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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 ...
Alexander Kalmynin's user avatar
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)...
Peter Mueller's user avatar
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 ...
Peter Mueller's user avatar
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 ...
Will Sawin's user avatar
  • 126k
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|...
Iosif Pinelis's user avatar
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 ...
Echo's user avatar
  • 1,328
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 ...
Thomas Kurbach's user avatar
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 $f(z) = z^3 - 1$.
Sam Nead's user avatar
  • 22.9k
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 ...
M.G.'s user avatar
  • 6,469
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 $2$-...
Gael Meigniez's user avatar
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 ...
GH from MO's user avatar
  • 90.6k
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 ...
Sébastien Loisel's user avatar
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 ...
Iosif Pinelis's user avatar
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 ...
Jacques Carette's user avatar
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) ...
Chad Brown's user avatar
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 \...
an_ordinary_mathematician's user avatar

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