Question: Evaluate assuming the closed contour is traversed in the positive direction: $\int_C\frac{dz}{z^4+1}$ where $C$ is the circle $x^2+y^2=2x$
My Thoughts: Considering the circle $C:=x^2+y^2=2x$, we can complete the square to see that $C:=(x-1)^2+y^2=1$ is the circle centered at $(1,0)$ of radius $1$. Considering the integral, we see that $f(z)=\frac{1}{z^4+1}$ has $4$ poles of order $1$ located at $z=e^{i\frac{\pi k}{4}}$ where $k=1,3,5,7$. So, the only poles that will be inside our contour is when $k=1,7$ (right?)
So, to compute the integral, we just compute the residue of $f$ at $e^{i\frac{\pi}{4}}$ and $e^{i\frac{3\pi}{4}}$, add them together, and multiply that sum by $2\pi i$, right? I am just wanting to make sure I am understanding this process correctly. Moreover, we are traversing the contour in the positive direction… if we, instead, were to traverse the contour in the negative direction, we would have to multiply our answer by $-1$, right?
One last note, I didn't post my final solution, because I haven't finished computing it yet, but computing those residues doesn't look too ugly (and, since they have order $1$, we can even use the "shortcut" instead of the statement of "residue of order $n$").
Best Answer
Assuming I didn't make some algebra mistake, here is my solution: As mentioned above, $C$ is the circle centered at $(1,0)$ of radius $1$, so we only need to compute the residues of $f:=\frac{1}{z^4+1}$ when $z=e^{i\frac{\pi}{4}}$ and $z=e^{i\frac{7\pi}{4}}$. Let $I$ be the value of the integral. Then (since the poles of are order $1$), $$ I = 2\pi i(\frac{1}{4z_1^3}+\frac{1}{4z_7^3}) \\ =2\pi i(\frac{1}{4e^{i3\pi/4}}+\frac{1}{4e^{i5\pi/4}}) \\ =\frac{\pi i}{-\sqrt{2}+i\sqrt{2}}+\frac{\pi i}{-\sqrt{2}-i\sqrt{2}} \\ =-\frac{\sqrt{2}\pi i}{2} $$