I'm trying to solve the following problem:  Given an analytic function $f: D \subset \mathbb{C} \to \mathbb{C}$, if $f^{(n)}(z) = 0$ for some $n \in \mathbb{N}$ and for all $z \in \mathbb{D}$, then $f$ is a polynomial of degree less than $n$.   I wanted to use induction to solve this problem. I managed to show that this is true for the base case of $f' = 0$. I then wanted to prove the inductive step holds. To do this, I define the function $F = f'$, so that $f^{(n+1)} = F^{(n)}$ and I can apply the induction hypothesis on $F$. By doing this I get that $$ F = \sum_{k=0}^{n-1} a_kz^k = \sum_{k=0}^{n-1} \alpha_k r^k \cos(k\theta) + i \sum_{k=0}^{n-1} \alpha_k r^k + \sin(k\theta) \tag{1} $$ for some $a_k \in \mathbb{C}$ (possibly equal to $0$), $r = |z|$ and $\theta = \arg(z)$. From here I just need to recover $f$ to finish the problem, but here's where I ran into trouble. The only way I know I can relate $f$ and $F$ is by using the Cauchy-Riemann equations, i.e., if $f = u(r, \theta) + iv(r, \theta)$, then \begin{align} F = f' &amp;= \frac{\partial}{\partial r}\left(\left[\cos(\theta)u + \sin(\theta)v\right] + i \left[\cos(\theta)v - \sin(\theta)u\right]\right)\tag{2}\\ &amp;= \frac{\partial}{\partial \theta}\left(\frac{1}{r}\left[\sin(\theta)v + \cos(\theta)u\right] + i \frac{1}{r}\left[\cos(\theta)v - \sin(\theta)u\right]\right)\tag{3} \end{align} Note: These last equations can be obtained from the polar C-R equations as in . After this, I tried combining equations $(1), (2)$ and $(3)$ to get a system of differential equations from where I could solve for $u$ and $v$ explicitly, and therefore solving for $f$ since $f = u + iv$. The problem I encountered was that I couldn't solve for the values of the constants of integration. As much as I tried I just kept getting expressions that I couldn't simplify. Is my approach correct? Or alternatively, does anyone know a simpler method I could use to prove this? Thank you!
93698
Suppose $f$ is entire and that in every power series $f(z)=\sum_{n=0}^\infty c_n(z-a)^n$ at least one coefficient is $0$. Prove that $f$ is a polynomial.
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!
6527
Assume $x \in \mathbb{R}$ and $(x + \frac{1}{x}) \in \mathbb{Z}$.   Use Induction to prove that for all $n \in \mathbb{N},~ (x^n + \frac{1}{x^n}) \in \mathbb{Z}$.  I'm not sure how to use the information given to prove the induction step. What I have so far is.  $P(n) : (x^n + \frac{1}{x^n}) \in \mathbb{Z}$  Base Case Then $(x^0 + \frac{1}{x^0}) = 2 \in \mathbb{Z}$  Induction Hypothesis: $x^k + \frac{1}{x^k}\in \mathbb{Z}$  Want to prove: $x^{k+1} + \frac{1}{x^{k+1}} \in \mathbb{Z}$  My first step was  $x^{k+1} + \frac{1}{x^{k+1}} = x \cdot x^k + \frac{1}{x} \cdot \frac{1}{x^k}$
123447
Let $f$ be a function holomorphic in an open set containing the closed unit disc $D(0,1)$, except at the point $z_0$ with $|z_0|=1$, where $f$ has an isolated singularity. If $a_n$ are the coefficients of the Taylor expansion of $f$ centered at the origin, show that   $$\displaystyle\lim_{n \to\infty}\frac{a_n}{a_{n+1}}=z_0$$  I considered first the case: $z_0$=simple pole. Then i can write the Laurent series for $f$ at $z_0$  $$f(z)=\frac{c}{z-z_0}+c_0+c_1(z-z_0)+\ldots$$ I put  $$g(z)=f(z)-\frac{c}{z-z_0}$$  Then $g$ is holomorphic in an open set containing the closed unit disc $\overline{D(0,1)}$ and we have $$g(z)=\sum a_n z^n+\frac{c}{z_0}\sum(\frac{z}{z_0})^n=\sum(a_n+\frac{c}{z_0^{n+1}})z^n$$ The power series of $g$ has radius of convergence $&gt;1$, hence i can substitute 1 in RHS of last equation, and i get a numeric convergent series. The necessary condition for convergence says that its general term has to be infinitesimal $$\displaystyle\lim_{n\to\infty}(a_n+\frac{c}{z_0^{n+1}})=0$$ from which follows $\frac{a_n}{a_{n+1}}\rightarrow z_0$  the question is:how can i adapt this argument to deal with cases where z_0 is not a simple pole?
21389
Is the following statement true?  Suppose, $ f:D\to \mathbb C $ is an analytic function where $ D $ is the unit disc of radius $ 1 $ around $0 $. Suppose, $ f $ is analytic on the boundary of $ D $ as well. Then prove that, if $ f $ is one-one on the boundary of $ D $, then $ f $ is one-one on $ D $.  P.S. I think I got a solution. But I was wondering if this problem has a trivial solution.  Here is my solution, since nobody has posted a solution yet:-  Suppose, $ \gamma $ is the boundary of $ D $(assuming $ \gamma(t)=e^{2\pi it} $). Then $ f\circ \gamma $ is a simple continuously differentiable curve. So by Jordan curve theorem, $ f\circ \gamma $ divides $ \mathbb C $ into 2 path connected regions $ A_1, A_2 $ such that if $ a\in A_1 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=2\pi i$ and if $ a\in A_2 $, then $\int_{f\circ \gamma} \frac {1}{z-a} dz=0$. So, $ f(z)=a $ for some $ a\in A_1,z\in D $ implies $ \int_{\gamma} \frac {f'(z)}{f(z)-a} dz=2\pi i $. So $ f(z)=a $ has only one solution in $ D $. Similarly if $ a\in A_2 $, then $\int_{\gamma} \frac {f'(z)}{f(z)-a}=0 $, so $ f(z)=a $ has no solution in $ D $. If $ f(z)=a $ for some $ z\in D, a\in f\circ \gamma $, then by open mapping theorem, $ f(z)=a' $ for some $ z\in D, a'\in A_2 $, which is not possible. So $ f $ is one-one in $ D $.
64934
Here, $Q$ is a polynomial with distinct roots $\alpha_1, \ldots, \alpha_n$ and $P$ is a polynomial of degree $&lt;n$. Once again, the task is to show  $$\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$$  For reference, this is page 32 exercise 2 in Complex Analysis by Ahlfors. I'm having a ton of trouble making connections between what is presented in the text and this problem. The section is on rational functions and is incredibly concise. There was never a mention of derivatives and their relationships to rational functions, so I'm guessing here that it just means that $Q'(a_k)$ is a polynomial of degree $n-1$.  I don't even know where to start here, but I do have a couple question which might motivate the proof. What does $P(\alpha_k)$ mean? I don't see what a root of $Q$ has to do with a completely different polynomial. What is the relationship between the roots of $Q$ and the roots of $Q'$? And finally, where does the derivative of $Q$ come in?
29274
Let $f: \mathbb{C} \to  \mathbb{C}$ be a continuous function such that $f^2$ and $f^3$ are both analytic. Prove that $f$ is also analytic.   Some ideas: At $z_0$ where $f^2$ is not $0$ , then $f^3$ and $f^2$ are analytic so $f = \frac{f^3}{f^2}$ is analytic at $z_0$ but at $z_0$ where $f^2$ is $0$, I'm not able to show that $f$ is analytic.
34513
I'm having trouble with the following exercise from Ahlfors' text (not homework)  "If $f(z)$ is analytic in a neighborhood of $\infty$ and if $z^{-1} \Re f(z) \to0$ as $z \to \infty$, show that $\lim_{z \to \infty} f(z)$ exists. (In other words, the isolated singularity at $\infty$ is removable)  Hint: Show first, by use of Cauchy's integral formula, that $f = f_1 +f_2$ where $f_1(z) \to 0$ for $z \to \infty$ and $f_2(z)$ is analytic in the whole plane.  My attempt:  $f$ is analytic in the exterior of some disk $\{|z|\geq R_0 \}$. For large enough $|z|$ the circle $C_1(z)$ around $z$ with radius one lies in the domain of analyticity, and we can apply Cauchy's integral formula:  $$f(z)=\frac{1}{2\pi i}\oint_{C_1(z)}\frac{f(\zeta)}{\zeta-z} \mathrm{d} \zeta=\frac{1}{2 \pi i} \oint_{C_1(z)}\frac{\Re f(\zeta)}{\zeta-z} \mathrm{d} \zeta+\frac{1}{2\pi} \oint_{C_1(z)}\frac{\Im f(\zeta)}{\zeta-z} \mathrm{d} \zeta. $$ I want to say that the last expression is exactly the decomposition $f_1+f_2$. Indeed, using Cauchy's estimate (and the fact about the real part) one can show that the first integral tends to zero for large $|z|$, but I can't understand how to define the other one for all $z \in \mathbb C$.  In fact, even if I have such a decomposition, I don't think that it will suffice: For instance $f(z)=f_1(z)+f_2(z)$ with $f_1(z)=0,f_2(z)=\exp(z)$ has an essential singularity at $\infty$.  Thank you for any directions.
110296
The hint is: Let F has as elements $a_1,...,a_n$. Note that if $$f_i(x)=c(x-a_1)...(x-a_{i-1})(x-a_{i+1})...(x-a_n)$$, then $f_i(a_j)=0$ for $i\neq j$ and the value $f_i(a_j)$ can be controlled by the choice of $c\in F$. Use this to show that every function on F is a polynomial function.  Using Fraleigh’s seventh edition of A First Course in Abstract Algebra. Need help with problem 22.31c. I’m completely lost. Thank you!
8689
There is an exercise in Function Theory of One Complex Variable by Greene &amp; Krantz that is very similar to a Proposition in the book, but I am having trouble getting to the conclusion.      Let $f : \mathbb{C} \to \mathbb{C}$ be a polynomial. Suppose further that $$\frac{\partial f}{\partial z} = 0$$   and $$\frac{\partial f}{\partial \bar z} = 0$$    for all $z \in \mathbb{C}$    Prove that $ f \equiv$ constant.    Now, part of the proposition (1.3.2) proves that if $p$ is a polynomial, $$p(z, \bar z) = \sum a_{lm}z^l \bar z^m$$ with $\frac{\partial f}{\partial \bar z} = 0$, then $\frac{\partial ^{l + m}}{\partial z^l \partial \bar z^m}p$ evaluated at zero is $l!m!a_{lm}$.  Is this the constant that $f$ equals? It doesn't seem like I've used the hypothesis, or reached the conclusion!  Maybe I should use the definition of the partial... $$\frac{\partial f}{\partial z} := 1/2 (\frac{\partial }{\partial z} - i \cdot \frac{\partial }{\partial z})f$$  Any hints would be greatly appreciated.
111707
Suppose $f$ is a function such that $f:\mathbb C \to \mathbb C$ such that $f(z)f(iz)=z^2$, then we have to show that $\forall z\in \mathbb C f(z)+f(-z)=0 $.   Actually I have solved this problem and have no doubt about it but actually this question came in MADHAVA-2020. So I posted it on this site.   I am not answering the question and inviting all users to answer it.  It is a nice problem.   I will answer it within a week, but I am letting everyone try.
111112
I've been asked to prove that if a polynomial $f\in \mathbb{Q}[x]$ has degree $n$, then the splitting field of $f$ has degree less than or equal to $n!$. That is, $[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb Q]\leq n!$, where $\alpha_i$ are the roots of $f$.   My approach to this question is to use the tower law, to arrive at the equation: $$[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb{Q}]=[\mathbb{Q}(\alpha_1,...,\alpha_n):\mathbb{Q}(\alpha_1,...,\alpha_{n-1})][\mathbb{Q}(\alpha_1,...,\alpha_{n-1}):\mathbb{Q}(\alpha_1,...,\alpha_{n-2})]\cdots[\mathbb{Q}(\alpha_1):\mathbb{Q}]$$  Clearly $[\mathbb{Q}(\alpha_1):\mathbb{Q}]\leq n$, since the minimum polynomial of $\alpha_1$ divides $f$.  I claim that in general: $[\mathbb{Q}(\alpha_1,...,\alpha_i):\mathbb{Q}(\alpha_1,...,\alpha_{i-1})]\leq n-i, \forall i= 1,2,...,n$. My question is, whether this claim is indeed true, and if so how would I prove it. I've tried using induction but I can't seem to make it work.  Also, if anyone has a proof for the original question without using my claim, then can you please post that too.
18635
Let $A \subset \mathbb{C} $ be an open set containing the closed unit disc. Let $f$ be an analytic function from $A$ to $\mathbb{C}$ such that $|f(z)|=1$ if $|z|=1$.  Does it follow that $f(z) = a z^{n} \frac{cz^{m}-b}{1-cz^{m}\bar{b}} $ for some $a,b,c \in \mathbb{C}$ s.t. $|c|=1$ $|a|=1$, $|b|&lt;1$ and some $n,m\ge 0 $?
66462
I am trying to solve the following two problems:  1) Prove that the functions $\cos(z)$, $\sin(z)$ are surjective over the complex numbers.  2) Find all $z \in \mathbb C$: $cos(z) \in \mathbb R$ and find all $z \in \mathbb C$: $\sin(z) \in \mathbb R$.  For 1),I've tried to prove it for the function $\cos(z)$ (I suppose the other one is analogue)  so I've used the fact that $\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}$.  Let $w \in \mathbb C$, I want to show there exists $z \in \mathbb C : f(z)=w$, i.e., $\dfrac{e^{iz}+e^{-iz}}{2}=w$, multiplying by $2$ and then by $e^{iz}$ yields $e^{2iz}+1=2we^{iz}$ iff $e^{2iz}-2we^{iz}+1=0$.   I don't know if this approach is the correct one but here I've replaced $e^{iz}$ by $x$, so the solutions of the equation would be the roots of the polynomial $p(x)=x^2-2wx+1$, by the quadratic formula, I get that $x \in \{\dfrac{2w+w_0}{2},\dfrac{2w-w_0}{2}\}$, where $w_0^2=4w^2-4$.  Then, $e^{iz} \in \{\dfrac{2w+w_0}{2},\dfrac{2w-w_0}{2}\}$. At this point I got lost, I would like to explicitly show that $z$ exists and I can't see existence directly from the fact that $e^{iz}=\dfrac{2w+w_0}{2}$ or $e^{iz}=\dfrac{2w-w_0}{2}$.  First I thought of taking logarithm of both sides of the equation ir order to solve for $z$, but this is not a legitimate operation unless $e^{iz} \in \mathbb R$.   I couldn't go any farther.  For point 2) I have no idea what to do, should I use the identity $\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}$?  I would appreciate some help with the two points (specially with point 2), at least in 1) I could do something).   Btw, Happy new year!
81785
Suppose that $[F(\alpha):F]$ is odd. Show that $F(\alpha)=F(\alpha^{2})$. Generalize this statement if possible.    Clearly, we always have that $F(\alpha^{2}) \subset F(\alpha)$ by definition of the field generated by $F$ and $\alpha$ (assuming that we have a field extension $K/F$ where $\alpha \in K$). However, how can we show the other inclusion? If the degree is odd, then it follows that $\alpha$ will solve a polynomial of odd degree in $K[x]$ (for the coefficients in $F$). What would be a way to proceed from here? (and as well to generalize the statement?)  I was thinking of using the formula: $[F(\alpha):F]=[F(\alpha):F(\alpha^{2})][F(\alpha^{2}):F]$ and maybe proceed from there.   Thanks for the help.
82172
Suppose $f(z)$ and $g(z)$ are analytic in domain $D$, $f(z)$ and $g(z)$ never vanish at any $z\in D$ and that     $$ \frac{f'(z_n)}{f(z_n)}=\frac{g'(z_n)}{g(z_n)} $$ at a sequence  of points $\{z_n\}$ converging to $z_0\in D$. Show that $f=Kg$ for some $K\in \mathbb{C}$.   My Solution:   As $f,g$ are analytic in $D$ and none of them have any zero in $D$ implies $\frac{f'}{f}$ and $\frac{g'}{g}$ are also analytic on $D$. As $\frac{f'(z_n)}{f(z_n)}=\frac{g'(z_n)}{g(z_n)}$ and $z_n\to z_0 \in D$, using the identity theorem, $\frac{f'}{f}\equiv \frac{g'}{g}$ on $D$.  Now,  \begin{align*} \frac{f'(z)}{f(z)}=\frac{g'(z)}{g(z)},\ \forall\ z\in D. \end{align*} Solving this differential equation gives me $f=Kg$ for some $K\in \mathbb{C}$. Now my question is after doing the integration we will get $\log(f)=\log(Kg)$, is it valid in the complex logarithm?
75575
I have two related problems that are causing me big trouble:  Let $\;a_1,...,a_n\in\Bbb R\;,\;\;a_i\neq 0\;\;\forall\;i\;$ , and let $\;b_1,...,b_n\in\Bbb R\;,\;\;b_i\neq b_j\;\;\forall\,i\neq j\;$ .  (1) Prove that the equation  $$a_1x^{b_1}+\ldots+a_nx^{b_n}=0$$  has at most $\;n-1\;$ different zeros in $\;(0,\infty)\;$ (Hint: Use induction)  (2) Prove that the equation  $$a_1e^{b_1x}+\ldots+a_ne^{b_nx}=0$$  has at most $\;n-1\;$ different zeros in $\;\Bbb R\;$ .  Now, the following is what I've done so far: for a differentiable function $\;f\;$ ,  I can prove that if $\;f'(x)=0\;$ has $\;n-1\;$ different zeros, then $\;f(x)\;$ has at most $\;n\;$ different zeros, using the mean value theorem, say.  But I can't see how to use induction in this case: if I differentiate in the first problem, I get $$0=f'(x)=a_1b_1x^{b_1-1}+\ldots +a_nb_nx^{b_n-1}=\frac1x\sum_{i=1}^na_ib_ix^{b_i}$$  ...and now?! How does induction kick in here? I though perhap something like: if we suppose WLOG $\;b_1&lt;b_2&lt;...&lt;b_n\;$ , then perhaps inducting on $\;\lfloor b_n\rfloor\;$, but it gets messy and blurry if $\;b_n&lt;0\;$...  Any help, hints are very much appreciated.