpha$, thus applying Cauchy's Integral Formula gives:

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

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

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

Thus adding up these four integrals gives:

$$\frac{-\pi i + \sqrt{2}\pi + \pi}{2\sqrt{2}}$$

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

Last edited: Dec 10, 2011
2. Dec 10, 2011

### susskind_leon

3. Dec 10, 2011

### Poopsilon

Shoot I actually wrote down the original integral wrong in my original post, either way though my decomposition was still screwed up, so here is the new decomposition, verified by wolfram:

$$\int_\alpha \frac{z^7 +1}{z^6 + z^2}dz = \int_\alpha zdz + \int_\alpha \frac{1}{z^2}dz +\frac{i}{2}\int_\alpha \frac{z+i}{z^2 - i}dz -\frac{1}{2}\int_\alpha \frac{iz + 1}{z^2 + i}dz$$

Unfortunately I'm still getting the wrong answer at the end! The new 3rd and fourth integrals are:

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

and

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

Which when added all up gives me $\frac{-\pi}{2}$. Man this integral is driving me a little bit nuts.

Last edited: Dec 10, 2011
4. Dec 10, 2011

### susskind_leon

5. Dec 10, 2011

### Poopsilon

The integral was written wrong in my original post, it should be $\int_\alpha \frac{z^7 +1}{z^6 + z^2}dz$. You still have the old integral in Wolfram.