Can the complex square root of $z\sin z$ be defined in a neighborhood of the origin? (I.e., including the origin)

Question

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,

Answer 1

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 $0$.

Answer 2

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 $$ \frac{d}{dz}\frac{F(z)^2}{z\sin z} = 0. $$