Obtain the series expansion of $f(z)=z^{1/3}$ at z=1 such that $1^{1/3}=\frac{-1+i\sqrt{3}}{2}$
The way I've done it is the following:
I need $1^{1/3}=e^\frac{i\arg{1}}{3}=e^{i2\pi/3}$, so any branch cut that gives $\arg1=2\pi$ makes it. Then, $z^{1/3}=\sum_{n=0}^{\infty}\frac{f^{(n)}(1)}{n!}(z-1)^n$, and $f^{(n)}(1)=\frac{1}{3}·(\frac{1}{3}-1)···(\frac{1}{3}-(n-1))·1^{1/3}$, which yields $$z^{1/3}=\frac{-1+i\sqrt{3}}{2}\sum_{n=0}^{\infty}{1/3 \choose n}(z-1)^n$$
In the end, is the same as using the binomial theorem, with $z^{1/3}=(1+(z-1))^{1/3}$, and multiplying by $1^{1/3}$ (otherwise the binomial will give $1^{1/3}=1$)
I think that what I did is correct, the thing is that my teacher told us that we could also have done it with the expansion for the logarithm, using that $z^{1/3}=e^\frac{\log{z}}{3}$ (then I need $\log{1}=2\pi i$), but I don't know how is it done:
I've tried to do the following $$z^{1/3}=e^\frac{\log{z}}{3}=e^{\frac{2\pi i}{3} + \sum_{n=0}^\infty\frac{(-1)^n(z-1)^{n+1}}{3(n+1)}}\stackrel{?}{=}e^{\frac{2\pi i}{3}}\sum_{k=0}^\infty\frac{(\sum_{n=0}^\infty\frac{(-1)^n(z-1)^{n+1}}{3(n+1)})^k}{k!}$$
I know this probably isn't the best way to do it, but I would like to understand it. The teacher won't solve any questions, and I don't know how to deal with $e^{\sum_{n=0}^\infty\frac{(-1)^n(z-1)^{n+1}}{3(n+1)}}$, any help would be appreciated.