If $P(z)=z^n+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+...+a_1z+a_o$ a polynomial who's coefficients are all real.   Explain why if $z_1$ is a root of $P$, then $\bar{z_1}$ is also a root.  And give an example as to why the contrary is false if the coefficients of $P$ are not all real.
26020
If an polynomial has a complex root, is it necessary that the conjugate of that root is also a root of the polynomial? I am not sure about this. Please comment on this.
19912
This is a question from Putnam and Beyond, and I have a question about the proof. The question is:  Show $x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ is irreducible over $\mathbb{Z}[X]$.  The proof first shows that the polynomial $P(x) = x^{p-1} + 2x^{p-2} + 3x^{p-3} + \dots + (p-1)x + p$ only has zeros of absolute value $&gt; 1$. Then, it assumes it can be factored into two polynomials, namely, $P(X) = Q(X)R(X)$, and since $p = P(0) = Q(0)R(0)$, one of $Q(0), R(0) = \pm 1$, so WLOG $Q(0)$ has a zero of absolute value $&gt; 1$ which contradicts the fact that $P(X)$ only has zeros of absolute value $&gt; 1$. I don't see why $Q(0) = \pm 1$ means that it has a zero with absolute value greater than one. (The following is the proof in question.) &quot;We will show that all the zeros of $P (x)$ have absolute value greater than $1$. Let $y$ be a complex zero of $P(x)$. Then $0 = (y − 1)P(y) = y^p + y^{p−1} + y^{p−2} +···+ y − p.$ Assuming $|y| ≤ 1$, we obtain $p = |y^p + y^{p−1}+ y^{p−2}+···+ y| ≤ \sum_{i=1}^p |y|^ i ≤  \sum_{i=1}^p 1 = p.$ This can happen only if the two inequalities are, in fact, equalities, in which case $y = 1$. But $P (1) &gt; 0$, a contradiction that proves our claim. Next, let us assume that $P (x) = Q(x)R(x)$ with $Q(x)$ and $R(x)$ polynomials with integer coefficients of degree at least $1$. Then $p = P (0) = Q(0)R(0)$. Since both $Q(0)$ and $R(0)$ are integers, either $Q(0) = ±1$ or $R(0) = ±1$. Without loss of generality, we may assume $Q(0) = ±1$. This, however, is impossible, since all zeros of $Q(x)$, which are also zeros of $P (x)$, have absolute value greater than $1$. We conclude that $P(x)$ is irreducible.&quot;
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 $p(z_1)&lt;p(z_0)$.
18508
I am reading "Linear Algebra Done Right". Right now I am stuck with the uniqueness part of Corollary 4.8, Existence and Uniqueness polynomial factorisation in the form of $p(z) = c(z-\lambda_1)(z-\lambda_2)...(z-\lambda_m)$.  It is stated that if there are two factorisations of $p$, $(z-\lambda_1)(z-\lambda_2)...(z-\lambda_m) =(z-\tau_1)(z-\tau_2)...(z-\tau_m)$. We can divide both sides by $(z-\lambda_1)$ as one of the $\tau$ must equal $\lambda_1$.  Here comes the confusing part for me: After division, we get $(z-\lambda_2)...(z-\lambda_m) = (z-\tau_2)...(z-\tau_m)$ for all $z \in C$ except for possibly $z=\lambda_1$. Actually the equation above must hold for all $z \in C$ because otherwise by subtracting the right side from the left side we would get a nonzero polynomial that has infinitely many roots.  I understand a finite polynomial can't have infinitely many roots, but need elaboration on bolded part, like why it suddenly(?) holds for $z=\lambda_1$, and how comes the theoretical nonzero polynomial has infinitely many roots.
13629
Find all entire functions $f(z)$ such that $|f(z)|=1$ for $|z|=1$  Hint: First show that $f(z)$ is a polynomial.  Clearly one can not use Cauchy Estimates to prove that $f(z)$ is a polynomial, the other way is to prove that $f(z)$ has a pole at infinity which I am not sure how to prove that, are there any other ways to prove that a given entire function is a polynomial ?  Thank you !
102611
Prove that there is no polynomial with integer coefficients $P(x)$ such that $P(7)=5$ and $P(15)=9$.  In general; how to know if there exists a polynomial with integer coefficients $P(a)=b$ and $P(c)=d$, where $a,b,c$, and $d$ are all integers?
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.
52164
Let the degree of the polynomials $P(z)=a_0+a_1z+a_2z^2+\cdots+a_nz^n$   $a_n\neq0$ and $Q(z)=b_0+b_1z+b_2z^2+\cdots+b_mz^m$ $b_m\neq 0$ be such that $m\geq n+2$. Show that   if all the zeros of $Q(z)$ are interior to a simple closed contour C,   then $$\int_C\frac{P(z)}{Q(z)} \, dz=0$$   I am lost on how to show it $P(z)$ and $Q(z)$ are polynomials, then the ratio is analytic in all the points except where $Q(z)=0$. If $\frac{P(z)}{Q(z)}$ be analytical in all the points on the contour or inside, so I could tell straight that the integral is zero, but that is not the case.
48676
Question: For every polynomial $P(x)$ of degree at n we have $$\sum_{i=0}^{n+1}(-1)^i\binom{n+1}{i}P(i)=0$$  Well I was inducting on degree of the polynomial, for $n=0$ its true and I got a bit confused in the inductive step, actually I cant figure the inductive step.   Is this property (if true) a famous one?  Thank you for your help.
115324
Consider the polynomial   $$ P(x) = x^{n} + a_{n-1}x^{n-1} + ... +a_{1} x + a_{0} $$  with only real roots and coefficients. If $R$ is the smallest bound such that $|x_{i}| \le R$ (it is the bound for the largest root), $P(x_{i})=0, i=0,1,...n-1$, then for every conjugate $\frac{1}{p} + \frac{1}{q} = 1, p,q &gt;1$, prove that  $$ R &lt; \left( 1 + A_{p}^{q} \right)^{1/q}, A_{p} = \left( \sum |a_{i}|^{p} \right)^{1/p}$$    Attempt  We have   $$ P(x_{0}) = P(x_{1}) = ... = P_(x_{n-1}) = 0 $$  sum all of them to get  $$ nx^{n} + a_{n-1} \sum x_{i}^{n-1} + ... + a_{1} \sum x_{i} + n a_{0} = 0$$    Let us start from easy case, $n=2$:  with only two roots $x_{0},x_{1}$ then we have $-x_{0} - x_{1} = a_{1}$, and $ a_{0} = x_{1} x_{0}$. We wish to prove   $$ R &lt;  \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p}   \right)^{q/p} \right]^{1/q} $$  ... Clearly we must have $R = max(|x_{1}|, |x_{0}|)$, WLG letit be $|x_{0}|$. Let us work backwards:  $$ |x_{1}+a_{1}| + |\frac{a_{0}}{x_{1}}| &lt; 2\left( 1 \right)^{1/p} \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p}   \right)^{q/p} \right]^{1/q} $$  $$ |x_{1}+a_{1}| + |\frac{a_{0}}{x_{1}}| &lt; \left( (2^{q})^{p} + (2^{q})^{p} \right)^{1/p} \left[1 + \left( |x_{1}x_{0}|^{p} + |x_{1} + x_{0}|^{p}   \right)^{q/p} \right]^{1/q} $$  How to continue..?
40105
Let $P(z)=a_0+a_1z+\cdots+a_nz^n$ be a polynomial whose coefficients satisfy $$0&lt;a_0&lt;a_1&lt;\cdots&lt;a_n.$$  I want to show that the roots of $P$ live in unit disc. The obvious idea is to use Rouche's theorem, but that doesn't quite work here, at least with the choice $f(z)=a_nz^n, g(z)=$ (the rest).  Any ideas?
64532
I have $P(x)$ a polynomial with degree $n$ ,$P(x) \ge 0$ for all $x \in$ real.  I have to prove that:     $f(x)=P(x)+P'(x)+P"(x)+......+P^{n}(x) \ge 0$ for all $x$.   I tried different methods to solve it but I got stuck.Any suggestion or advice is welcomed.
90180
Are there some books or articles regarding "polynomials" with non-integer (real) exponents, i.e., $$f(x)=a_1x^{e_1}+a_2x^{e_2}+\dots+a_nx^{e_n},$$ where $e_1,e_2,\dots$ are any real numbers (and $x$ being also real)?  I am mostly interested in theorems regarding the location of the roots, number of the roots, bounds, etc., of such "polynomials". Thanks.
55879
The following is an exercise (Exercise #3 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics.  Let $P(x)$ be a nonzero polynomial with real coefficients.  Show that the following two conditions are equivalent:  There exists a nonzero polynomial $Q(x)$ with real coefficients such that all coefficients of $P(x) Q(x)$ are nonnegative.  There does not exist a real number $a &gt; 0$ such that $P(a) = 0$.    That the first item implies the second is straightforward (since if $a$ is a positive real root of $P(x)$ then it is a positive real root of $P(x)Q(x)$, but since $P(x)Q(x)$ is nonzero with all coefficients nonnegative, this is impossible).  However, I can't seem to find a way to prove that the second item implies the first.  I would very much like to see a proof if someone can provide it.
88800
Let $P\in\mathbb{R}[X]$ be a positive polynomial.  I want to show that there exists $A,B\in\mathbb{R}[X]$ so that $P=A^2+B^2$    $\displaystyle P=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^{2p}(x-b_k)=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^p(x^2-2\Re(b_k)+|b_k|^2)$ where $a$ is the leading coefficient, $a_k$ are the real rooks, $b_k$ are the complex roots.  What can i do now ? I know that the identity $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ should come in handy, but I can't figure out :  1) How to show that the product real roots is a square  2) How to use the above identity for the complex roots.  If I can solve those two points, then I can solve the problem.
45145
Is it true that  $$P_n(x):=1+ \sum_{m=1}^n\dfrac{x^m}{m!}$$ has no real root for even $n$ and exactly one real root for odd $n$? I can only prove that the polynomial cannot have any multiple roots. Am stuck , Please Help.
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?