Placement of singularities in the residue theorem

Question

Why do the singularities in Cauchy's residue theorem have to be within the contour, and why do they still count if they're not on the path of integration, like I'd suspect for real integrals? Sorry if this is basic but I haven't found it anywhere else.

Answer 1

I find homotopy to the most intuitive explanation: suppose you have two closed contours $\gamma_1$ and $\gamma_2$ and a function $f(z).$ If you can smoothly deform $\gamma_1$ into $\gamma_2$ without "crossing over" a singularity of $f,$ then $$\oint_{\gamma_1}f(z)\,dz=\oint_{\gamma_2}f(z)\,dz.$$ So, for example, if you can smoothly (continuously) deform a closed contour down to a single point (not a singularity), then it would make sense the integral would be zero, right? So in that sense, whatever singularities start out inside your contour have to stay inside that contour, or you change the value of the integral.