Verifying an anti-derivative of a complex function 
Let $f(z) = z^{1/3}$ be defined with a branch choice such that it is continuous with $(-1)^{1/3}=-1$ on the complex plane with a branch cut at $Re(z)>0$. If $f(z)$ has an anti-derivative, then we can make it by $F(r,\theta)=\displaystyle \int^{re^{i\theta}}_{-1}w^{1/3}dw$, so that $F(r,\theta)=F(z)$. Verify that $F(z)$ is the desired anti-derivative function so $f(z)$ so that it satisfies the anti-derivative theorem. (Theorem below).


I've shown that $\displaystyle \int^{re^{i\theta}}_{-1}w^{1/3}dw = \frac{3}{4}[w^{4/3}]^{re^{i\theta}}_{-1}=\frac{3}{4}((re^{i\theta})^{4/3}-1).$ So $F(r,\theta) = \displaystyle\frac{3}{4}(z^{4/3}-1)=F(z)$.
I'm unsure what is required of me to show that $F(z)$ is the desired anti-derivative function. Do we simply use the FTC? or do  I look at the multi-valuedness of $(-1)^{1/3}$?


 A: While your $z^{1/3}$ is well defined (albeit somewhat nonstandard) it remains unclear what you mean by $z^{4/3}$, and your calculations are obscure. Of course the final result qua typographical picture is what we guessed all along from our experience with the real $x^{1/3}$.
Denote the standard domain for ${\rm Log}$ and principal values of fractional powers, i.e., the complex plane with the negative real axis removed, by $\Omega$, and your domain, where the full right half plane seems to be removed, by $\Omega^-$. The standard definition of the third root is then given by
$${\rm pv}\bigl(z^{1/3}\bigr):=e^{{\rm Log}(z)/3}\qquad(z\in\Omega)\ .$$
It follows that we can write your $f$ as
$$f(z)=-e^{{\rm Log}(-z)/3}\qquad(z\in\Omega^-)\ .$$
This gives $f(-1)=-1$ and 
$$f^3(z)=-e^{{\rm Log}(-z)}=z\qquad(z\in\Omega^-)\ ,\tag{1}$$
as it should. Taking the derivative in $(1)$ we obtain
$$3f^2(z)\>f'(z)\equiv1\qquad(z\in\Omega^-)\ .$$
Use this to compute
$$\bigl( z\>f(z)\bigr)'=f(z)+{z\over 3f^2(z)}=f(z)+{f(z)\over3}\>{z\over f^3(z)}={4\over3}f(z)\ .$$
This shows that $z\,f(z)$ is almost the required primitive of $f$, so that we now can write
$$F(z)={3\over4}z\,f(z)+C\qquad(z\in\Omega^-)$$
for the general primitive. If you want $F(-1)=0$ you arrive at
$$F(z)={3\over4}\bigl(z\,f(z)-1\bigr)\ .$$
