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It looks false. Note that $1-\cos x<1+\cos x-2\cos^2 x=\cos x-\cos 2x$ for $|x|\in (0,\pi/2)$. Therefore if say $|2f(t)|\in (\pi/10,\pi/3)$ for almost all $t$, the real part of $\int F^2(t)(1-F^2(t))$ …

Yes it can. If $z$ is a root of $Q(z)$ outside $C$, then $\sum a_k/(z-z_k)=0$, therefore (take the complex conjugate) we get $\sum a_k(z-z_k)/|z-z_k|^2=0$, but all summands in LHS belong to the same h …

Denote $f(z)=z+e^z$. On the real line $f$ is strictly increasing and takes all values, so, it has the unique real root. Also, $f(\bar z)=\overline{f(z)}$, so, other roots are partitioned onto pairs of …

I like the proof with Riemann sums along the arithmetic progressions: since the integral converges and the function $\log |e^{it}-1|$ is piecewise monotone, it equals $$\int_0^{2\pi} \log|e^{it}-1|=\l …

It equals $2a\Re \int_{-\infty}^\infty \frac{\log(t+i)}{t^2+a^2}dt$. We think that $\log$ is defined in the upper half-plane. Then the contour may be closed to an upper halfplane, since integrated fu …

If $\Re z>0$, the series $\sum_n \binom{z}n$ converges absolutely (by Raabe test, for example), thus we may replace $x$ to 0 and just need to check that this sum equals $2^z$. This follows from anothe …

If you have it for $n=2$, just sum up over all pairs $(z_i,z_j)$ with $i<j$ and divide by $n-1$. As for the proof for $n=2$, yours is quite ok for me, and the proof by math110 is especially elegant, …

I think, yes. Without loss of generality, all $b_i$'s and $d$ are equal to 1. Assume that the vector $c=(c_1,\dots,c_n)$ does not lie in a convex hull of vectors $(wa_{i,1},\dots,wa_{i,n})$ for all $i …

Another heuristic explanation. Denote $m_1,m_2,\dots$ all mutually non-equivalent non-zero not-invertible elements of $\mathbb{Z}$ or $\mathbb{Z}[i]$ (equivalent elements are those whose ratio is inve …

For $P(x)=(1-x)f(x)=(x^{n+1}+1)-2x^n$ you may apply a version of Rouché's theorem or argument principle: choose a very small arc $a1b$ of the unit circle so that $a$ is above the real line, and $b$ be …

I am afraid that this is too good to be true. Consider the polynomial $P(z)=2z+2$. There are two points on the unit circle, $\alpha$ and $\bar{\alpha}$, for which $|P(\alpha)|=|P(\bar{\alpha})|=1$. Al …

Let's try to understand in which sense this equality $$\sum_{k=-\infty}^{\infty}k^nz^k=\left(z\frac{d}{dz}\right)^n\sum_{k=-\infty}^{\infty}z^k=\left(z\frac{d}{dz}\right)^n 0=0$$ may be understood. O …

We use a standard notation $[x^n]h(x)$ for a coefficient of $x^n$ in the series of $h(x)$. Assume that $t$ is a real root of $P_n$. First of all, note that $$0=a_0+a_1t+\dots+a_nt^n=t^n[x^n]\frac1{f …

Write the equation in the form $1=x^{-1}+x^{-2}+x^{-3}+x^{-4}+cx^{-5}:=f(x) $. If $|\beta|\geqslant \alpha$, the RHS has absolute value at most $f(\alpha) =1$ with equality if and only if $|\beta|=\al …

Yes, it may be simplified. The text below is not probably the shortest way, but it explains how to calculate many similar sums. We start with $$\sin^2 x=-\frac14(e^{ix}-e^{-ix})^2=-\frac14e^{-2ix}(e^ …