18
votes
Accepted
Can the equation $1+z+z^2=z^n$ for natural $n$ have multiple complex roots $z$?
Suppose that $z$ is our multiple root, $n\geq 4$. Since $z$ is a root of equation $(n-2)z^2+(n-1)z+n=0$, we have $|z|^2=\frac{n}{n-2}$. Indeed, the roots are non-real and conjugate, so the other root ...
17
votes
Can the equation $1+z+z^2=z^n$ for natural $n$ have multiple complex roots $z$?
As I was typing essentially Alexander Kalmynin's argument, his answer popped up ...
Therefore, I now offer a different argument: The roots of $p_n(z)$ are algebraic integers, while the roots of $(n-2)...
7
votes
Accepted
Can the equation $1+z+z^q=z^n$ have multiple complex roots $z$?
The roots are always simple, this follows without further reasoning from the old paper On the irreducibility of certain trinomials and quadrinomials by Ljunggren. Set $f(z)=z^n-1-z^p-z^q$. Then ...
6
votes
Bounding minimal absolute value of a point on a complex algebraic variety
No (for one interpretation of the question). There does not exist a continuous function $b $ from the set of systems of polynomial equations to $\mathbb R^{>0} \cup \{\infty\}$ that takes a finite ...
6
votes
Can the equation $1+z+z^2=z^n$ for natural $n$ have multiple complex roots $z$?
This is a slight modification of Alexander Kalmynin's answer. For $n\ge3$. let $z$ be a multiple root of $p_n$. Then $0=p_n'(z)=1+2z-nz^{n-1}$ and, as shown by Alexander Kalmynin, $|z|>1$. So, $n|z|...
5
votes
Accepted
Log-convexity of determinant
Your claim can be deduced from the case $n=1$.
Let $H$ be the Hilbert space you start with.
Let $H^{\otimes n}=H\otimes\dots\otimes H$ be the $n$-th tensor power. Equip it with the usual inner product ...
4
votes
Accepted
Real analytic subvariety in complex manifold which is complex outside of its singular set
There is a paper by Hans-Jörg Reiffen called "Fastholomorphe Algebren" (1970) where he apparently proves something like this. Here is a link to the springer publication. I have not read the ...
3
votes
Accepted
Jordan curve boundaries of Fatou components
An answer of "no" is provided by the Julia set for Newton's method applied to $f(z) = z^3 - 1$.
3
votes
Accepted
Residue calculation for Eulerian expansion of the cotangent
The residue approach to the partial fraction expansion of $\cot(z)$ is explained in Freitag & Busam's "Complex Analysis", Prop. III.7.13, and probably in other books as well.
Here is a ...
3
votes
Accepted
Finding a real-analytic diffeomorphism
The answer is positive. First, the fact that $U_1$ is simply connected implies that every connected component $S_i$ ($1\le i\le n$) of its smooth boundary $\partial U_1$ is diffeomorphic to the $2$-...
2
votes
Sign of real part and imaginary part zeta function at 1/2-x+iy and 1/2+x+iy
Let $\rho=1/2+iy$ be the first zero of $\zeta(s)$. Then $\zeta'(\rho)$ has nonzero real and imaginary parts, and
$$\zeta(\rho+h)\sim\zeta'(\rho)h\qquad\text{as}\qquad h\to 0.$$
That is, if $x>0$ is ...
2
votes
Sum of holomorphic squares?
I can't answer my own question, as Igor Khavkine points out, it may be too hard. For the paper I'm writing, I was able to find a suitable "square root" to my analytic function. This is what ...
2
votes
Accepted
Is this a valid method of extending convergence of the Maclaurin series for $\frac{x}{x+1}$ from $|x|<1$ to $\Re(x)>-1$?
In \eqref{5}, the inner sum is obviously $x(1-x)^{n-1}$. So, the limit in \eqref{5} (equal $\dfrac x{x+1}$ indeed) exists if and only if $|1-x|<2$.
In \eqref{6}, using the substitution $k=K-n-j$ in ...
1
vote
Dense orbits for a rational map
Even for polynomials, the answer is No: See the question on Smooth Julia Sets for links to papers with more details.
Note that for rational functions, there is an additional case, as sometimes the ...
1
vote
Inequality between coefficients of a polynomial and its supremum
I've recently been looking into this problem. Here's what I've came up with to show the existence of such a $\kappa$. Let me know if there are any issues. (I gave the same answer on the cross post)
...
1
vote
Pair of positive harmonic functions with negative inner product in Drury-Arveson space
I will try to prove that such functions do not exist. Suppose that $f,g$ are positive (pluri)harmonic functions in the pluri harmonic Drury Arveson space $\mathcal{H}DA_d$ such that $ \langle f ,g \...
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