I want to determine a branch of logarithm such that $f(z)=L(z^3-2)$ is analytic at $0$. I am not really sure how to find a branch but I will explain few things I tried.
Since $z^3-2$ maps $0$ onto $-2$, what needs to be done is to find a branch of logarithm which is analytic at $-2$.So if L is a branch of logarithm, by the chain rule we have that $L(z^3-2)$ is analytic at $0$ since $z^3-2$ is analytic at $0$. I just don't know how to proceed from here. Any help will be appreciated. Thanks
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Define $g(z)=z^3-2$. The chain rule asserts that $f$ will be differentiable at each point $z$ for which $g(z)$ lies in the domain consisting of all points of the complex plane except those lying on the nonpositive real axis (the branch cut of Log ($z$)). Thus, $f$ will not be analytic at the points where $z^3-2$ is negative or zero if $L$ is the principal value of the logarithm. At the point zero, $g(z)=-2$, which is a negative real number. This means we must use a branch of the logarithm that has its branch cut not along the negative reals.
We can specify the logarithm branch $L(z)=\mbox{Log }|z|+i\arg_0(z)$ where $\arg_0(z)$ is the branch of arg(z) in the interval ($0$, $2\pi$), which solves the problem.
You can define a logarithm on any simply connected subset of $\mathbb{C}\setminus \{0\}$, so this can for instance be $\mathbb{C} \setminus \mathbb{R}^+$.
This logarithm $L$ is a holomorphic function, so composing it with the holomorphic $z \mapsto z^3-2$ gives a holomorphic function $f$. Holomorphic functions are analytic on their domain, so $f$ will be analytic at $0$ as well as at any point on which it is defined.