If $f(z)$ is not an entire function, then $g(z) = (f(z))^2$ cannot be an entire function.”

From this statement why would saying "Let $f(z) = \sqrt z$, assuming that it is a branch that takes $−1$ to $i$. Then it is not analytic on the branch cut, but $g(z) = (f(z))^2 = z$ is obviously an entire function." Not be a proper example to disprove this statement/ be a incorrect counterexample?

I understand that for $\sqrt z$ to be a incorrect counterexample means it is not analytic on the branch -1 to I, therefore not entire.

However I thought $\sqrt z$ would be analytic throughout the whole branch cut from -1 to i. Since on the complex plane,

$\lim_{z \to -1} \sqrt z$ exists as it would approach $i$ in that case, and in the other case $\lim_{z \to i} \sqrt z$ it approaches $\sqrt i$ meaning that at least on that branch cut in the complex plane it is continuous everywhere, therefore the partial derivatives exist everywhere on the branch cut meaning it is analytic, but this assumption seems to be incorrect. What am I missing or not understanding here.

Precisely, I don't understand why $\sqrt z$ would not be a proper counterexample to the statement.

  • $\begingroup$ And maybe I’m misunderstanding your branch cut, but $\lim_{z\to i }$ seems completely irrelevant if the negative numbers are your branch cut. $\endgroup$ May 30 at 5:38
  • $\begingroup$ (Comment edited/re-posted.) You are wrong that $\lim_{z\to-1}\sqrt{z}$ exists. It is different when approached from the upper half-plane and the lower half-plane. $\endgroup$ May 30 at 5:46

Your counterexample is essentially correct, although when one talks about a "branch" of the square root function, one generally means an open set $U \subseteq \mathbb{C}$ together with a holomorphic function $f : U \to \mathbb{C}$ such that $(f(z))^2 = z$ for all $z \in U$. In this case, that is not what we want.

In fact, all you need to know is that for all $z \in \mathbb{C}$, there exists a $p \in \mathbb{C}$ such that $p^2 = z$. Then there must be some "square root function" $f$ which satisfies $(f(z))^2 = z$ for all $z$. But this function cannot be entire because it cannot be differentiable at $0$, even though its square is entire.

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    $\begingroup$ Let me add that if $f$ is continuous and $f^{2}$ is entire then $f$ is entire. $\endgroup$ May 30 at 4:45
  • $\begingroup$ @KaviRamaMurthy That's good to know and quite interesting, as it means that under Brouwer-style intuitionism, a function's square being entire would make the function itself entire (at least assuming the proof is constructive). Is there a nice proof of this? I'm not familiar with the theorem. $\endgroup$ May 30 at 4:46
  • $\begingroup$ That is quite easy to prove. Write down the definition of the derivative of $f^{2}$, factor $f(z+h)^{2}-(f(z))^{2}$; you will see that $f$ is differentiable at $z$ if $f(z) \neq 0$. Now use the fact that zeros of $f$ are isolated to show that $f$ has a removable singularity at $z$ if $f(z)=0$. $\endgroup$ May 30 at 4:52
  • $\begingroup$ How about why it wouldn't be analytic on the branch cut -1 to i, why would it not be considered analytic on that cut? If the function is continuous on a branch cut wouldn't that mean that it must also be analytic on the cut? I'm just not quite understanding why this particular example wouldn't work as a counterexample to the statement. $\endgroup$ May 30 at 4:59
  • $\begingroup$ More generally, if $f$ is continuous in an open set $U$, and analytic except possibly on a straight line segment, then $f$ is analytic in $U$. This is easy to prove using Morera's theorem. $\endgroup$ May 30 at 5:33

Your example IS a counterexample to the statement above. Yes, it is possible to have a function that cannot be made entire by analytic extension such that the square is an entire function. The square root is precisely such an example.

We can make this more concrete. We will define "a" square root function to be based on the polar representation of each point in $\mathbb{C}$; i.e., it is well known that given a complex number $z\not=0$ there exist unique $r\in\mathbb{R}^+$ and $\theta\in[-\pi,\pi)$, such that $z=re^{i\theta}$. Hence, we define $f(0)=0$ and for $z=re^{i\theta}\not=0$,


We see that $f(z)$ is defined and exists for every value in $\mathbb{C}$. Moreover, $f(z)^2=z$ for every value in $\mathbb{C}$. However, $f(z)$ (while defined at every negative real value) is not continuous at any value on the negative real axis. This is because a limit from the 2nd quadrant down towards a point and a limit from the 3rd quadrant up towards a point, will not agree. Explicitly, if $a\in(-\infty,0)$, then

$$\lim_{\epsilon\to 0}f(a+\epsilon i)=\lim_{\epsilon\to 0}\sqrt{-a-\delta_\epsilon}e^{i(\pi-\gamma_\epsilon)/2} = \sqrt{-a}e^{i(\pi/2)} = i\sqrt{-a} $$


$$\lim_{\epsilon\to 0}f(a-\epsilon i)=\lim_{\epsilon\to 0}\sqrt{-a-\delta_\epsilon}e^{i(-\pi+\gamma_\epsilon)/2} = \sqrt{-a}e^{i(-\pi/2)} = -i\sqrt{-a}$$

These two values are not the same value and hence no general limit can exist at the value of $a$. This means $f$ fails to be continuous at $a$ and means that $f$ fails to be analytic at $a$. Everywhere, (besides $z=0$ and the explicitly chosen interval endpoints of $[-\pi,\pi)$) the function is analytic. Moreover we could have chosen any interval we wanted, thus creating a new "branch cut"; hence there are many possible square root functions that can be defined.


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