Complex fundamental theorem of calculus & $\int_γ z^n \,dz$

Question

State the complex fundamental theorem of calculus. Show that if n $\neq$ -1 is an integer and γ is a smooth closed curve not passing through z = 0, then

$$\int_γ z^n \,dz= 0$$

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I know that the theorem states that:

let F be a holomorphic and γ be a smooth curve. Then $$\int_γ F'(z) > \,dz= F(γ(b)) - F(γ(a))$$ in particular, if γ is a closed curve, then $\int_γ F' \,dz= 0$.

Then, isn't $\int_γ z^n \,dz$ obviousely equals to 0 due to the last sentence of the theorem as γ given is closed curve regardless of n$\neq$-1 being integer and curve passing through z=0?

Answer 1

No, because the statement should actually be:

Let $f$ be a holomorphic function and let $\gamma\colon[a,b]\longrightarrow D_f$ be a continuous curve. If $f$ has a primitive $F$, then$$\int_\gamma f(z)\,\mathrm dz=F\bigl(\gamma(b)\bigr)-F\bigl(\gamma(b)\bigr).$$

If $f(z)=z^n$ with $n\in\mathbb{Z}\setminus\{-1\}$, then $f$ has a primitive: $F(z)=\frac{z^{n+1}}{n+1}$. But if $n=-1$, $f$ doesn't have a primitive and, in general, the integral is not $0$. For instance, if $[a,b]=[0,2\pi]$ and $\gamma(t)=e^{it}$, $\int_\gamma z^{-1}\,\mathrm dz=2\pi i$.