If f is an entire function satisfying f(z+1)=f(z) and f(z+i)=f(z) then f must be constant throughout the complex plane. I would appreciate if someone helps me know how to prove the above statement.
826
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.)
83190
Let $f$ be an entire function s.t $f(\mathbb{C}) \cap B_R(z_0) = \varnothing$ for some $z_0$ and some $R$. Then $f$ is constant.  I guess since the image of the whole plane isn't dense then $f$ doesn't have an essential singularity at infinity. Now I have to exclude it has a polar singularity at infinity..
104711
A problem asks to show that an entire function on $\mathbb{C}$ with positive real part must be a constant. I spoke to a professor, and asked why not just use the Picard theorem. He said that we should try to aim the solution at the level of the problem, and that Picard was a little too high-powered for this problem.  How would I solve it in the absence of Picard's theorem? A related question: suppose we have an isolated singularity, near which $\Re (f)$ (alternately, $\Im (f)$) is bounded. How can we show the singularity is removable?  Yet another related problem: why is a positive harmonic function on Rn a constant? Mean value property seems not to be the way...
43334
If $f$ is non-constant and entire, prove that there exists a $z_0 \in C$ such that $f(z_0)$ is a positive real number, without appealing to Picard's theorem.  The obvious approach to this would be to modify a proof of Picard's little theorem to try and prove this, but all of the proofs I have found of little Picard's seem to be higher level than I have - I'm trying to solve this using theorems from a quarter long course in complex. I would really appreciate a hint in the right direction....
31147
Let's say I have a complex function like $$f(z)=z+e^{\frac{1}{z-1}}+\sin\left(\frac{1}{z-2}\right)$$ which is holomorphic in $\mathbb{C}\setminus \left \{ 1,2 \right \}$ Then there exist entire functions $$f_{0}(z)=z$$ $$f_{1}(z)=e^{z}$$ $$f_{2}(z)=\sin(z)$$ Such that, the equality $$f(z)=f_{0}(z)+f_{1}\left(\frac{1}{z-1}\right)+f_{2}\left(\frac{1}{z-2}\right)$$ holds for all points except for $z=1$ and $z=2$ which are essential singularities. My question is, can this be extended in general? Precisely put, let's say I have a holomorphic function $f(z)$ except for a finite number of essential singularity points $a_{1},\ldots,a_{n}$ in the complex plane. Is it true in general, that there exist $n+1$ entire functions $f_{0}(z),\ldots,f_{n}(z)$ such that: $$f_{i}(z) \text{ is holomorphic in }\mathbb{C}\setminus \left \{ a_{i} \right \} \text{ for }i=1,\ldots,n $$ $$f(z)=f_{0}(z)+f_{1}\left(\frac{1}{z-a_{1}}\right)+\ldots+f_{n}\left(\frac{1}{z-a_{n}}\right)$$
89554
Suppose that $f(x)$ is continuous on $(0, +\infty)$ such that for all $x &gt; 0$,$f(x^2) = f(x)$. Prove that $f$ is a constant function. My attempt is to show that for any point $a \neq b$ , we have $f(a)=f(b)$. But I have no idea on how to get this. Anyone can help?
115632
Question  Let $f$ be entire and suppose there exists $\lambda_1,\lambda_2 \in \mathbb{C}$ linearly independent over $\mathbb{Q}$ such that $f(z+\lambda_1)=f(z)=f(z+\lambda_2)$. Then $f$ is constant.  Attempt  A similar question was asked  but I don't think the answer is satisfactory. I want to (hopefully) show that for any $\epsilon&gt;0$ I can find integers $m_1,m_2$ such that $|m_1\lambda_1+m_2\lambda_2|&lt;\epsilon$. Then I will be able prove the result as follows:  Let $z_0\in \mathbb{C}$. Then $g(z)=f(z)-z_0$ has a zero at $z_0$ and this zero must be isolated. Consequently we can find an $\epsilon&gt;0$ such that the epsilon ball about $z_0$ contains only one zero ($z_0$) of $g(z)$. However, by the above, we can find $m_1,m_2$ so that $|m_1\lambda_1+m_2\lambda_2|&lt;\epsilon$ and by assumption $g(z_0+m_1\lambda_1+m_2\lambda_2)=g(z_0)=0$ which is only possible if $f$ had been constant.
21472
Let $f:[0,1] \to \mathbb{R}$ be a continuous function such that $f(x)=f(x^2)$ for all $x \in [0,1]$.  Any hint/idea for proving that $f$ has to be constant?
48580
Let $f$ be an entire function such that $\mid f(z) \mid \leq A \mid z \mid$ for all $z$, where $A$ is a fixed number. Show that $f(z)=a_1z$, where $a_1$ is a complex constant.
52230
Show that if $f$ is analytic in a domain $D$ and $|f(z)|$ is constant in $D$, then the function $f(z)$  is constant in $D$.     Here is what I try so far.  So we are given that $|f(z)|$ is constant in $D$, so that means $|f(z)|^2 = u^2 + v^2$ is also constant.  If either $u = 0$, or $v = 0$, then $f(z)$ is constant.  $$\frac{\partial|f(z)|^2}{\partial x} = 2u \frac{\partial u}{\partial x} + 2v \frac{\partial v}{\partial x} = 0$$  $$\frac{\partial|f(z)|^2}{\partial y} = 2u \frac{\partial u}{\partial y} + 2v \frac{\partial v}{\partial y} = -2u \frac{\partial v}{\partial y} + 2v \frac{\partial u}{\partial x} = 0 $$  So what do I do to finish this proof? Would I have to use the Cauchy Riemann equation?
54076
Let $f$ and $g$ be entire functions such that $f^n+g^n=1$, where $n\geq 3$ is an integer. Prove that $f$ and $g$ are constant.  I suppose I should somehow prove that either $f$ or $g$ is bounded so that I can apply Liouville's Theorem, but I don't see how. I tried setting the derivative of the left hand side equal to zero and work with that but that did not seem to work.
16886
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 $f(z) =0$?
98713
Let $f$ be an entire function on the complex plane.      Is the radius of convergence of $f$ around any point $z_0$ infinite? If so, why?   Thank you.
11247
Suppose $f:(a,b) \to \mathbb{R} $ satisfy $|f(x) - f(y) | \le M |x-y|^\alpha$ for some  $\alpha &gt;1$ and all $x,y \in (a,b) $. Prove that $f$ is constant on $(a,b)$.   I'm not sure which theorem should I look to prove this question. Can you guys give me a bit of hint? First of all how to prove some function $f(x)$ is constant on $(a,b)$? Just show $f'(x) = 0$?
76444
Prove if $f$ and $g$ are entire and $e^f+e^g=1$, then $f$ and $g$ are constant.  I believe the simplest way would be to use Louiville's theorem by using Pick's theorem but I am not sure on how to go about this.
6711
Let $f(z)$ be an entire function s.t $|f(z)|\leq M|z|$ ; $M&gt;0$ is a constant then $f(z)$ Is polynomial of degree 1   My try $f(z)$ is entire so it can be written as a $\sum a_n z^n$  $$|\sum a_n z^n|\leq \sum |a_n||z|^n\leq M|z|$$  So $n=1$  Seems that the move $\sum |a_n||z|^n\leq M|z|$ is problematic
86744
Let $f$ be an entire function. Suppose $f(z)$ is real if and only if $z$ is real. Prove that $f$ has at most $1$ zero. How to prove? Totally I have no idea... Please give the solution in detail since I do not understand.