Can the complex square root of $z\sin z$ be defined in a neighborhood of the origin? (I.e., including the origin) Edit: on a second thought, I don't think it's possible since
$$ f(z) = \sqrt {z\sin z} = e^{\large \frac{1}{2} \log z}e^{\large \frac{1}{2} \log\sin z}$$
$$e^{\large \frac{1}{2} (\ln|z| + iArg(z))}e^{\large \frac{1}{2} (\ln|\sin(z)| +iArg(sin(z))}$$
is undefined at the origin -- we'd get a $ln|sin(0)| = ln|0|$ factor.  What do you think?  Thanks,
The problem statement is:
Consider the function $ f(z) = \sqrt {z\sin z}$ . Can $f(z)$ be defined near the origin as a single valued analytic function?
What will be the radius of convergence of the power series expansion of $f$ around $z=0$? 
My thoughts:
It is tempting to say "no", since $z=0$ seems like a branch point of $f(z)$.
But a closer look at the function 
$$ f(z) = \sqrt {z\sin z} = e^{\large \frac{1}{2} \log z}e^{\large \frac{1}{2} \log\sin z}$$
shows that we need to pick two branch cuts.  And if we choose both cuts to  be $\mathbb R^- \cup \{0\}$, i.e., choose the principal branch of the logarithm for both factors, then each factor is discontinuous and not defined on the negative real axis (plus the origin), and each jumps by a factor of $e^{\frac{1}{2} 2 \pi i} = e^{i\pi} = -1$.  
However, considering both factors together, we get a total "jump" of $e^{2\pi i} =1$, and I think that now $f(z)$ has been made single-valued and analytic on the negative real axis, including the origin, and thus we get an analytic continuation onto the negative axis and the origin, so that $f(z)$ can in fact be defined near the origin as a single-valued analytic function.
What do you think?
This seems a bit too generous, though.  I feel like I have claimed that this function is entire, which I am almost certain cannot be true...because of the complex logarithm used to define $f(z)$.
Any ideas are welcome.
Thanks,
 A: HINT:
$$z \sin z = z^2 \left( 1 + (\frac{\sin z}{z} -1) \right) $$
and $\phi(z) = \frac{\sin z}{z} -1$ is entire, taking value $0$ at $0$, so one can define $\sqrt{1 + \phi(z)}$ for $z$ close enough to $0$.
A: Yes. Consider the problem in $0 < |z| < \pi$. You want
\begin{align}
        F(z) &= \exp\left\{\frac{1}{2}\log(z\sin z)\right\} \\
      &= C\exp\left\{\frac{1}{2}\int_{\pi/2}^{z}\frac{\frac{d}{dz}(w\sin w)}{w\sin w}dw\right\} \\
      &= C\exp\left\{\frac{1}{2}\int_{\pi/2}^{z}\frac{\cos w}{\sin w}+\frac{1}{w} dw\right\}.
\end{align}
The integral is any path from $\pi/2$ to $z$ that remains in $0 < |z| < \pi$.
Define
$$
      C = \sqrt{\frac{\pi}{2}\sin\frac{\pi}{2}}=\sqrt{\frac{\pi}{2}}.
$$
Then $F(\pi/2)^2=\pi/2 \sin(\pi/2)$.
The reason that the answer is "yes" is that
$$
                   \frac{1}{2}\oint_{|z|=\pi/2}\frac{\cos w}{\sin w}+\frac{1}{w}dw = 2\pi i,
$$
which ensures that the definition of $F$ does not require a branch cut. The function $F$ is holomorphic in $0 < |z| < \pi$ by its definition, and it remains bounded near $0$, which means that it has a removable singularity at $0$. The function $F$ is what you want because $F(\pi/2)=\sqrt{\pi/2\sin(\pi/2)}$, and because
$$
  \frac{d}{dz}\frac{F(z)^2}{z\sin z}
    = 0.
$$
