Proving/refuting the complex functions: $\sqrt{z(1-z)}$ and $\sqrt{z}\sqrt{1-z}$ are the same in $\mathbb{C}\backslash((-\infty,0]\cup[1,\infty))$ The exercise presents both functions, where the square root is defined by using the principal logarithmic branch:
$$f(z)=\sqrt{z}=e^{\frac{1}{2}\operatorname{Log}(z)}$$
The question is:

Are both function: $\sqrt{z(1-z)}$ and $\sqrt{z}\sqrt{1-z}$ the same in $\mathbb{C}\setminus((-\infty,0]\cup[1,\infty))$?

If I take different complex numbers and plug them into the functions, I end up with the same result (I'm not able to find an example to disprove the claim). Therefore, my intuition is that the functions are the same. However, I find myself stuck proving that:
$$ \sqrt{|z(1-z)|}e^{\frac{1}{2}i(\arg_{-\pi}(z) + \arg_{-\pi}(1-z))} = \sqrt{|z(1-z)|}e^{\frac{1}{2}i\arg_{-\pi}(z(1-z))}$$
for every $z$ in the given domain. It all boils down to proving:
$$ \arg_{-\pi}(z) + \arg_{-\pi}(1-z) = \arg_{-\pi}(z(1-z))$$
Neither $z=x+iy$ nor $z=re^{\theta}$ seem to help me reach a conclusion.
I would appreciate it if someone could help in proving/disproving the claim.
 A: We first prove that both $\log(z(1-z))$ and $\log(z) + \log(1-z)$ are analytic on $\Omega = \mathbb{C} \setminus ((-\infty, 0] \cup [1, \infty))$.

*

*Since the principal branch cut is $(-\infty, 0]$, the claim is trivial for $\log(z) + \log(1-z)$.


*$\log(z(1-z))$ can only fail to be differentiable at the points satisfying $z(1-z) \in (-\infty, 0]$. However,
\begin{align*}
z(1-z) \in (-\infty, 0]
&\quad\iff\quad z^2 - z \in [0, \infty) \\
&\quad\iff\quad (z - \tfrac{1}{2})^2 \in [\tfrac{1}{4}, \infty) \\
&\quad\iff\quad z - \tfrac{1}{2} \in (-\infty, -\tfrac{1}{2}] \cup [\tfrac{1}{2}, \infty) \\
&\quad\iff\quad z \in (-\infty, 0] \cup [1, \infty).
\end{align*}
So it follows that $\log(z(1-z))$ is analytic on $\Omega$.
Now, on $\Omega$, we find that
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}z} \left( \log z + \log(1-z) - \log(z(1-z)) \right)
= \frac{1}{z} - \frac{1}{1-z} - \frac{1-2z}{z(1-z)} = 0.
\end{align*}
Since $\Omega$ is a domain, this implies that $\log z + \log(1-z) - \log(z(1-z))$ is constant. Then by plugging values to $z$, such as $z = \frac{1}{2}$, we can verify that the constant value is exactly $0$. Therefore
$$ \log z + \log(1-z) - \log(z(1-z)) = 0 $$
and hence $\sqrt{z}\sqrt{1-z} = \sqrt{z(1-z)}$ on $\Omega$.
A: $p×q$ musst never rotate over $i^2$ to avoid
$\sqrt{p}\sqrt{q}\ne\sqrt{pq}$
as the principal square root function maps
$ℂ=|ℝ|i^{(-2...2]}\to\sqrtℂ=|ℝ|i^{(-1...1]}$
and in the case of $z(1-z)$ we can be sure that at least one of both factors is positive in its real part and that in the complex number plane the factors form an isosceles triangle around apex $o=\frac12$ whose moving leftwards to $o=0$ reduces both angles and thus their sum to less than $i^2$ such that together with the generic case that both factors are real you arrive at
$\forall z\in ℂ: \sqrt{z}\sqrt{1-z}=\sqrt{z-z^2}$ for these particularly benign factors.
