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 $f(z) =0$?


2 Answers 2


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.

  • $\begingroup$ so in your b> part : zeros of $f^2$ are same as the zeros of $f$ and zeros of $f$ are isolated (as zeros of $f^2$ are isolated ). so $f$ is analytic possibly at these zeros and then use riemanns thrm $\endgroup$
    – jim
    Feb 18, 2013 at 14:09
  • $\begingroup$ Dear jim: yes, precisely. $\endgroup$ Feb 18, 2013 at 14:12
  • $\begingroup$ sorry for not accepting your answer i upvoted it thanks for the explanation $\endgroup$
    – jim
    Feb 18, 2013 at 14:13
  • 2
    $\begingroup$ +1 for using the derivative wrt $\,z\,$ notation, which I didn't want to use because I thought it'd make things harder to understand to the OP, but no doubt renders a neater proof. $\endgroup$
    – DonAntonio
    Feb 18, 2013 at 14:26
  • 1
    $\begingroup$ Dear DonAntonio, I am sure you knew Wirtinger's notation. The decision to use more advanced notation/concepts is always a very problematic one and your choice is very reasonable. I find it optimal that readers now may select themselves the version they are more comfortable with. $\endgroup$ Feb 18, 2013 at 14:50


$$f(x,y)=u(x,y)+iv(x,y)=:u+iv\Longrightarrow f^2=u^2-v^2+2uvi$$

Since $\,f^2\,$ is analytic then the Cauchy-Riemann equations apply here:




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 $\,f=u+iv\,$) ?

  • $\begingroup$ if $u$ and $v$ are both not $0$ then $Ax=0$ has nonzero solution which implies $\det(A) =0\implies (u_x-v_y)^2+(u_y+v_x)^2=0$ from this cr equation follows. here $A= \begin{bmatrix} u_x-v_y & u_y+v_x \\ u_y+v_x & u_x-v_y \\ \end{bmatrix}$ $\endgroup$
    – jim
    Feb 18, 2013 at 13:45
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    $\begingroup$ The only delicate point is precisely to prove analyticity at the points where $u=v=0$, i.e. at the points where $f$ vanishes. [And that is the reason why I decided to answer this question: in order that users have a possible solution also for that case on record. I also wanted to advertise the power of Wirtinger's notation] Anyway, +1 for DonAntonio. $\endgroup$ Feb 18, 2013 at 13:56
  • $\begingroup$ Hi, I also have the same question so don't want to ask it again. Wondering how you "take it from here now". I dont get any cancellation from adding or subtracting the two equations so how can I deduce the terms in the parentheses are zero? $\endgroup$
    – user24907
    Apr 18, 2018 at 20:37
  • $\begingroup$ I think.. this is wrong answer because we don't know $u, v$ are partially differentiable fucntion. we just know that $f^2$ is analytic $\endgroup$
    – hew
    Jul 19, 2019 at 7:55
  • $\begingroup$ @hew You seem to be missing the point: never in the answer I did assume $\;f\;$ is analytic: I only used the given data that $\;f^2\;$ is analytic and thus the CR equations apply to it...and from this I deduce equations from which we can have the CR equations for $\;f\;$ . $\endgroup$
    – DonAntonio
    Jul 19, 2019 at 8:30

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