I'm working on this problem  "Find all entire functions $f(\frac{1}{p})=\frac{1}{1+p}$ for all prime $p$."  My approach is using identity theorem. But in this case it does not seem good. We have $$f(1/p)=\frac{1/p}{1+1/p}$$ So naturally, I set $g(z)=f(z)-\frac{z}{z+1}$. Consider the plane excluded a circle around $z=-1$. Then $g$ is analytic in that region. And $g=0$ on the set of $\{1/p\}$ which has a limit point. So $g=0$ on the constructed given. But how do I extend it to $-1$? Or I can't?
123954
Let $U$ be an open connected neighborhood of $0$ in $\Bbb{C}$ and $f$ a holomorphic function on $U$. The exercise is to prove that if $f(\frac{1}{n})=\frac{1}{n+1}$ for all $n\ge 1$ then $f(z)=\frac{z}{z+1}$ on $U$.  So I did it using the function $g:\Bbb{C} \setminus\{-1\}\to \Bbb{C}$ such that $g(z)=\frac{z}{1+z}$ so $g$ is holomorphic on $\Bbb{C} \setminus\{-1\}$, now as $U$ is open we can find a disc $D(0,r)\subset U$ for $r$ so that $r&lt;1$. Therefore, using analytic continuation we have $f=g$ on $D(0,r)$. My main problem is now to prove that $U\subset\Bbb{C} \setminus\{-1\}$. Perhaps, I can say that if $-1\in U$ we can find a path from $0$ to $-1$ but not sure how can I continue.
115106
Let $f(z)=\frac{1}{(z-1)(z-2)}$ and $g(z)=\frac{z}{(z-1)(z-2)}$. Let $\gamma$ be the circle of radius 4 centered at the origin and traveled once in the counterclockwise direction. Let $A=\{ z \in \mathbb{C}: |z|&gt;3 \}$  1) I need to show there is no analytic function $G: A \to \mathbb{C}$ such that $G'(z)=g(z)$ for all $z$ in $A$.   2) I also need to show that there is an analytic function $F:A \to \mathbb{C}$ such that $F'(z)=f(z)$ for all $z$ in $A$.  Do I have to integrate to find such function?
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.
59712
In my book I have a theorem that goes something like the following     Let $P(x)$ be $Q(x)$ polynomials such that $\deg(Q) \geq \deg(P) + 2$.   Then \begin{align*}  \int_{-\infty}^{\infty} \frac{P(x)}{Q(x)}  e^{iax} \,\mathrm{d}x  = 2 \pi i \frac{a}{|a|} \sum_{k=1}^{m} \mathrm{Res}\left[  \frac{P(x)}{Q(x)} , z_k \right] \end{align*} where $a$ is a real   constant and $z_1,\,\ldots\,,z_m$ are the singularities to $P(x)/Q(x)$ in   the upper halfplane if $a&gt;0$ and the lower halfplane if $a&lt;0$.   I am having a bit of problems understanding this. I know the reason why we have to switch contour is that $e^{iz}$ blows up. But I really can not see why or how it blows up.   Take the canonical example for why this theorem is useful  $$   J = \int_{-\infty}^\infty \frac{e^{iz}}{z^2+1} $$    Now the clue here is to show that the integral along the curve tends to zero, and then use the residue theorem. To show that the integral tends to zero I did this \begin{align*}  \left| \int_{C_1} \frac{e^{iz}}{1+z^2} \,\mathrm{d}z\right|  \leq \int_{C_1} \left| \frac{e^{iz}}{1+z^2} \right| \,\mathrm{d}z   \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{\left|1+z^2\right|} \right)       \int_{C_1} |e^{iz}| \,\mathrm{d}z \end{align*} where the $ML$-inequality was used in the last inequality  On the circle with radius $R$ we have $|e^{iz}|=R$, and we can use the inequality $|a+b|\leq|a|-|b|$ to simplify further \begin{align*}  \left|\int_{D_R} \frac{e^{iz}}{1+z^2}\,\mathrm{d}z \right|  \leq \sup_{z = R e^{i \theta}} \left( \frac{1}{|z|^2-1}\right)  \int_{D_R} R \,\mathrm{d}z  \leq \pi \frac{R}{R^2-1} \end{align*} which tends to zero as $R \to \infty$ as wanted. but if one instead had $e^{-iz}$ then the theorem states that one has to use the lower half plane. But I do not see where my calculations err if one persists in using the contour in the upper half plane? What goes wrong, and why does it go wrong? $|e^{-iz}|$ should still be $R$ on the semi-circle.   How can one formally show that the integral diverges when picking the contour as a semi circle in the lower half plane?
52254
Let $f$ be an entire function and let for each $a\in \mathbb R$, there exists at least one coefficient $c_n$ in $f(z)=\sum\limits_{n=0}^{\infty}c_n(z-a)^n$, which is zero. Then    $f^{(n)}(0)=0$ for infinitely many $n\geq 0$ $f^{(n)}(0)=0$ for every $n\geq 0$ $f^{(2n+1)}(0)=0$ for every $n\geq 0$ There exists $k\geq 0$ such that $f^{(n)}(0)=0$ for all $n\geq k$   We know that $c_n=\frac{f^{(n)}(a)}{n!}$ for all $n\in \{0,1,2\ldots\}$. Thus for $a=0$, $c_n=\frac{f^{(n)}(0)}{n!}$. By hypothersis, atleast one $c_n=0$. After that I could not do anything. Please help!
97979
Is there an analytic function $f$ on $B(0,1)\subset\mathbb{C}$ such that $f(1/n)=e^{-n}$ for $n=2,3,4,...$? I know the following doesn't work:  Let $g(z)=\exp(-1/z)$. Then, $f=g$ on a sequence with a limit point in $B(0,1)$ and so $f=g$ on $B(0,1)$. Since $g$ is not $\mathbb{C}$-differentiable at $0$, neither is $f$ and so such a function cannot exist.  This is not the solution because you cannot use the identity principle with a non analytic function like $\exp(-1/z)$ is not analytic at $z=0$. Any help using the identity principle another way?
12758
This is part of a problem, 7.22c in "Ireland and Rosen" (self-study).  In the prior problem it is shown that for prime $p$ and $\alpha\in F_{p^n}$, $f(x)=(x-\alpha)(x-\alpha^p)(x-\alpha^{p^{2}})\cdots(x-\alpha^{p^{n-1}})$ is in $\mathbb{Z}/p{\mathbb{Z}}[x]$.          Show there exists an $\alpha\in F_{p^n}$ such that Tr$(\alpha)=\alpha +\alpha^p+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}\neq 0$ for some $\alpha\in F$.      So far my approach is that with $f(x)\in \mathbb{Z}/p{\mathbb{Z}}[x]$, the coefficients are in $\mathbb{Z}/p{\mathbb{Z}}$. And by the symmetric polynomial functions, the Tr$(\alpha)$ as defined is a coefficient, it follows it is in $\mathbb{Z}/p{\mathbb{Z}}$.   Thus I would think that Tr$(\alpha)\neq 0$ is equivalent to $p\nmid \alpha +\alpha^p+\alpha^{p^2}+\cdots+\alpha^{p^{n-1}}$  If this is correct I would please appreciate help as to how to show it.  Otherwise, I would appreciate further help.  Thanks
69053
Show there is no analytic function in $\{|z|&lt;1\}$ such that $f(0)=1$ and $|f(z)|\ge 1+|z|^2$. I guess something went wrong along the way and I would like your assistance.   $f$ does'nt get a maximum in the domain, because if it did, it would be constant in the domain, meaning $|f(z)|=1\ge 1+|z|^2$ which doesn't hold for all z. Since $f$ isn't zero in the domain, ${1\over f}$ is analytic in the domain. but $|{1\over f(z)}|\le {1\over 1+|z|^2}\le 1={1\over f(0)}$, which indicates a local maximum making ${1\over f}$ constant, contradiction the fact the $f$ isn't. What am I really lacking here?
36898
Suppose that $f : \mathbb{C} \to \mathbb{C}$ is entire. Let $\Omega = \{ z : \frac{1}{2} &lt; |z| &lt; 2\}$. Furthermore, suppose that $ \mathcal{F} = \{f(kz) : k \in \mathbb{C}\}$ is a normal family of analytic functions defined on $\Omega$; the family $\mathcal{F}$ is normal on $\Omega$ iff each sequence from $\mathcal{F}$ has a subsequence converging uniformly on compact subsets of $\Omega$. Prove that $f$ is a polynomial.  I'm studying for my qualifying exam in complex analysis, and this question appears on an old exam. Usually, when I want to show that some entire $f$ is a polynomial, I just try to show that there exists some $N$ so that $f^{(k)}(0) = 0$, all $k \ge N$, but I'm having trouble getting that strategy to work here. I've been looking at the sequence $f(\frac{z}{n})$ trying to make something work, but have not had any success.   I would prefer to receive a hint or two, and then I'll try to post my own solution. Thanks in advance!
100033
Let, $p$ be a polynomial in $1$-complex variable. Suppose all zeros of $p$ are in the upper half plane $H=\{z\in \mathbb C|\Im(z)&gt;0\}. $ Then , which are corrct ?   $\Im\left\{\frac{p'(z)}{p(z)}\right\}&gt;0$ for $z\in \mathbb R$. $\Re \left\{i\frac{p'(z)}{p(z)}\right\}&lt;0$ for $z\in \mathbb R$. $\Im\left\{\frac{p'(z)}{p(z)}\right\}&gt;0$ for $z\in \mathbb C$ with $\Im (z)&lt;0$. $\Im\left\{\frac{p'(z)}{p(z)}\right\}&gt;0$ for $z\in \mathbb C$ with $\Im (z)&gt;0$.   I know that , since all roots of $p(z)$ lie in the upper half plane so, all roots of $p'(z)$ lie in the same upper half plane , from Luca's theorem.  But from there how we can say about real part or imaginary part of $\frac{p'(z)}{p(z)}$ ?
91615
Let $C$ be a regular curve enclosing the distinct points $ω_1,ω_2,...ω_n$ and let $p(ω) = (ω −ω_1)(ω −ω_2) \cdots (ω −ω_n)$. Suppose that $f (ω)$ is analytic in a region that includes $C$. Show that   $$P(z) =  \frac{1}{2 \pi i} \int_C \frac{f(ω)}{p(ω)} . \frac{p(ω) -p(z)}{ω -z} dω$$  is a polynomial of degree $n-1$ and $P(ω_i) = f(ω_i)$.
98833
"Find an $f \in [0,1]$ such that $f \in L^1$ but $f \not\in L^p$ for any $p &gt; 1$."  I've thought about doing something like $$f(x) = \frac{1}{x}$$ where $|f|^p = \frac{1}{x^p}$ doesn't converge when $p &gt; 1$.  But this function isn't itself in $L^1$.  Could someone please give me a hint for how to solve this problem?  I wish there were a situation where you had convergence on the closed half disc $[0,1]$ and divergence on $(1, \infty)$, rather than my current predicament where I have convergence on the open half-disc $[0, 1)$ and divergence on $[1, \infty)$.
121165
(To clarify, I would just like a hint. Please do not give me the answer to this problem. ) The solution to the following problem has really evaded me here:  Problem: Assume that $f$ is entire and that $\lim_{z \to \infty} f(z)/z  = 0.$ Prove that $f(z)$ is constant.  My Thoughts and Work So Far: We know that proving that $f'(z) =  0$ or that for some fixed $c \in \mathbb{C}$, $f(z) = c$ for all $z\in \mathbb{C}$. My first approach was to use Liouville's Theorem; If I could could show that $f$ is bounded then I am done. Since $\lim_{z \to \infty} f(z)/z  = 0$, for all $\varepsilon &gt; 0$ there exists a $N \in \mathbb{C}$ so large that if $z \geq N$ then $|f(z)/z| \leq \varepsilon$. Thus, if $C_R$ is the circle of radius $R$ centered at the point $z$, then, as long as z is large enough by Cauchy's Inequality $$ |f'(z)| \leq \frac{1}{2\pi i} \oint_{C_R} \frac{f(\zeta)}{(\zeta - z)^2}\ d\zeta \leq \bigg | \frac{1}{2\pi i} \bigg | \oint_{C_R} \bigg | \frac{f(\zeta)}{(\zeta - z)^2} \bigg | d \zeta \leq \frac{1}{2\pi}  \frac{|\zeta|\varepsilon 2\pi R}{R^2} = \frac{|\zeta|\varepsilon}{R}.  $$  Now taking the limit as $R \to \infty$ (which to me says, "let our circle around our point z dilate to an infinite radius so that it covers all of $\mathbb{C}$)  $$|f'(z)| \leq \lim_{R \to \infty} \frac{|\zeta|\varepsilon}{R} = 0.$$  Thus $f'(z) = 0$ and $z$ was arbitrary, so $f$ must be constant.   Why I Think Im Wrong: I say $z$ was arbitrary, but really it is "any $z \geq N$" which really isn't all that arbitrary.   This is where I am stuck. Am I right, wrong, close, or totally lost? Any hints would be great.  Edit: I am very sorry but I accidentally posted this before I was done typing the problem.
33426
This is a question from an old qualifying exam I am trying to solve.  Suppose $p$ and $q$ are nonconstant polynomials. Prove that the range of the function $f:\mathbb{C} \rightarrow \mathbb{C}$ defined by $f(z)=p(z)e^{q(z)}$ is all of $\mathbb{C}$.  My assumption is that I will need to use Picard's Little Theorem.  I know that $e^{q(z)}$ omits only $0$ and $p(z)$ has range all of $\mathbb{C}$.  However, I cannot see how to properly use this information. Any hint or suggestion would be very appreciated. Thanks!
110270
Problem: Let $f(z)$ be a meromorphic function on the complex torus $\mathbb C/\Lambda$ that as a function on $\mathbb C$ has no zeros and no poles on $\partial P$, the boundary of the fundamental parallelogram $P$. Show that  \begin{equation}                 \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz \in \Lambda \, . \end{equation}  Thoughts: By the Residue Theorem  \begin{equation} \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz = \sum_{z_0 \in \text{ Int }P} v_{z_0}(f)z_0. \end{equation}  I don't know why the latter is on the lattice.  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.
108112
We begin by assuming that a non-constant complex polynomial $$p(z)=a_n z^n+a_{n-1}z^{n-1}+...+a_1 z+a_0,$$ where $a_0 \neq 0$, doesn't vanish in $\mathbb{C}$. Therefore $1/p$ is also an analytical function everywhere. Note that $(p(z)-a_0)/z$ is a polynomial and therefore, due to Cauchy's theorem, the following integral must be zero.  $$\int_{S_r^1}\frac{(p(z)-a_0)/z}{p(z)}dz=\int_{S_r^1}\frac{a_nz^{n-1}+a_{n-1}z^{n-2}+...+a_1}{a_nz^n+a_{n-1}z^{n-1}+...+a_1 z+a_0}dz=0.$$  On the other hand  $$\int_{S_r^1}\frac{a_nz^{n-1}+a_{n-1}z^{n-2}+...+a_1}{a_nz^n+a_{n-1}z^{n-1}+...+a_1 z+a_0}dz$$  $$=\int_{0}^{2\pi}\frac{a_n(re^{i\theta})^{n-1}+a_{n-1}(re^{i\theta})^{n-2}+...+a_1}{a_n(re^{\theta})^n+a_{n-1}(re^{i\theta})^{n-1}+...+a_1 (re^{i\theta})+a_0}(ire^{i\theta})d\theta$$  $$=\int_{0}^{2\pi}i d\theta - \int_{0}^{2\pi}\frac{a_0 i}{a_n(re^{i\theta})^n+a_{n-1}(re^{i\theta})^{n-1}+...+a_1 (re^{i\theta})+a_0}d\theta$$ $$=2\pi i -\int_{0}^{2\pi}\frac{a_0i}{p(re^{i\theta})}d\theta.$$  Let $r \rightarrow \infty$ and the integral approaches value $2\pi i$ - a contradiction. Therefore $1/p$ can't be entire and $p$ must vanish somewhere in the complex plane.  EDIT: I have changed this post and deleted the idea that the integral $\int_{\partial D_r}\frac{(p(z)-a_0)/z}{p(z)}dz=0$ could be evaluated with the fundamental theorem of calculus. Now, as it stands, this proof is essentially the same as the one that Mr. José Carlos Santos points out in his comment.