Consider $f(z)=\sqrt{z\sin z}$. Can $f(z)$ be defined near the origin as a single valued analytic function?

How do we choose the branch cut. The answer is here http://math.nyu.edu/student_resources/wwiki/index.php/Complex_Variables:_1999_September:_Problem_4 but this is not comprehensible

  • $\begingroup$ Yes, if you restrict "near the origin" to defined by $|z|<R$, for $R<\pi$ since there are branch point singularities at $z=n\pi$ for all non-zero integer values of $n$ (i.e. the zeros of $\sin z$). $\endgroup$
    – Mark Viola
    Jul 15, 2015 at 3:45

1 Answer 1


Write $z \sin z = z^2 g(z)$, where $g(z) = \frac{\sin(z)}{z}$. Since $g$ has a removable singularity at $0$ by defining $g(0) = 1$, we can define a single valued square root of $g(z)$ in a neighborhood of $0$. Thus we get $f(z) = z \,\sqrt{g(z)}$.

  • $\begingroup$ +1, but perhaps you should explicitly say "we can define a single valued square root of $g(z)$ in a neighborhood of $0$" because the completed $g$ is near $1$ when $z$ is in the neighborhood of zero. The branch cut is then defined at the negative real axis (away from $1$) but is not needed. $\endgroup$ Jul 15, 2015 at 3:36
  • $\begingroup$ The branch cuts will have to take into account where ever $g(z) = 0$. So the sensible thing would be to take a branch cut at $(-\infty,-\pi] \cup [\pi,\infty)$. $\endgroup$ Jul 15, 2015 at 3:52
  • $\begingroup$ @StephenMontgomery-Smith thanks for the answer. Can you explain more why we can define a single valued square root of $g$ when we have removable singularity. Also, for what you said above, we must move the branch cut to $(-\infty, -\pi]$ and $[\pi, \infty)$ to ensure the analyticity right? $\endgroup$
    – nerd
    Jul 15, 2015 at 14:48
  • $\begingroup$ It's not that the singularity is removable - its also that the value at that point is non-zero. Then you see that there is a neighborhood $N$ of $0$ such that $g(N)$ is in a small enough neighborhood of $1$ that the square root can be defined in a single valued manner. $\endgroup$ Jul 15, 2015 at 15:17
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    $\begingroup$ @User001 I am not taking the square root, I am taking a square root. Also, if $z$ is complex, then the square roots of $z^2$ are $\pm z$. The formula $\sqrt{z^2} = |z|$ is only valid if $z$ is real. $\endgroup$ Dec 21, 2015 at 2:08

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