For a rational number $p > 1$. We know that the function $z^p$ is holomorphic on $\mathbb{C} \setminus \mathbb{R}^-$ (excluding $z = 0$). Is there an analytic continuation of the function $z^p$ at zero ?

Thank you.


If $p\in\mathbb N$, then yes, obviously.

Otherwise, the answer is negative. So, you have $p=\frac mn$, with $m,n\in\mathbb N$, $n>1$ and $\gcd(m,n)=1$. Suppose that you there was an analytic continuation to a larger set containing $0$. Let$f$ be that continuation. Then $f^n(z)=z^m$. Therefore, $f(0)=0$. Let $a_1z+a_2z^2+\cdots$ be the Taylor series of $f$ at $0$ (there is no $a_0$ here, since $a_0=f(0)=0$). Then$$(a_1z+a_2z^2+\cdots)^n=z^m.$$But if you expand $(a_1z+a_2z^2+\cdots)^n$, you get a power series which begins with ${a_1}^nz^n$ and therefore $n=m$. This cannot happen, since $n>1$ and $\gcd(m,n)=1$.

  • $\begingroup$ Simple and precise. Your answers are always amazing! $\endgroup$ – JustDroppedIn May 10 '18 at 12:57
  • $\begingroup$ @JustDroppedIn Thank you. $\endgroup$ – José Carlos Santos May 10 '18 at 12:57

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