Show $f$ is analytic if $f^8$ is analytic

Solution 1:

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 $f(z)$.