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de Branges's theorem / Bieberbach's conjecture says that if $f$ is injective on the unit disc then $|a_n| \leq n$. Then if $f$ is injective on the ball of radius $r$ than by rescaling $|a_n| \leq n /r …

Is your function entire? An entire algebraic function is just a polynomial function. So you know that it's a sum of terms of that type. (Unless it might have poles, in which case it's a rational fun …

No. Take a convex set and use an exponential function $e^{2\pi i z/c}$, where $c$ is the difference between two points in the set. The image of the set is no longer simply connected, and thus cannot b …

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 val …

As SashaP already pointed out, this is false. In fact it fails very badly: The space of non-constant holomorphic maps from $X$ to $\operatorname{Sym}^2 Y$ can already contain infinitely many connected …

The $k$-Hodge conjecture is false for any $k$ containing an irrational real $\alpha$. Indeed we may clearly assume $\alpha$ is negative. Consider the elliptic curve $E = \mathbb C / \langle 1, \sqrt{ …

Residues satisfy $$\operatorname{res}_x(fdg) + \operatorname{res}_x(gdf) = \operatorname{res}_x(d (fg)) =0$$ which tells you that they are antisymmetric, and many antisymmetric pairings in math arise …

This is not a solution, but could be partially helpful in one. Let $n$ be the number of zeroes of $p$. Then a counterexample occurs if and only $f^{(k)}$ for some $k$ has a zero of order $k(n+1)$. P …

The way you would want to produce that vector bundle is gluing the trivial vector bundle $\mathbb H \times \mathbb C^r$ together along the maps: for each $g \in \Gamma$, the map sending $(x,y)$ to $( …

$$\int_x^\infty \{ u\} u^{-s-1} du = \int_x^\infty \left( \{ u\}-\frac{1}{2}\right) u^{-s-1} du + \frac{1}{2} \frac{x^{-s}}{s}$$ and by integration by parts, $$\int_x^\infty \left( \{ u\}-\frac{1}{2} …

If you know them as algebras over $M(SL_2(\mathbb Z))$, almost exactly the procedure you describe works. Since the modular forms with weight a multiple of $12$ are sections of powers of an ample line …

Let $\Gamma$ be a non-algebraic analytic curve. Let $z_1,\dots, z_n$ be points of $\Gamma$. Let $X_{n,d}$ be the set of algebraic curves of degree $d$ that contain $z_1,\dots, z_n$. Now consider a ran …

1 Sort of. It's not totally obvious what "the eigenvalues of the curvature" are because the curvature is not a matrix, but a matrix valued in 2-forms. Also, there's the $1/2\pi i$ thing you mention. O …

Here is a way to construct such threefolds. Take $E \to \mathbb P^1$ a non-isotrivial elliptic surface. Take $S$ another surface, say a general type surface. Any rational function on $S$ gives a ratio …

We have $$ \frac{d}{dx}\left ( f(x) e^{h(x)}\right) = \left( \frac{ f'(x)}{f(x)} + h'(x) \right) \left( f(x) e^{h(x)}\right).$$ Thus $$ \sum_n a_n n x^{n-1} = \left( \frac{ f'(x)}{f(x)} + h'(x) \rig …