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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
If $F(0)=1$, then $\log F(x)$ may be defined uniquely and is holomorphic in any open connected and simply connected set $S$ containing the point $x=0$ as long as this open set avoids all points $x$ for which $F(x)=0$. After all, in one such an open set - the disk around the origin - the logarithm may be calculated as the Taylor expansion $$\log F(x) = (F(x)-1) -\frac{(F(x)-1)^2}2 + \frac{(F(x)-1)^3}{3} - + \dots $$ The radius of the convergence is the minimum of the values $|x_0^{(i)}|$ where $x_0^{(i)}$ are all points such that $F(x_0^{(i)})=0$.
It doesn't matter that $\log F(0) = 0$. The logarithn is just equal to zero at the origin but because we're not calculating another logarithm of the logarithm, it doesn't hurt. Near the origin, $\log F(x)$ may be approximated by $F(x) -1$. The full series is above.
The way we define the Logarithm is the inverse of the exponential.
If $z= re^{i\theta}$,then $ln(z)=ln(r)+i\theta$.
Now , a problem that is there is, with this equation alone , logarithm is multi-valued i.e for every z there are infinitely many values for the logarithm . For $\theta$ differing by a value of $2\pi$ , one will get the same value. It isn't injective. This leads us restricting its domain so that each point(z) corresponds to only one value. This is referred to as a branch cut. Check the image from Wiki,
So, for log(z), simply remove any ray joining 0 and infinity.(Restricting complete rotation(0 to $2\pi$!). Also, see in the graph what happens. The principle log removes one particular ray, the negative real axis, I guess. This is only a convention, I guess.
Now, coming back to your question, you want log(iz)(PS:-The $iz$ only changes the labels on the axes) to be analytic in some given region. All you need to do is to remove a ray(not in the given region) to make it analytic in the given region.
Best Answer
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.