7

Let $f : \mathbb{C} \to \mathbb{C}$ be continuous. If $f^2$ and $f^3$ are analytic prove that $f$ is analytic at every point of $\mathbb{C}$.

if $f^2$ has no zero then $f=f^3/f^2$ and then it is analytic.but if $f^2$ has zero then how can I able to proceed.help me please.

user67634
  • 119

3 Answers3

5

Suppose that $f^2$ and $f^3$ have zeroes of order $n$ and $m$ at $z_0$, so that $f(z)^2=(z-z_0)^n g(z)$ and $f(z)^3=(z-z_0)^m h(z)$ near $z_0$, with $g$ and $h$ analytic functions such that $g(z_0)\neq0$ and $h(z_0)\neq0$.

Then $f(z)^6=(f(z)^2)^3=(z-z_0)^{3n} g(z)^3$ and $f(z)^6=(f(z)^3)^2=(z-z_0)^{2m} h(z)^2$. It follows that for $z$ near but different from $z_0$, $$(z-z_0)^{3n-2m}=\frac{h(z)^2}{g(z)^3}.$$ Since $hg$ is not zero near $z_0$, the right hand side here is as analytic near $z_0$ and does not vanish there, so that $3n=2m$. In particual, $m=\frac32n>n$ and the quotient $f^3/f^2$ is analytic at $z_0$.

Mariano Suárez-Álvarez
  • 135,076
1

If $f^2$ has a zero, then $f^3/f^2$ will have removable singularities at those zeros since $f^3$ will have the same zeros with at least the same multiplicity.

Clayton
  • 24,751
  • does this argument work also if the zero of $f^2$ is essential? – Ittay Weiss Apr 02 '13 at 02:54
  • 2
    what is an essential zero? – Mariano Suárez-Álvarez Apr 02 '13 at 02:54
  • Sorry, I'll clarify. By essential zero I mean a situation like Cauchy's function, having Taylor expansion identically zero, yet the function is not zero in an interval. I meant to ask what happens if $f$ has an essential zero. For entire functions, having an essential zero means the function is identically zero. This answers my doubts too. My doubts arose since this argument does not use the continuity of $f$. – Ittay Weiss Apr 02 '13 at 03:03
  • what should be the case if it has essential singularity? – user67634 Apr 02 '13 at 03:26
  • @user67634: As $f^3=f\cdot f^2$, it will not have an essential singularity but only removable singularities. – Clayton Apr 02 '13 at 03:29
1

At nonzero points this is easy.

At zero points of $f$, look at $f^6$. Since $f^6 = (f^2)^3$, it is analytic and its zeroes have multiplicity divisible by $3$. But since $f^6 = (f^3)^2$, its zeroes have multiplicity divisible by $2$. Therefore its zeroes have multiplicity divisible by $6$ and you conclude that $f$ is analytic, since $\displaystyle f = \frac {f^6} {f^2 \cdot f^3}$.

Edit: Mariano Suárez-Alvarez's answer is the same, but more detailed.

Yoni Rozenshein
  • 6,817