Complex integral of the function $f(z)=\dfrac{1}{z^4+1}$

Question

I must calculate this integral $$\int_C\dfrac{dz}{z^4+1}$$ , where $C$ is the circle $x^2+y^2=2x$. My result is $\int_C\dfrac{dz}{z^4+1}=-\dfrac{\pi}{\sqrt{2}}$ , but my book "A collection of problems in Complex Analysis(Dover)" give me as resul $-\dfrac{i\pi}{\sqrt{2}}$.

Answer 1

Your contour is the circle $$(x-1)^2+y^2=1$$ Your integrand has poles at $z=\pm\sqrt i, \pm i\sqrt i $.

Note that $\sqrt i=\exp(i{\pi\over4})={1\over\sqrt 2}(1+i)$. Interior of your circle only intersects $1$st and $4$th quadrant. $-\sqrt i, i\sqrt i$ are in $3$rd and $2$nd quadrant respectively. So we don't need to worry about them. The other poles are inside your circle. $$ \text{Res}_\sqrt i\left({1\over z^4+1}\right)=\text{Res}_\sqrt i\left({1\over (z^2+i)(z^2-i)}\right)={1\over2i\cdot2\sqrt i}=-{i\over 4\sqrt i}\\ \text{Res}_{-i\sqrt i}\left({1\over z^4+1}\right)=\text{Res}_{-i\sqrt i}\left({1\over (z^2+i)(z^2-i)}\right)={1\over(-2i\sqrt i)\cdot(-2 i)}=-{1\over 4\sqrt i} $$
So your integral becomes $$ -{2\pi i\over4\sqrt i}(1+i)=-{2\pi i\over{4\over\sqrt 2}(1+i)}(1+i)=-{i\pi \over\sqrt2} $$