Let $f(z)=(z^3-1)^{1/2}$, find a branch of the logarithm that makes $f(z)$ holomorphic inside the unit disc and satsfies $f(0)=i$ Let $f(z)=(z^3-1)^{1/2}$, find a branch of the logarithm that makes $f(z)$ holomorphic inside the unit disc and satsfies $f(0)=i$
I tried factoring out a $z^3$ and writing $z^{\frac{3}{2}}e^{\frac{1}{2}\operatorname{Log}\left(1-\frac{1}{z^3}\right)}$ and then we have a branch cut when $1-1/z^3 = -p$ for $p<0$. BUT I have this $z^{\frac{3}{2}}$ term which I don't know how to deal with.
Can anyone tell me how you would approach this problem instead of my approach?
Thanks.
 A: The ideal situation is to find a branch defined on $\mathbb{C}\backslash S$, where $S$ is some ray from the origin to infinity. Let's first decompose $f$ as a composition of simpler functions. $f = (c\circ b\circ a)(z)$ where:
$$
\begin{align}
a(z) &= z^3 \\
b(z) &= z-1 \\
c(z) &= z^\frac{1}{2},\,\text{whenever it makes sense taking the square root.}
\end{align}
$$
First, $a$ sends the unit disk to itself, $b$ translates the image of $a$ to the left by $1$. So $b\circ a$ maps the unit disk ($D_1(0)$) to $D_1(-1)$.
Remember that, giving a branch of the logarithm is equivalent to giving a branch of the argument, and that
$$
\log z = \log|z| + i\arg z.
$$
So,
$$
\begin{align}
z^\frac{1}{2} :&=\exp\left(\frac{1}{2}\log z\right)\\
&= \exp\left(\frac{1}{2}\log|z|\right)\exp\left(\frac{i}{2}\arg z\right) \\
&=|z|^\frac{1}{2} \exp\left(\frac{i}{2}\arg z\right)
\end{align}
$$
Considering $\arg z \in (0,2\pi)$, for $f(0)=(-1)^\frac{1}{2} = i$ we must have:
$$
\frac{\arg (-1)}{2} = \frac{\pi}{2} \iff \arg(-1) = \pi.\qquad (\bigstar)
$$
So one of the rays you're looking for is the non-negative real axis, i.e, the branch of the logarithm defined in $\mathbb{C}\backslash [0,\infty)$.
$(\bigstar)$ In fact, we could've chosen any branch of the argument as long as the ray didn't intersect with $D_1(-1)$.
