Given
$$f(x,y)=\frac{2}{i(1-y)-x}$$
I have to integrate $f$ over the origin-centered circle of radius 4. I see that
$$f(x,y)=-\frac{2x+i2(1-y)}{x^2+(1-y)^2}$$
There is a singularity in $-i$ so I use the residues theorem. I also know I can integrate over any other closed curve which contains the singularity and obtaining the same result.
Is it the same if I integrate the function:
$$g(z)=-\frac{2z}{|z|^2}$$
over the unit circle?
How can i find the residues?
What does it mean to find the primitive of a complex function?
Please, suggest some resource to study this topic. Thank you.
By setting $z=(x,0)$, we get that $$f\left ( x,0 \right )=\frac{2}{i-z}=f\left ( z \right )$$
We see that the function has a pole at $z=i$, thus $$\oint_{\left | z=4 \right |} f\left ( z \right )dz=2 \pi i Res_{z=i} f\left ( z \right )$$ Knowing that we can use the following formula to calculate the residue at a first order pole : $$Res_{z=z_k}\frac{f\left ( z \right )}{g\left ( z \right )}=\lim _{z\rightarrow z_k}\frac{f\left ( z \right )}{g^{'}\left ( z \right )}$$ It is now clear that $$\oint_{\left | z=4 \right |} f\left ( z \right )dz=-4 \pi i$$ Note that you can calculate the antiderivative(in the usual, real sense) only when a function has no poles(or other singularities) in $\mathbb{C}$, which is really not the case here, meaning that the integral IS dependent on the path of integration( think of non-conservative fields),