Fix nonzero ω1,ω2 ∈R. Suppose that f is an entire function which satisﬁes $$f(z + ω1) = f(z + iω2) = f(z)$$ for all $z ∈\Bbb C$. Prove that f must be constant.   My first and immediate thought upon seeing this problem is to use Liouville's theorem somehow, because if I can prove that the function's $n$th derivative is entire and bounded for some fixed positive integer $c ∈\Bbb C$, then the function is constant throughout the plane.   So can I say, by the Cauchy Integral formula, that since we know that f is entire, all of its derivatives are entire? Thus, by the conditions of Liouville, everything follows from there?
Let $f: \Bbb C \to \Bbb C$ be an entire (analytic on the whole plane) function such that exists $\omega_1,\omega_2 \in \mathbb{S}^1$, linearly independent over $\Bbb R$ such that: $$f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.$$Prove that $f$ is constant.   The intuition seems clear to me, we have the three vertices of a triangle given by $0$, $\omega_1$ and $\omega_2$. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be $f(0)$, fine. Throwing values for $z$ there, I have found that $$f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.$$  I don't know how to improve the above for, say, rationals (at least). Some another ideas would be:   Checking that $f' \equiv 0$. I don't have a clue of how to do that. Write $w = a\omega_1+b\omega_2$, with $a,b \in \Bbb R$, do stuff and conclude that $f(w) = f(0)$. This approach doesn't seem good, because I only have a weak result with integers above. Finding that $f$ coincides with $f(0)$ on a set with an accumulation point. This seems also bad: the set on with $f$ coincides with $f(0)$ by which I found above is discrete.    Nothing works and this is getting annoying... And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea.  (On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.)
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Suppose we are in free space so we have no gravity so Lagrangian is equal to kinetic energy.   $$ L = T = \int_M \rho(x)(x\cdot \Omega_B)^2 d^3x $$  where $\rho$ is density, $\Omega_B$ is angular velocity bivector in reference body frame and $M$ is the rigid body.  The action principle says(if I haven't messed up something) that   $$ 0 = \frac{d}{ds}\bigg|_{s=0}\int_0^T \int_M \rho(x)[x\cdot (\Omega_B(t)+s \delta \Omega(t))]^2 d^3x $$  for every $\delta \Omega$ with zero ends i.e. $\delta \Omega(0)=\delta \Omega(T) = 0$. I should get these equations:  $$ I(\Omega_B) - \Omega \times I(\Omega_B)=0 $$  where  $$ I(\Omega) = \int_M \rho(x) (x\wedge (x\cdot \Omega)) d^3x. $$  But I can't see where I would get the time derivative of $\Omega_B$ from the action principle. Can someone tell me what I'm doing wrong? I guess there is some trouble with that $\delta \Omega$ can't be completely arbitrary. 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