If $f^2$ is analytic in $\Omega$ then so is $f$-Problem in proof Understanding

Question

Problem in understanding the following proof in my book:

If $f^2$ is analytic in $\Omega$ then so is $f$ where $f$ is continuous in domain $\Omega$

• Zeros of $f$ are isolated .

Let $z_0$ be a zero of $f$ then $f(z_0)=0\implies f^2(z_0)=0$.Now zeros of an analytic function are isolated hence zeros of $f^2$ are also isolated. And so zeros of $f$ are also isolated.Done

• $f$ has a derivative at each point at each point where it does not vanish.

Unable to understand this fact

• Using Riemann's singularity theorem we have that $f$ is analytic in $\Omega$.

I am unable to understand that how the fact that If $f$ has a removable singularity at $z_0$,then $f$ is bounded in a ngbd of $z_0$ proves that $f$ is analytic in $\Omega$.

Answer 1

Hint: Suppose $f(a) \ne 0.$ We know that

$$\lim_{h\to 0}\frac{f(a+h)^2 - f(a)^2}{h} = (f^2)'(a).$$

Write the numerator as $(f(a+h) - f(a))(f(a+h) + f(a))$ to see $f'(a) = (f^2)'(a)/(2f(a)).$