This is from Gamelin's book on Complex Analysis.

Problem: Show that if $f(z)$ is continuous on a domain $D$, and if $f(z)^8$ is analytic on $D$, then $f(z)$ is analytic on $D$. (I assume the intention is that $D$ is nice, i.e. open and connected)

I'm not exactly sure how to approach this. I'm guessing it has something to do with zeroes of $f$. For example, around non-zeroes, $f$ is analytic since $z^{1/8}$ is nonzero in a neighborhood. Or something along those lines. The details are eluding me.

Any help would be appreciated!

  • 1
    $\begingroup$ It's true that you can prove a function analytic by proving that every point has a neighborhood in which it's analytic. So you can immediately reduce to the case of a small disk around a zero of $f$. Can you then show that the order of the zero of $f(z)^8$ there must be a multiple of $8$? $\endgroup$ – Greg Martin Jun 11 '14 at 23:15
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    $\begingroup$ Does this work? Say $z_0$ is a zero of $f$. We have $f(z)^8 = (z-z_0)^n g(z)$ for some $n$, where $g$ is analytic. Then $f = (z-z_0)^{n/8} g^{1/8}$, where $g^{1/8}$ is analytic (since $z_0$ isn't a zero of $g$). Since $f$ is continuous, the limit as $z\to z_0$ of both sides must be $0$, but this isn't true if $n/8$ isn't an integer. $\endgroup$ – Sameer Kailasa Jun 11 '14 at 23:23

Suppose $f(z) \neq 0$. Then $${{f(z+h)^8 - f(z)^8}\over{h}} = {{f(z+h) - f(z)}\over{h}}\left(f(z+h)^7 + f(z+h)^6f(z) + \dots + f(z)^7\right).$$ Using continuity, we see the expression in parentheses is nonzero for small $h$ and has limit $8f(z)^7$ as $h \to 0$. Dividing by this expression and taking limits, we get $${{\left(f(z)^8\right)'}\over{8f(z)^7}} = f'(z).$$ If $f(z)$ has some zeros in the domain, then as $f(z)^8$ is holomorphic so its zeros are isolated as well as those of $f(z)$, since zeroes of $f(z)$ and $f(z)^8$ are the same set. Now $f(z)^8$ is locally bounded at those points so $f(z)$ is also locally bounded at those points. Applying removable singularity, we can conclude that $f(z)$ is also holomorphic at all zeros of $f(z)$.

  • $\begingroup$ So in the last paragraph, you know that $f(z)$ is holomorphic on $D\setminus f^{-1}(0)$, and so now you just have to show it's also holomorphic at its zeroes? Can't we directly say that $f$ is continuous, therefore it's bounded at those points? $\endgroup$ – Jack M Jun 13 '14 at 8:19

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