Let $p(z) = z^n+ a_{n-1} z^{n-1}+...+a_0,$ where $a_i \in \mathbb{C} $. If $|p(z)| \leq 1$ for all $z$ with $|z| \leq 1$, then show that $p(z)=z^n, \forall z\in \mathbb{C}$. Now by maximum modulus principle , we can deduce that $|p(z)| \lt1 \ for \  |z| \lt 1 $. Then in particular if we take $a_0=0$ then also $p(0)=0$. So by Schwarz's Lemmas we have $|p(z)|\lt |z|$. I can only deduce this far, and still confused about the usability of my deductions. How can I proceed further.
123396
Let $p(z)=z^n+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+...+a_0$ be a complex polynomial satisfying $|p(z)| \leq 1$ for $|z|\leq 1$, then prove that $p(z)=z^n$  I tried with Cauchy integral formula for $n$th derivative to get $$n!=\frac{n!}{2\pi}\int_0^{2\pi}p(e^{i\theta})e^{-in\theta}d\theta$$ implies $$1\leq \frac{1}{2\pi}\int_0^{2\pi}|p(e^{i\theta})e^{-in\theta}|d\theta$$ and thereby $|p(e^{i\theta})e^{-in\theta}|=1$. (since $|p(e^{i\theta})e^{-in\theta}|$ is a non negative real continuous function $\leq 1$)  ie., $|p(z)|=|z^n|$ for $z=e^{i\theta}$, i stucked here and how can we procced?
29374
Let $f\in H(B(0,1))$, $f(0)=0$, $f(B(0,1))\subset B(0,1)$. Suppose there are 2 different points $z_1$,$z_2\in B(0,1)$, s.t. $|z_1|=|z_2|$, $f(z_1)=f(z_2)$. Show that $|f(z_1)|=|f(z_2)|≤|z_1|^2=|z_2|^2$.    I have used the Schwarz-Pick Proposition to prove that $|f(z_1)|=|f(z_2)|≤|z_1|=|z_2|$ directly, but that is not enough. Thank you for help!
46357
I gave an exercise to a student the other day, thinking I had a simple solution for it, but it seems that my solution was just bullshit (at least, not conclusive). Well, at least, I warned her I wasn't sure of my solution :-)  Here it is : $a\in[-1,1]$, show that all roots of polynomial $P=X^{n+1}-aX^n+aX-1$ have modulus $1$ (I try different ways of expressing the problem, tell me which one is better english...).  It is easy to see that    roots of $P$ are $1$, $-1$ for some special cases, and complex not real roots $z_i$ such that $\overline{z_i}$ is also a root, as $P$ is kind of reciprocal, $z$ root implies $\frac1z$ is also a root$.   All the other things I tried are not conclusive. For example, $a\in[-1,1]$ may be translated as $a=\cos\theta$ for some $\theta\in\mathbb R$, but I don't see how to use this. I tried some rewriting, but nothing seems to work. I tried to work on the modulus of a root, tried the relations roots-coefficients...  I'm quite stuck here. Could you help me please ?  Thanks.
122769
This is a homework problem that I wasn't able to solve, and I feel a little silly about it (the assignment has been submitted). The problem:     Let $P(z) = 1 + 2z + 3z^2 + \cdots + nz^{n-1}$. By considering $(1-z)P(z)$, show that zeros of $P(z)$ are contained inside the unit disc.   This is from Chapter 1 of Bak's Complex Analysis. A solution should only use algebraic manipulations of complex numbers (in other words, this problem is very easy to solve by employing the Gauss-Lucas theorem, but that's not "on the table" -- complex differentiation hasn't been defined in the scope of what's permissible).  What I've done:  $$\begin{align*} (1-z)P(z) &amp;= 1+z+z^2+\cdots+z^{n-1}-nz^n \\  &amp;= \frac{z^n-1}{z-1}-nz^n. \end{align*}$$ Let $z$ be a zero of $P(z)$; then it is a zero of $(1-z)P(z)$. Consequently, $$\begin{align*} nz^n &amp;= \frac{z^n-1}{z-1} \\ nz^n(z-1) &amp;= z^n-1 \\ nz^n(z-1)+1 &amp;= z^n \end{align*} $$  Then $$|z^n| = |z|^n = |nz^n(z-1)+1|.$$ From the triangle inequality, $$ \begin{align*} |z|^n &amp;\le |nz^n(z-1)| + |1| \\ |z|^n &amp;\le n|z^n||z-1| + |1| \\ n|z|^n(|z-1|) &amp;\le 1 \\ |z|^n &amp;\le \frac{1}{n|z-1|} \end{align*} $$  At this point, I am stuck. I feel like I went down the wrong path, or maybe I just made an arithmetical error, but I can't find where.
69762
If $P(A_n) \rightarrow 0$ and $\sum_{n=1}^{\infty}{P(A_n^c\cap A_{n+1}})&lt;\infty$ then $P(A_n \text{ i.o.})=0$.  How to prove this? Thanks.
86441
The question is  Let $f $ be analytical and not constant in $\{z\in\mathbb C:|z|\leq 1\}$. If there is some $z_0\in \{|z|\leq 1\}$ such that $$|f(z_0)|=\max_{|z|\leq 1} |f(z)|\,,$$ then there must be $|z_0|=1 $ and $f’(z_0)\neq 0$.  The first part is just the maximum principle, but for the derivative, I have no idea how solve it. Looking forward to some hints or answers, thank you.
94833
Each of the polynomial of the form $p(z)=a_0+\dots+a_{n-1}z^{n-1}+z^n$ satisfies the inequality $\sup\left\{\,|p(z)|\,\big\vert\,|z|\le 1\,\right\}\ge 1$   Is this statement true or false that we have to find. well MMP says that sup will be attained at $|z|=1$ so when $|z|=1$ we have $|p(z)|=|a_0+a_1\dots+ a_{n-1}+1|\le |a_0|+\dots+|a_{n-1}|+|1|$ I can not conclude more.plz help.
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 $\leq 4$
67562
Let g be an entire function such that   $2g(z)=g(\frac{z}{2}) + g(\frac{z+1}{2})$ holds $\forall z\in \mathbb{C}$   the I want to prove that $ g $ is constant.  Hint: Apply maximum modulus principle on $cl(B_r (0))=${$ |z| \leq r $}$ $  also   if $g$ has no zeros and  $cg(z)=g(\frac{z}{2}) g(\frac{z+1}{2})$  holds $\forall z\in \mathbb{C}$ and for some $c\in \mathbb{C}$ prove that $g(z)=ce^{\frac{-b}{2}}e^{bz}$ for some $b\in \mathbb{C}$   I have no idea how to prove this statement . Any Hints leading to solutions  are appreciable
3134
I'm having trouble with a question from an exam I'm studying for:  " Prove that all roots of the polynomial $\sum\limits_{k=0}^n \frac{z^k}{k!}$ (for $n\geq 1$) are in the annulus $\{z: \frac{n}{e}&lt;\vert z\vert&lt;2n \}$. Hint: use the inequality $n^ne^{-n}&lt;n!$"  These kind of questions are usually solved by using Rouché's theorem, so I assume that's the case here as well, but I'm having trouble choosing the right functions to use and bounding their norms correctly. I feel like I should be comparing $e^z$ and $\left(e^z - \sum\limits_{k=0}^n \frac{z^k}{k!}\right)$ on the circle with radius $\frac{n}{e}$ since e^z has no roots there, so maybe that will imply the polynomial also has no roots there. And then maybe using some other argument to show that all the polynomial's roots are inside the larger disk of radius $2n$.
107737
Assume $0 &lt; a_0 \leq a_1 \leq \cdots \leq a_n$, show that equation   $$ p_n(z) = a_0 z^n + a_1 z^{n-1} + \cdots + a_{n-1} z + a_n = 0 $$   does not have root in $|z| &lt; 1$. Here $z$ is complex.   I started by considering the case in which $n = 2$, and tried to use the triangle inequality, while no progress was made. Any hints will be appreciated.
80929
For a polynomial $p(z)=a_0+\ldots+a_nz^n$, prove there exist $R&gt;0$, such that if $|z|=R$, then $|p(z)|\geq |a_n|R^n/2$.  I get $|p(z)|\geq |a_n|R^n$, which makes that factor of $1/2$ useless. So I must have done something wrong.  $$|\frac{\partial^n{f}}{\partial z^n}(P)|\leq \frac{\sup_{z\in \overline{D}(P,r)} |f(z)|k!}{r^k}$$. This is the Cauchy estimate.   $$|\frac{\partial^n{p}}{\partial z^n}(z)| = |a_n n!|$$ $$|\frac{\partial^n{p}}{\partial z^n}(0)|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$ $$|a_n n!|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n$$ $$R^n |a_n|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|$$  There exist a $z_0$, such that $|z_0|&gt;0$ and $$\sup_{\zeta\in \overline{D}(0,|z_0|)} |p(\zeta)| = |p(z_0)|$$.  This is saying there exist a disk centered at the origin, such that $|p|$ is maximum on some point on the boundary.   Let $R=|z_0|$, we have $R^n |a_n|\leq |p(z_0)|$  Edit: I know what I got wrong. $z_0$ is just one point on the circle $|z| = R$. One can chose another point $z_1$ on the circle, such that $p(z_1)&lt;p(z_0)$.
19100
Let $a,b,c,x,y,z&gt;0$ and $x+ y+ z= 1$. Show that $$ax+ by+ cz\geq  a^{x}.b^{y}.c^{z}$$ I have found the generalization of this inequality: For 2 sets of numbers $\left \{ a_{i} \right \}, \left \{ b_{i} \right \}&gt; 0$ and $\sum_{1}^{n}b_{i}= 1$, we have: $$\sum_{1}^{n}\left ( a_{i}b_{i} \right )\geq  \prod_{1}^{n}\left ( a_{i}^{b_{i}} \right )$$ First solution Simply compare the logs: as the natural logarithm is a concave function, one has $$\ln\bigl(a^x b^y c^z)=x\ln a+y\ln b+z\ln c\le \ln(ax+by+cz)$$ for all non-negative $x,y,z$ such that $x+y+z=1$. How about another solution? I really hope to see that. Thanks!
55241
Doing a discrete math review and am trying to solve problem 1.6 in the text found here:  - I believe I've gotten parts (a) and (b) correctly, but (c) is a bit tricky for me. Would appreciate a review of a/b and a hint for c if possible. Restatement here:  For n = 40, the value of polynomial $p(n)::=n^2 + n + 41$ is not prime, as noted in Section 1.1. But we could have predicted based on general principles that no nonconstant polynomial can generate only prime numbers.  In particular, let $q(n)$ be a polynomial with integer coefﬁcients, and let $c::=q(0)$ be the constant term of q.  (a) Verify that $q(cm)$ is a multiple of c for all $m \in Z$  Proof: Since q is a polynomial, it will be of the form: $q(x) = a_nx^n+a_{n-1}x^{n-1} + ... + a_1x + c$  If $x = cm$, then notice that cm will be inside every term of the polynomial, and since the final term is c, c can be factored out of every term. Hence, q(cm) is a multiple of c for all $m \in Z$.  b) Show that if q is nonconstant and c > 1, then as n ranges over the nonnegative integers, $N$, there are infinitely many $q(n) \in Z$ that are not primes.  Proof: Continuing from the result found in (a), it is easy to see that for all $m \in Z$, there will be an infinite number of multiples of c that will be generated by q. Since c > 1, these multiples are guaranteed not to be prime.  c) Conclude that for every nonconstant polynomial, q, there must be an $n \in N$ such that $q(n)$ is not prime. Hint: Only one easy case remains.  c = 0 is trivially easy to prove using the (b) above.  I assume the "easy" case is referring to c = 1, but in this case I'm not sure how to continue, since the result of (a) and (b) don't apply: I can't use them since adding 1 to any even number may make it prime. If the result of the terms not including c is odd, then adding 1 to that result makes it even and it is not prime. However, if the terms add up to an even number, I don't see a way of using the knowledge I have so far to prove conclusively that that number + 1 will NOT in fact be prime.
4812
Problem: $(\forall a\in\mathbb{Z^+})(m\in\mathbb{Z^+}\to a^m\equiv a^{m-\phi(m)}\pmod{m})$  My work: Start by letting $m=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$. If $(a,p_i)=1$ for some integer $i$, then we have that  $$ a^{\phi(p_i^{a_i})}\equiv 1\pmod{p_i^{a_i}} $$ by Euler's theorem. Since $\phi(p_i^{a_i}) \mid \phi(m)$, we have that $p_i^{a_i}\mid (a^{\phi(m)}-1)$ if $(a,p_i)=1$.   Beyond this, I am stuck at the moment.
74920
Suppose that $0\neq x\in\mathbb{R}$ and $x + \frac1x\in\mathbb{Z}$. Prove that, for all  $n\ge1$, $x^n + \frac1{x^n}\in\mathbb{Z}$.  I can't figure out and understand the question. Can you give me some hints ?
54315
Could someone advise me on how to approach this problem:  without  the use of argument principle ?    Here is my attempt: Suppose $f(z)$ has two zeroes at $z=a.$ Let $f(z) = \sum^{\infty}_{n=0}a_n(z-a)^n, \forall z. \ $   Then $f(z)=(z-a)^2 \left(\dfrac{a_{0}}{(z-a)^2} +\dfrac{a_1}{z-a}+a_2+a_3(z-a)+...\right)$   $\implies a_0=a_1=0.$  $\implies f(z)= (z-a)^2\left(a_2+a_3(z-a)+a_4(z-a)^2+...\right)$  $\implies .... ?$  Thank you.