Non-real complex numbers are roots of polynomials with positive real coefficients
$begingroup$
Prove this:
For every $zinmathbb{C}$ with nonzero imaginary part, there exists a polynomial $p(x)=a_0+a_1x+a_2x^2+cdots + a_nx^n$ such that $a_k>0$ for every nonnegative $kleqslant n$, with $p(z)=0$.
For instance, $i$ is a zero of $1+x+x^2+x^3.$
I could reduce the problem to $z=a+i.$ That is, the statement is proved if the following can be shown:
For every $ainmathbb{R},$ there exists a polynomial $p(x)=a_0+a_1x+a_2x^2+cdots + a_nx^n$ such that $a_k>0$ for every nonnegative $kleqslant n$, with $p(a+i)=0$.
But I could not show this. Perhaps, the general statement is easier to prove! Any help will be appreciated.
complex-analysis polynomials
edited Dec 21 '18 at 13:20
Did
248k23224462
asked Dec 21 '18 at 11:12
ArvindArvind
885
$endgroup$
add a comment |
$begingroup$
Prove this:
For every $zinmathbb{C}$ with nonzero imaginary part, there exists a polynomial $p(x)=a_0+a_1x+a_2x^2+cdots + a_nx^n$ such that $a_k>0$ for every nonnegative $kleqslant n$, with $p(z)=0$.
For instance, $i$ is a zero of $1+x+x^2+x^3.$
I could reduce the problem to $z=a+i.$ That is, the statement is proved if the following can be shown:
For every $ainmathbb{R},$ there exists a polynomial $p(x)=a_0+a_1x+a_2x^2+cdots + a_nx^n$ such that $a_k>0$ for every nonnegative $kleqslant n$, with $p(a+i)=0$.
But I could not show this. Perhaps, the general statement is easier to prove! Any help will be appreciated.
complex-analysis polynomials
edited Dec 21 '18 at 13:20
Did
248k23224462
asked Dec 21 '18 at 11:12
ArvindArvind
885
$endgroup$
add a comment |
$begingroup$
Prove this:
For every $zinmathbb{C}$ with nonzero imaginary part, there exists a polynomial $p(x)=a_0+a_1x+a_2x^2+cdots + a_nx^n$ such that $a_k>0$ for every nonnegative $kleqslant n$, with $p(z)=0$.
For instance, $i$ is a zero of $1+x+x^2+x^3.$
I could reduce the problem to $z=a+i.$ That is, the statement is proved if the following can be shown:
For every $ainmathbb{R},$ there exists a polynomial $p(x)=a_0+a_1x+a_2x^2+cdots + a_nx^n$ such that $a_k>0$ for every nonnegative $kleqslant n$, with $p(a+i)=0$.
But I could not show this. Perhaps, the general statement is easier to prove! Any help will be appreciated.
complex-analysis polynomials
edited Dec 21 '18 at 13:20
Did
248k23224462
asked Dec 21 '18 at 11:12
ArvindArvind
885
$endgroup$
Prove this:
For every $zinmathbb{C}$ with nonzero imaginary part, there exists a polynomial $p(x)=a_0+a_1x+a_2x^2+cdots + a_nx^n$ such that $a_k>0$ for every nonnegative $kleqslant n$, with $p(z)=0$.
For instance, $i$ is a zero of $1+x+x^2+x^3.$
I could reduce the problem to $z=a+i.$ That is, the statement is proved if the following can be shown:
For every $ainmathbb{R},$ there exists a polynomial $p(x)=a_0+a_1x+a_2x^2+cdots + a_nx^n$ such that $a_k>0$ for every nonnegative $kleqslant n$, with $p(a+i)=0$.
But I could not show this. Perhaps, the general statement is easier to prove! Any help will be appreciated.
complex-analysis polynomials
complex-analysis polynomials
edited Dec 21 '18 at 13:20
Did
248k23224462
asked Dec 21 '18 at 11:12
ArvindArvind
885
edited Dec 21 '18 at 13:20
Did
248k23224462
asked Dec 21 '18 at 11:12
ArvindArvind
885
edited Dec 21 '18 at 13:20
Did
248k23224462
edited Dec 21 '18 at 13:20
Did
248k23224462
edited Dec 21 '18 at 13:20
Did
248k23224462
248k23224462
asked Dec 21 '18 at 11:12
ArvindArvind
885
asked Dec 21 '18 at 11:12
ArvindArvind
885
asked Dec 21 '18 at 11:12
ArvindArvind
885
885
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We start from the simple observation: if $a< 0$, then
$$
p(z) = z^2 -2az+a^2+1
$$ satisfies the condition. If $ageq 0$, we can find $nin mathbb{N}$ such that $Re[(a+i)^n]< 0.$ Let $b = (a+i)^n$. Then,
$$
(z^{2n} -2Re[b]z^n +|b|^2)(z^n+z^{n-1} +cdots +1)
$$ satisfies the condition.
edited Dec 21 '18 at 13:36
J.G.
27.1k22843
answered Dec 21 '18 at 13:31
SongSong
13.9k633
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We start from the simple observation: if $a< 0$, then
$$
p(z) = z^2 -2az+a^2+1
$$ satisfies the condition. If $ageq 0$, we can find $nin mathbb{N}$ such that $Re[(a+i)^n]< 0.$ Let $b = (a+i)^n$. Then,
$$
(z^{2n} -2Re[b]z^n +|b|^2)(z^n+z^{n-1} +cdots +1)
$$ satisfies the condition.
edited Dec 21 '18 at 13:36
J.G.
27.1k22843
answered Dec 21 '18 at 13:31
SongSong
13.9k633
$endgroup$
add a comment |
$begingroup$
We start from the simple observation: if $a< 0$, then
$$
p(z) = z^2 -2az+a^2+1
$$ satisfies the condition. If $ageq 0$, we can find $nin mathbb{N}$ such that $Re[(a+i)^n]< 0.$ Let $b = (a+i)^n$. Then,
$$
(z^{2n} -2Re[b]z^n +|b|^2)(z^n+z^{n-1} +cdots +1)
$$ satisfies the condition.
edited Dec 21 '18 at 13:36
J.G.
27.1k22843
answered Dec 21 '18 at 13:31
SongSong
13.9k633
$endgroup$
add a comment |
$begingroup$
We start from the simple observation: if $a< 0$, then
$$
p(z) = z^2 -2az+a^2+1
$$ satisfies the condition. If $ageq 0$, we can find $nin mathbb{N}$ such that $Re[(a+i)^n]< 0.$ Let $b = (a+i)^n$. Then,
$$
(z^{2n} -2Re[b]z^n +|b|^2)(z^n+z^{n-1} +cdots +1)
$$ satisfies the condition.
edited Dec 21 '18 at 13:36
J.G.
27.1k22843
answered Dec 21 '18 at 13:31
SongSong
13.9k633
$endgroup$
We start from the simple observation: if $a< 0$, then
$$
p(z) = z^2 -2az+a^2+1
$$ satisfies the condition. If $ageq 0$, we can find $nin mathbb{N}$ such that $Re[(a+i)^n]< 0.$ Let $b = (a+i)^n$. Then,
$$
(z^{2n} -2Re[b]z^n +|b|^2)(z^n+z^{n-1} +cdots +1)
$$ satisfies the condition.
edited Dec 21 '18 at 13:36
J.G.
27.1k22843
answered Dec 21 '18 at 13:31
SongSong
13.9k633
edited Dec 21 '18 at 13:36
J.G.
27.1k22843
edited Dec 21 '18 at 13:36
J.G.
27.1k22843
edited Dec 21 '18 at 13:36
J.G.
27.1k22843
27.1k22843
answered Dec 21 '18 at 13:31
SongSong
13.9k633
answered Dec 21 '18 at 13:31
SongSong
13.9k633
answered Dec 21 '18 at 13:31
SongSong
13.9k633
13.9k633
add a comment |
add a comment |
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