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I’m doing exercises about branch-cut. It really confuses me about the difference between, for example, $\sqrt{z(z-1)}$ and $\sqrt{z(1-z)}$, because the solutions are seemingly the same all the time!

It would be greatly appreciated if you could kindly tell me what I might have missed.

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When looking for branches, we look at where the function is discontinuous. This happens in the following cases. When $z(z-1)<0$ we see that $0<z<1$ is part of some branch. When $z(1-z)<0$ we see that $z>1$ is part of some branch.

Algebear
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  • How could $z(z-1)<0$? That is $\sqrt{z(z-1)}$. I’m stuck, could you please help me? –  Apr 28 '18 at 12:32
  • You're looking at cases where normally (in $\mathbb{R}$) the functions is not defined. That is, when the whole thing under the square root is strictly smaller than 0 (or with the logarithm function when the part inside the log is smaller than or equal to zero). – Algebear Apr 28 '18 at 15:38
  • Almost in every case of these kind of Complex functions, you have to reduce to a logarithmic function. Note that: $\sqrt{z(z-1)}=(z(z-1))^{\frac{1}{2}}=e^{\log((z(z-1))^{\frac{1}{2}})}=e^{\frac{1}{2}\log(z(z-1))}$, so therefore you look at the branch for $z(z-1)$. – Algebear Apr 28 '18 at 15:47
  • Thank you, so the two branches for $\sqrt{z(z-1)}$ are $(-\infty,0] $and $[1, +\infty]$. Is that right? –  Apr 28 '18 at 15:52
  • We know that $f(z)=\sqrt{z(z-1)}$ is defined $\forall z\in \mathbb{R^{-}}$ (for all $z$ on the negative real axis), because e.g. $f(0)=\sqrt{0}=0$, $f(-1)=\sqrt{(-1)*(-2)}=\sqrt{2}$ and so on. Note that for example $f(0.5)=\sqrt{-1/4}$, hence it's a trouble point (evil point). So $[0,+\infty)$ is the sufficient branch to cut away all real numbers $\geqslant0$. – Algebear Apr 28 '18 at 16:07