complex analysis fundental theorem of caculus

Question

Can anyone please explain how $$\int \frac{1}{(z-2)^3}dz $$ evaluated about the closed continuous path $$1+3e^{i2t\pi}$$ is 0 by the fundamental theorem of calculus?

Answer 1

Generally, if $f$ is holomorphic in a domain $D \subset \mathbb C$ and has an antiderivative $F$ in $D$ (a holomorphic function $F:D \to \mathbb C$ such that $F' = f$), then

$$ \int_\gamma f(z) \, dz = F(\gamma(b)) - F(\gamma(a)) $$

for every curve $\gamma : [a, b] \to D$. If $\gamma $ is a closed curve in $D$ then $\gamma(a) = \gamma(b)$ and therefore $\int_\gamma f(z) \, dz = 0$.

In your case, $ f(z) = (z-2)^{-3} $ has the antiderivate $F(z) = -\frac 12 (z-2)^{-2}$ in $D = \mathbb C \setminus \{ 2 \}$ and $\gamma :[0,1]→ \mathbb C,\gamma(t)=1+3 e^{2\pi i t}$ is a closed curve in $D$.