Proving/refuting the complex functions: $\sqrt{z(1-z)}$ and $\sqrt{z}\sqrt{1-z}$ are the same in $\mathbb{C}\backslash((-\infty,0]\cup[1,\infty))$

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 $\Omega$.