If $f^2$ is an analytic function then so is $f$ I want to show the following:
If $f(z) $ is a continuous  function on a connected open subset of the complex plane and $f(z)^2$ is an analytic function, then $f(z)$ is analytic.
Clearly if $f(z) \neq 0$ then $$\frac{f(z+h)-f(z)}{h}=\frac{f(z+h)^2-f(z)^2}{h}.\frac{1}{f(z+h)+f(z)}$$
which shows $f^\prime(z) $ exists if $f(z)\neq0$.
What do I do when $f(z) =0$?
 A: Suppose
$$f(x,y)=u(x,y)+iv(x,y)=:u+iv\Longrightarrow f^2=u^2-v^2+2uvi$$
Since $\,f^2\,$ is analytic then the Cauchy-Riemann equations apply here:
$$(u^2-v^2)'_x=(2uv)'_y\;\;\;,\;\;\;(u^2-v^2)'_y=-(2uv)'_x$$
But
$$(u^2-v^2)'_x=2uu_x-2vv_x\;\;\,\;\;(2uv)'_y=2u_yv+2uv_y\\(u^2-v^2)'_y=2uu_y-2vv_y\;\;,\;\;-(2uv)'_x=-2u_xv-2uv_x$$
So equalling:
$$I\;\;\;\;uu_x-vv_x=\;\;\;\;uv_y+vu_y\Longleftrightarrow\;\, u(u_x-v_y)-v(u_y+v_x)=0\\II\;\;\;\;uu_y-vv_y\;=-uv_x-vu_x\Longleftrightarrow u(u_y+v_x)+v(u_x-v_y)=0$$
Can you take it now from here (i.e., check the CR equations for $\,f=u+iv\,$) ?
A: a) Since $f^2$ is holomorphic, the Cauchy-Riemann criterion (expressed in Wirtinger's wonderfully concise notation) $\partial f^2/\partial \bar z=2f\partial f/\partial \bar z=0$, shows that $f$ is holomorphic at all points $z$ where $f(z)\neq 0$, since there $\partial f/\partial \bar z=0$. [This is the part solved by Don Antonio in more classical notation] 
b) Since $f^2$ is holomorphic its zeros are isolated and so are those of $f$ (they are the same!)
But  $f$ is locally bounded at those potential singularities, because $f^2$ is, and so Riemann's theorem on removable singularities permits you to conclude that the singularity is bogus and that actually $f$ is holomorphic also at the zeros of $f$ .  
Conclusion:  $f$ is holomorphic everywhere on its domain.
