This particular question was part of a multiple choice question asked in my quiz. I contradicted other $3$ options but this one is true and I have no idea how to prove it.  Assume that $f$ is entire and satisfying $Im(f(z)) &gt; 4$ for all $z \in \mathbb{C}$. Then show that $f$ is constant.  Kindly shed some light on which result I should use to solve this problem.
122614
$f$ is an entire function, and it satisfies $f(\mathbb{C}) \subseteq \{z \in \mathbb{C} \mid \operatorname{Im} z &gt; 0\}$. Show that $f$ is constant.   I want to take advantage of the Liouville's Theorem, but I just can't figure out the relationship between its image part with its module.
66307
Suppose $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire function. Let $g:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function, which has no zeros.    I have shown that $\vert f(z) \vert \leq \vert g(z) \vert$ for all $z\in\mathbb{C}$ implies $f(z)=Cg(z)$ for some constant $C\in\mathbb{C}.$  I have to decide if this also holds if $g$ is allowed to have zeros. Help?
101791
Let $f$ be an entire function with $|f(z)|\le 100\log|z|,\forall |z|\ge 2,f(i)=2i, \text{ Then} f(1)=?$  I have no idea how to solve this one!  $g(z)={f(z)\over \log|z|}$ Then Can I say $g$ is constant by Liouville Theorem?
79968
Suppose $f$ is entire and    $$\iint_\mathbb{C}|f(z)|^2dxdy &lt; \infty$$Prove that $f\equiv 0.$   So far I have:  Suppose $f$ is bounded.  Then $f$ is constant by virtue of Liouville and so the conclusion is obvious.  Thus, assume $f$ non-bounded.  Then  $$\Big|\iint_\mathbb{C}f^2(z)dA\Big| \leq \iint_\mathbb{C}|f(z)|^2dxdy &lt; \infty$$ We can parameterize \begin{align*} \iint_\mathbb{C}f^2(z)dA &amp;= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;d\theta\;rdr \\ &amp;= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;ire^{i\theta}[-i\frac{1}{r}e^{-i\theta}]d\theta\;rdr \\ &amp;= \int_0^{\infty}\oint_{C_R}f^2(z)\;[-i\frac{1}{z}]dz\;rdr \\ &amp;= \int_0^{\infty}2\pi r dr\frac{1}{2\pi i}\oint_{C_R}\frac{f^2(z)}{z}dz \\ \\ &amp;= 2\pi f^2(0)\int_0^\infty rdr  \end{align*}  Whence f(0) = 0.  And I have no idea where to go from there.  The "entire" bit seems to suggest Liouville but I have already dealt with the bounded case.  Please advise ...
101432
I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times.  However, I can't see how to proceed.  I went down what I thought were the typical paths and just reached dead ends.  Does someone maybe have a hint?  I think that with the appropriate nudge I should be able to solve this question.  If not, I'll edit and ask if someone knows how to solve the whole thing.  Here's the question:  Let $f: \mathbf R \to \mathbf R$ be such that $f$, $f'$, $f''$, and $f'''$ exist and are continuous on $\mathbf R$ and satisfying $f(-3)=-1$, $f(0)=0=f'(0)$, and $f(3)=8$.  Prove that there exists $\xi \in (-3, 3)$ such that $f'''(\xi) \ge 1$.
63777
Let $f$ be entire, $|f(z)|\leq 3|z|^4+1$ prove $f$ is a polynomial with deggre $\leq 4$  Similar questions have been asked here, but non answer satisfies me, so I tried a way I get, I just want someone to verify this. $f^{(4)}(z_0)=\frac{4!}{2\pi i} \int_c\frac{f(z)}{(z-z_0)^5}dz$, $|f^{(4)}(z_0)|\leq 4! \frac{3|z|^4+1}{|z|^5|(1-\frac{z_0}{|z|^5})|}|z|=4!\frac{3+\frac{1}{|z|^4}}{|(1-\frac{z_0}{|z|^5})|} $, where $|z|$ comes from the length of the line (instead of $r$) leting $|z|\rightarrow \infty$ we get $|f^{(4)}(z_0)|\leq 72$ since $z_0$ is random from Liouville's theorem $f^{(4)}(z)$ is constant, it follows $f$ is a polynomial with deggre $\leq 4$
74290
I am solving previous year question paper some competitive exam. Give me some hint to solve the following problem.  Let $f$ be an entire function. Suppose for each $a \in \mathbb{R} $ there exists at least one coefficient $c_n$ in $f(z) = \sum_{n=0}^{\infty} c_n (z-a)^n$ which is zero. Then,  a) $f^{(n)}(0) = 0$ for infinitely many $n \ge 0$  b) $f^{(2n)}(0)=0$ $ \forall n \ge 0$  c) $f^{(2n+1)}(0)=0$  $ \forall n \ge 0$  d) $f^{(n)}(0) = 0$ for all sufficiently large n.  Thanks in advance.
22884
Assuming $a&lt;b$, function f(x) is continous at [a,b], and we have  $\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0$  Prove that $\exists \ x_1,x_2,x_3\text {(different from each other)}  \in(a,b)$ satisfying $f(x_1)=f(x_2)=f(x_3)=0$.  I have proved the existence of one zero point by differential mean value theorem, but have no idea about going on.  Thanks for any solution or hint.
71236
Show that if $f(z)$ is an entire function such that it takes real values on the lines $Im(z)=0$ and $Im(z)=a$ (for some $a \in \mathbb{R}_{\neq 0})$, then $f$ is periodic with period $2 i a$.    So here is my thought process: First of all by reflection principle $f(\bar{z})=\overline{f(z)}$ so if our function $f$ is given in the strip $0 \leq Im(z) \leq a$, then its values on $-a \leq Im(z) \leq a$ follow by reflection principle. Then by imposing $f(z)=f(z+2 ia)$ we can intuitively see that this will define an entire function by looking at how the strips $na \leq Im(z) \leq 2na$ seamlessly stack on top of each other. Is this logical? And how would one prove this analytically?
81335
If $f:\mathbb{C}\longrightarrow\mathbb{C}$ is an entire function such that $f(z)\neq w$ for all $z\in \mathbb{C}$ and for all $w\in [0,1]\subset \mathbb{R}$, how to prove that $f$ is constant (without using Picard's little theorem).   Any hint would be appreciated.
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.  $f(z)=u(x,y)+iv(x,y)$, $\bar{f}(z)=u(x,y)-iv(x,y)$
8166
This is from an old qualifying examination question.  If f is holomorphic in the unit disk $D$ and $|f(z)|&lt;1$ for all $z\in D$. Suppose also that $f$ has two distinct fixed points in $D$ then $f(z)=z$ for every $z\in D$  I know that I have to use the Schwarz's lemma and may be make use of Möbius transformations. I tried setting $g(z)=\phi_a\circ f\circ \phi_{-a}$. But that does not seem to work because I don't see why $|g(z)|=|z|$ for some non zero $z$.   Any helpful hints are greatly appreciated.  Edit: Of course the $a$ above is one of the fixed points.
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!
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.
47269
Suppose we have entire functions $f$ and $g$ that satisfy $e^{f(z)} + e^{g(z)} = 1 $ for all complex values $z$.  Show that $f$ and $g$ are constant.
78113
I was wondering whether anyone would be so kind as to briefly check my proof? I am supposed to prove the statement without using any theorems which would render the proof trivial.     If $\displaystyle\int_a^b f$ exists then $f$ has infinitely many points of continuity in $[a,b]$   For the sake of contradiction suppose $f$ has finitely many. It suffices to show that there is an interval $[u,v]\subset [a,b]$ over which $f$ is not Riemann integrable. Pick any $[u,b]\subset [a,b]$ such that $f$ is discontinuous everywhere in $[u,v]$. For any finite partition $D=\{x_1\cdots x_n\}$ of $[u,v]$ let:  $$s(f,D)=\sum_i (x_{i+1}-x_i)\inf_{x\in[x_{i},x_{i+1}]}f\quad\text{and}\quad S(f,D)=\sum_i (x_{i+1}-x_i)\sup_{x\in[x_{i},x_{i+1}]}f$$ We prove $\sup_D s(f,D) &lt;\inf_D S(f,D)\;(1)$. To achieve this consider an arbitrary chain: $$D_0\subset D_1\subset \cdots$$ Pick any $I_0\subset I_1\subset \cdots$ where $I_k=[u_k,v_k]$ and $I_k$ is an interval of the partition $D_k$. Without loss of generality $\bigcap_i I_i=z$. Given that $f$ is discontinuous everywhere, there exists $\epsilon&gt;0$ such that for any $\delta&gt;0$ there exists $y$ such that $|z-y|&lt;\delta$ yet $|f(z)-f(y)|\geq \epsilon$.   This immediately implies $\displaystyle\sup_{I_n} f&gt;\inf_{I_n} f$ and hence $\displaystyle\sup_{D_n} s(f,D)&lt;\inf_{D_n} S(f,D)$.   This is valid for any sequence $D_n$ so claim $(1)$ holds, contradicting the integrability of $f$.
5270
Let $f \in C[0,1]$ and $f(0)=f(1)$.   How do we prove $\exists a \in [0,1/2]$ such that $f(a)=f(a+1/2)$?  In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+\frac{1}{n})$.  For any other non-zero real $r$ (i.e not of the form $\frac{1}{n}$), there is a continuous function $f \in C[0,1]$, such that $f(0) = f(1)$ and $f(a) \neq f(a+r)$ for any $a$.  This is called the Universal Chord Theorem and is due to Paul Levy.  Note: the accepted answer answers only the first question, so please read the other answers too, and also this answer by Arturo to a different question:     This is being repurposed in an effort to cut down on duplicates, see here: .  and here: .