Fundamental theorem of calculus and complex integration

Question

I am teaching myself complex integration, and unfortunately my text book has left me confused as to when I can apply the Fundamental theorem of calculus for complex integration.

Consider the following on the unit circle (centered at $0$ with radius $1$)

$$\oint cosec^{2}\left ( z \right )$$

i believe this integral to be $0$ because the antederivative is $cot(z)$ and is defined on the unit circle, and $cosec(z)$ is continuous on the unit circle.

The fact that the unit circle contains $0$, on which neither $cosec(z)$ nor $cot(z)$ is defined is irrelevant. The only thing that matters is that the statements above hold true on the path itself.

consider now

$$\oint \frac{1}{z}$$ also on the unit circle

is the sole reason the FTOC fails to apply , that the antiderivative $ln(z)$ is not well defined at $z=1$ , assuming a branch cut $[0,+infinity]$ ?

and therefore if we took the same integral on the unit circle minus ${1}$, would the integral exist and be $= 0$ ?

Thank you

Answer 1

Actually, the integral of $\csc^2{z}$ does indeed vanish over the unit circle. The reason is that the behavior of $\csc{z}$ as $z \to 0$ is

$$\csc{z} = \frac1{z} + \frac16 z + O(z^3) $$

Thus,

$$\csc^2{z} = \frac1{z^2} + \frac13 + O(z^2)$$

Because the coefficient of $1/z$ in $\csc^2{z}$ is zero, the integral about the unit circle (or any closed loop about the origin) of $\csc^2{z}$ is zero.

Answer 2

If $f,F$ are holomorphic in an open set $U$ and $F'=f$ in $U,$ then $\int_\gamma f(z)\,dz = 0$ for every closed contour in $U.$ Proof: Because $(F\circ\gamma)' (t) = f(\gamma (t))\gamma '(t),$$$\int_a^b f(\gamma (t))\gamma '(t)\,dt = F(\gamma (b)-F(\gamma (a)) = 0.$$ The OP is certainly right on this. Poles, residues, simple connectedness, Cauchy's Theorem etc. are not a factor here.