# are $f$ and $g$ bounded function?

$f$ and $g$ are entire function such that $|f^2+g^2|=1$ Then which of the following are correct?

1. $f$ and $g$ are constant.

2. $f$ and $g$ are bounded.

3. $f$ and $g$ have no zeroes on unit circle.

4. $ff'+gg'=0$

Well What I do is let $h(z)= f^2(z) + g^2(z)$ then clearly $h(z)$ is bounded entire hence constant by Liouville theorem, hence (4) is correct right?

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Your answer for 4 is correct. Can you think of any examples of (nonconstant, analytic) functions satisfying $f^2+g^2=1$? – Jonas Meyer Jun 5 '12 at 8:30
I gotta know: who is Laluvulu? – user641 Jun 5 '12 at 8:32
@Jonas:yes $f(z)=\sqrt{z}$ and $g(z)=\sqrt{1-z}$, steve, I am lalu and my pet's name is vulu ;) :) directsum of me and my pet is laluvulu :) – Un Chien Andalou Jun 5 '12 at 8:34
@Mex: Those are not entire functions. Trig identities are more useful here. – Jonas Meyer Jun 5 '12 at 8:39
@SteveD Laluvulu, beats me! – AD. Jun 5 '12 at 8:40

The first option is wrong since for instance, $f(z) = \cos(z)$ and $g(z) = \sin(z)$.
The second option is wrong since for instance, $f(z) = \cos(z)$ and $g(z) = \sin(z)$.
The third option is wrong since for instance, $f(z) = \cos(\pi z)$ and $g(z) = \sin(\pi z)$. Note that $g(1) = 0$.
The fourth option is correct since $f^2 + g^2$ is again entire and by Liouville's theorem, we have that $f^2 + g^2 = c$, where $c$ is a constant such that $\lvert c \rvert = 1$. Hence, taking the derivative gives us $ff' + gg' = 0$.
@DaveRadcliffe Yes. A constant with magnitude $1$. – user17762 Jun 5 '12 at 9:14