Suppose an entire function $f$ exists with the properties: $$f(z) = f(z+1)$$  $$f(z) = f(z+i)$$ For all $z \in \mathbb{C}$. Show that $f$ is a constant.  I think I must inspect the unit square and show that it characterizes the whole function (rigorously i suppose), and show that $f$ is bounded in that unit square and hence $f$ is bounded, and by Liouville's theorem $f$ is then constant.   I am not sure how to show f is bounded in the unit square, is it possible for an analytic function to be unbounded/have its absolute value get arbitrarily large at any point?
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.)
31626
Suppose that $f$ is holomorphic on $\mathbb C$ and that $\Re(f(z))\ge 0$ for all $z$. Show that $f$ is constant. [Hint: consider $e^{−f(z)}$.]   My thoughts: If $\Re(f(z))\ge 0 $ holds, then $e^{−f(z)}$ is a bounded holomorphic function (do I need to prove this or is it obvious?) So then by Liouville's Theorem $e^{−f(z)}$ is constant.   But then I'm not sure how to rigorously go from this $\exp(−f(z))$ back to $f(z)$.   Could anyone help me piece this together please?  Thanks
85733
Let $f$ be an entire function such that $f\circ f$ has no fixed points. Prove that $f$ is a translation   $$z\mapsto f(z)=z+b \qquad (b\neq 0)$$   Firstly, we prove that there exists a constant $c\in \mathbb{C}\backslash \{0,1\}$ such that  $$f(f(z))-z=c(f(z)-z) $$ applying Picard's little theorem. If $c=0$, then $f(f(z))=z$, so $f\circ f$ has a fixed point (absurd). If $c=1$, then $f(f(z))=f(z)$, so $f$ is the identity $f(z)=z$ and of course it has fixed point (absurd). Then,  $$F(z)=\frac{f(f(z))-z}{(f(z)-z)}$$ is an entire function which does not take the values 0 and 1 so, by Picard's little theorem, it must be constant.  Also, I've proved that $f'\circ f$ is a constant function. Let's see this. Differentiating  $$f(f(z))-z=c(f(z)-z) $$ we have $$f'(z)f'(f(z))-1=cf'(z)-c$$ $$f'(z)[f'(f(z))-c]=1-c$$ Again, the entire function  $$G(z)=f'(z)[f'(f(z))-c]$$ does not take the values $0$ and $1$ so, by Picard's Little Theorem, then is constant.   However, I don't know how to prove this problem. Any help would be appreciate.
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?
120428
How is the function $e^z$ the only function that is entire and has the property $f'(z) = f(z)$ and that $f(0) = 1$ but how would you exactly prove it?  where $e^z = e^x(\cos y + i\sin y)$
10437
So the question asks me to prove that an entire function with positive real parts is constant, and I was thinking that this might somehow be related to showing an entire bounded function is constant (Liouville's theorem), but are there any other theorems that might help me prove this fact?
122647
The problem is:  Suppose that $f(z)$ is continuous on a domain $D$ that contains the unit circle, and that $f(z)$ satisfies: $$\vert f(e^{i\theta})\vert \leq M \; \forall \theta \, \in [0, 2\pi )$$ and $$\bigg \vert \int _{\vert z \vert=1} f(z) \text{d}z \bigg \vert =2\pi M $$ show that there exists a $c\in \mathbb{C}$ such that $f(z)=c \bar{z}$ for all $\vert z\vert=1$.  I've tried lots of different approaches such as defining a function $g(z)=\frac{f(z)}{\bar{z}}$ and trying to show that $g'(z)=0$ so that $g$ is constant on the circle, or trying to use some bounding theorems for analytic functions to show that $g$ is constant but this requires $g$ to be analytic. Some other methods were using the Cauchy-Riemann equations but I couldn't find any reason why $g$ should be analytic.  Any help would be greatly appreciated, thanks!
68326
is it true if $f,\overline f$ are both analytic in a domain $\Omega$ then $f$ is constant? I am not able to  find out what property of holomorphic map I need to apply. please help.Thank you.  $f(z)=u(x,y)+iv(x,y)$, $\bar{f}(z)=u(x,y)-iv(x,y)$
13160
How can I prove the following assertion?    Let $f$ be a holomorphic function such that $|f|$ is a constant.  Then $f$ is constant.    Edit: The more elementary the proof, the better. I'm working my way through a complex analysis workbook, and by this exercise, the only substantial thing covered has been the Cauchy-Riemann equations.
24447
I know I'm going to use Liouville's Theorem, but my main question is why is $1/|f(z)|$ entire as well if $f$ is entire? Is this just a basic property: if $f$ is entire, then $1/f$ is entire? Thanks for the help.
94122
I have to solve this problem but i have no idea how to do ...Thanks!  Let be $f:\mathbb{C}\rightarrow \mathbb{C}$ a holomorphic function. Show that if $f\circ f=f$ then f is either constant or $f(z)=z \ \forall\ z\in\mathbb{C}$
75614
Here is the question: Let $f:\mathbb{D}\rightarrow\mathbb{C}$ be an analytic function.  Suppose that there exists $\epsilon&gt;0$ such that for every $\theta$, $|\theta|&lt;\epsilon$, and $\lim_{z\rightarrow e^{i\theta}}f(z)=0$.  Prove that $f$ is identically $0$ on $\mathbb{D}$. My thoughts:  Well, honestly, I am quite stumped here.  I was thinking I could let $f$ be analytic in some other region (can I even do this?), say the lower half plane and suppose $\lim_{z\rightarrow x}f(z)=0$ whenever $-1\leq x\leq1$, by assumption.  Maybe then I could show that $f$ is identically $0$ on the bigger region, and then consider $f$ on the intersection of those two regions, and then ultimately making my way towards using the uniqueness theorem to show what I want... but I am quite stuck on how to do that... any thoughts, suggestions, etc. are greatly appreciated!  Thank you.
59610
Let $\Bbb D$ be the open unit disc and $f:\Bbb D\to\Bbb C$ be an analytic function such that $|f(z)|\le |f(z^2)|$, for all $z\in\Bbb D$. Prove that $f$ is constant.   Here is my proof:   For any $0&lt;r&lt;1$, consider the closed disc $B=\{z\in \Bbb D:|z|\le r\}$. Then the given condition implies that $|f(z)|\le |f(z^{2^n})|$, for all $n\in\Bbb N$ and for $z\in B$. Since for $z\in B$, $|z^{2^n}|\le r^{2^n}\to 0\implies z^{2^n}\to 0$. So by continuity, $|f(z)|\le |f(0)|$, for all $z\in B$. So by the Maximum modulus theorem, $f$ is constant on $B$. By the uniqueness theorem, $f$ is constant on $\Bbb D$. Is this correct? Let me know if there is any fallacy.
32672
Suppose that $f$ is analytic on some region $R\in\mathbb{C}$. If Im$(f)$ = $k\cdot$Re$(f)$ for some nonzero constant $k\in\mathbb{C}$, then show that $f$ is constant on $R$.  I know that if $f'(z)=0$ for all $z$ in some region $R$, then $f(z)$ is constant. However, I'm not sure how to apply this to the question at hand. I also think the Cauchy-Riemann equations could be helpful, but again, I'm not sure how to apply them to the question.  Any guidance would be greatly appreciated. Thank you!
93399
I've been thinking how to prove that an analytic function $f$ is a constant if the absolute value of $f$ is a constant, but I haven't figured it out yet.  What I was thinking is to use Cauchy-Riemann equations, but it didn't work well...  If this is not true, I would like to know the counterexample...  Here is what I tried:  $$|f|=|u+iv|=\sqrt {u^2+v^2}$$  Thus $u^2+v^2$ is a constant.  (1)    $\displaystyle u\frac {\delta u}{\delta x}+v\frac {\delta v}{\delta x}=0 $  (2)    $\displaystyle u\frac {\delta u}{\delta y}+v\frac {\delta v}{\delta y}=0 $  Plug Cauchy Riemann into (2).  $$\displaystyle -u\frac {\delta v}{\delta x}+v\frac {\delta u}{\delta x}=0 $$  and I'm stuck here...
29355
I need to prove that if $f,g$ are analytic in the unit disk, and $|f|^2+|g|^2=1$ for all $z$ in the unit disk, then $f,g$ are constant.  This is an exercise question so it should not be very hard, but I don't know where to start. Any hint is appreciated.
81233
Let $f:\mathbb R \to \mathbb R$ be a function such that for any irrational number r, and any real number x we have $f(x)=f(x+r)$. Show that f is a constant function.   It's easy to see any constant function satisfies the original properti... But I don't see how to show this is the only solution.