analytic nature of a function Let $f:\mathbb C\rightarrow \mathbb C$ be a continuous function.If $f^2$ and $f^3$ are analytic is it true that $f$ is analytic?
Tried it using definition of analytic function but not yielding anything.Neither found any counter examples.
 A: If $f^2 = 0$ then clearly $f=0$ is analytic.
Now assume that $f$ is not identically zero, hence neither are $f^2, f^3$.
We have $|f(z)|^2 = |f(z)^2| $ and similarly for $f^3$.
Then we have (i) $f$ is bounded on any ball and (ii) $f(z) = 0 $ iff
$f^2(z) = 0$ iff $f^3(z) = 0$.
If $f(z) \neq 0$, we have $f(z) = { f^3(z) \over f^2(z)}$ hence $f$ is analytic for all $z$ such that $f(z) \neq 0$.
Now suppose $f(z_0) = 0$. Since $f^2,f^3$ are analytic, there are integers $m_2,m_3$ and analytic functions $g_2,g_3$ such that $g_2(z_0) \neq 0$,
$g_3(z_0) \neq 0$ and we can write $f_2(z) = (z-z_0)^{m_2} g_2(z)$,
$f_3(z) = (z-z_0)^{m_3} g_3(z)$.
Since $f^3 = f \cdot f^2$, we have, for nearby $z \neq z_0$, that
$(z-z_0)^{m_3} g_3(z) = f(z) (z-z_0)^{m_2} g_2(z)$ and hence
$f(z) = (z-z_0)^{m_3-m_2} {g_3(z) \over g_2(z) }$. Since $f$ is bounded near
$z_0$ we see that $m_3 \ge m_2$, and since $f(z_0) = 0$, we see that
$m_3 > m_2$, hence $f$ is analytic in a neighbourhood of $z_0$.
Note that there is no a priori requirement that $f$ be continuous.
