A stable polynomial is one with zeros in the upper half plane or lower half plane. Interlacing polynomials are polynomials with only real zeros, where between every two zeros of one polynomial lies a zero of the other polynomial, in the sense that they can be ordered from least to greatest. Two interlacing polynomials automatically […]

$$\int_{\gamma(0;2)}\frac{e^{i\pi z/2}}{z^2-1} \, dz$$ Using the residue calculus i got $$-2\pi$$But the answer is $$=i$$ I must be wrong at this, but shouldn’t the answer have $\pi$ at least since the integral already requires $2\pi i \cdot\text{residue }$?

How to show the transition between $(4)$ to $(5)$ in here?

The question suggests that I proceed by contradiction, and exploit the fact that $ |z| $ is nowhere differentiable (rather than using C-R equations). I’m struggling to find a way to find a contradiction. Any help would be appreciated. Thanks

I need to prove this equality of integrals…but i dont know how to begin, so if anyone can give an idea… Let f a continuous function on $\overline{D}=\{z : |z|\leq 1\}$. Then: $$\overline{\int_{|z|=1} f(z) dz} = – \int_{|z|=1} \frac{\overline{f(z)}}{z^2} dz$$ And then ihave to calculate that $$\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}dz$$ for every a.

Let $f$ be a meromorphic function in a neighborhood of the closed unit disk $\bar{\mathbb{D}}$. Suppose that $f$ is holomorphic in $\mathbb{D}$ and $$ f(z) = \sum_{n=0}^\infty a_n z^n $$ for $z \in \mathbb{D}$. Prove that if $f$ has a pole on the unit circle $\mathbb{T}$ then the above power series diverges at any $z […]

Euler showed: \begin{equation} B_{2 k} = (-1)^{k+1} \frac{2 \, (2 \, k)!}{ (2 \, \pi)^{2 k}} \zeta(2 k) \end{equation} for $k=1,2, \cdots$. We could from here find $\zeta(2k)$ in terms of the even Bernoulli coefficients $B_{2k}$. How can we derive the equivalent representation by using the Euler Maclaurin formula ? Thanks. Update Here is what […]

I’m working in a problem that involves the equation $$ w(z)=\sqrt{1-z^{2}} \,\, . $$ I already know that there’re two branch points in this equation, namely $\pm 1$, so there’s a Riemann surface covering the domain of the function where the branch cut is from the $-1$ to $1$, as shown in the figure below. […]

I did the first parts : $$\omega= (\cos \pi + i \sin \pi)^\frac{1}{5} \implies \omega^5 = \cos \pi + i \sin \pi=-1$$ $\omega=-1$ is a root so : $$\omega^5-1= (\omega+1) (\omega^4-\omega^3+\omega^2-\omega+1)=0$$ $$\omega^4-\omega^3+\omega^2-\omega=-1$$ $$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} – (\cos \frac{4\pi}{5} + i \sin \frac{4\pi}{5})$$ $$\omega-\omega^4=\cos \frac{\pi}{5} + i \sin \frac{\pi}{5} – (-\cos \frac{\pi}{5} + […]

Could any one tell me precisely how to compute the automorphism group of an annulus say $r<|z|<R$? Thank you!

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