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!
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 $f(z)$.