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6

HINT: $$2y=-\sqrt{x^2+y^2}$$ But $\displaystyle\sqrt{x^2+y^2}\ge0\implies y\le0$


6

Absolute convergence in the right half-plane (standard for all Dirichlet series, but if you're unsure, you can compare via $$\zeta(\sigma)<\zeta(2)\quad\forall \sigma >2$$ by just comparing term-by-term) Then you take the limit inside, giving you $$\lim_{\sigma\to\infty}\zeta(\sigma)=\sum_{n=1}^\infty ...


5

Hint: $$ \frac{d}{dt}\tan^{-1}(t) = \frac{1}{1+t^2} $$ $$ d f(x,y) = \frac{\partial f(x,y)}{\partial x} dx + \frac{\partial f(x,y)}{\partial y} dy $$


5

For real $x>1+\frac{1}{n-1}$ then $$x^n = (1+(x-1))^n> 1+n(x-1) = x+(n-1)(x-1)> x+1$$ Now if $z^n+z+1=0$ then $\|z\|^n = \|z+1\| \leq \|z\|+1$. So, with $x=\|z\|$, we have $x^n\leq x+1$, so $\|z\|=x\leq 1+\frac 1{n-1}$.


5

Proof of (*) Adding the four finite sums, $$\sum_{k=1}^{n}H_{k}=(n+1)H_{n}-n,$$ $$\sum_{k=1}^{n}\ln{n}=\ln{n!},$$ $$\sum_{k=1}^{n}\gamma=\gamma\,n,$$ $$\sum_{k=1}^{n}\frac{1}{2k}=\frac12H_{n},$$ gives us a representation of the $n$-th partial sum for the infinite series. Writing the infinite series as the limit of partial sums, we get: $$\begin{align} ...


4

Let $w_1, w_2 \in \Bbb C - \gamma$. We have \begin{align} \left|H(w_2) - H(w_1)\right| &= \left|\int_\gamma \frac{h(z)(z - w_1) - h(z)(z - w_2)}{(z - w_2)(z - w_1)}dz\right| \\ &= \left|\int_\gamma \frac{h(z)(w_2 - w_1)}{(z - w_2)(z - w_1)}dz\right| \\ &= \left|\int_\gamma \frac{h(z)}{(z - w_2)(z - w_1)}dz\right| \left|w_2 - w_1\right|. \tag{*} ...


4

Let $w = \frac{z+1}{z-1}$. Then you have a power series in $w$, centered at $0$. Find its radius of convergence, call that $R$. Then find which $z$ correspond to $\lvert w\rvert < R$. The map $z \mapsto \frac{z+1}{z-1}$ can be explicitly inverted.


3

Think of a limit $$\lim_{n\to \infty} \sum_{k=1}^{\infty}\frac1{k^n}=\lim_{n\to \infty}\left(1+\frac1{2^{n}}+\frac{1}{3^n}+\ldots\right)=1$$ Each member of the series goes to $0$ except the first one.


3

In the big Montel theorem, and often otherwise, a normal family is a family $\mathcal{F}$ of meromorphic or holomorphic functions, such that every sequence $(f_n)$ of functions in $\mathcal{F}$ has a subsequence that converges uniformly on compact sets to a meromorphic/holomorphic function, or to the constant function $z \mapsto \infty$. Your $(f_n)$ ...


3

No, there is no such hope. For a simple example, take $f(z)=1/z$. In fact, an annulus is a domain of existence for $H^\infty$ as well as $A$, i.e. there exists bounded resp. continuous up to the boundary holomorphic functions on the annulus which can't be extended holomorphically across any boundary point. More concretely, let $g$ be your favorite ...


3

Notation first: $\mathfrak{S}_0(\mathbf{f}) \neq \mathfrak{G}_0(\mathbf{f})$. It's $S_0$, not $G_0$. Not that it matters, of course. What matters is that Ahlfors here isn't concerned with abstract Riemann surfaces, but with very special Riemann surfaces that come with a nice projection to $\mathbb{C}$. What does he mean, saying $\theta: G_0(f) = ...


3

Hint: Consider $w=z-1$. The integral becomes $$ \oint_{|w-i|=5/4}\frac{\log(1+w)}{w^2}\mathrm{d}w $$ Recall the power series for $\log(1+w)=w-\frac{w^2}{2}+\frac{w^3}{3}-\dots$ Since $w=0$ is inside the contour, we can apply the Residue Theorem. Comment: There is nothing wrong with your approach using Cauchy's Integral Formula. We are both actually ...


3

By the root test we have $$\left|2^n\frac{(4z-8)^n}n\right|^{1/n}\xrightarrow{n\to\infty}2|4z-8|<1\iff|z-2|<\frac18\iff z\in B\left(2,\frac18\right)$$ so the radius is $\frac18$.


3

Let $ \displaystyle f(z) = \frac{\log z}{z^{2}+z+1}$ and integrate around a keyhole contour where the branch cut for $\log z$ is placed on the positive real axis. As the radius of the little circle goes to $0$ and the radius of the big circle goes to $\infty$, $ \int f(z) \ dz$ will vanish along both circles. You can use the ML inequality to show this. So ...


3

Look at $e^{1/z}$ around $z = 0$. This function approaches every point in the complex plane (take the limit along the real axis, positive or negative or along the imaginary axis). Dividing by and adding a meromorphic function does not change this, so there is no singularity that can be characterized in the normal way with Laurent series, residues, etc. This ...


2

Here are a couple of tricks and general plans of approach I know: If $x\in\Bbb R$ and for all $\epsilon>0$ we have that $|x|\leq\epsilon$, then $x=0$. I think of this fact as being Real Analysis in a nutshell. Never forget that if $A\subseteq\Bbb R$ is bounded above and $M=\sup A$, then for all $\epsilon>0$ there is a $y\in A$ such that $M< ...


2

$$\begin{align} \frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy &=\frac{1}{(\frac{x}{y})^2+1}\frac{1}{y}dx+ \frac{1}{(\frac{x}{y})^2+1}(-\frac{x}{y^2}) dy\\ &=\frac{1}{(\frac{x}{y})^2+1}(\frac{1}{y}dx+(-\frac{x}{y^2})dy)\\ &=\frac{1}{(\frac{x}{y})^2+1}d(\frac{x}{y})\\ &= d\left(\tan^{-1}\left(\frac{x}{y}\right)\right) \end{align}$$


2

Your ratio is incorrect. You should have: $$\frac{c_n}{c_{n+1}}=\frac{2^{3n}(n+1)}{2^{3(n+1)}n}=\frac{2^{3n}(n+1)}{2^{3n+3}n}=\frac{n+1}{2^3n}$$


2

Some hints that might help you along the way: Start off by writing the expression as $$ \oint_{C} \frac{z^4}{1 + z^8}dz $$ where $C$ is the contour. Now find the residues: $$ 1 + z_n^8 = 0, z_n \in C $$ Then the integral value is $$ 2 \pi i \sum Res_{z=z_n}f(z) $$


2

You might be interested in an alternative method using the Beta function. Note that the integrand is even, so \begin{align} \int_{-\infty}^{\infty}\frac{x^4}{1+x^8}dx &=2\int_{0}^{\infty}\frac{x^4}{1+x^8}dx\\ \end{align} Use the substitution $x=u^{1/8}$, then $dx=\frac{1}{8}u^{-7/8}du$. The integral becomes \begin{align} ...


2

Hint: For the radius of convergence, a useful criterion here is the "up to the first singularity" criterion.


2

What you are missing is that $$f''(a) = \oint_{\partial C_1(0)} \frac{\sin^2 z}{(z-a)^3}\,dz$$ only gives you the second and higher derivatives of $f$, you have $$(f(z) - \pi i\sin^2 z)'' \equiv 0.$$ So the difference is a polynomial of degree $\leqslant 1$, $$f(z) = \pi i\sin^2 z + a + bz.$$ The initial conditions yield $f(0) = a$ and $f'(0) = b$. ...


2

Because $f$ is continuous on $D$, $f$ is bounded near $p$ and hence $p$ is a removeable singularity (see here for instance: http://en.wikipedia.org/wiki/Removable_singularity). Because of this, Goursat's theorem and the Residue Theorem give the same result.


2

The assumption implies that $\operatorname{Im}(f)$ is zero on the bounding curve, and therefore vanishes on the interior by the maximum modulus principle. From this it follows that $f$ is constant. At worst you have a technical hiccup with your particular curve being only piecewise smooth.


2

I don't know if that is Cauchy's method, but one way to establish $g\equiv 0$ without differentiating is to use a functional equation: $$2\pi \cot (2\pi z) = \pi \frac{\cos^2(\pi z) - \sin^2(\pi z)}{\sin (\pi z)\cos (\pi z)} = \pi \cot (\pi z) + \pi\cot \left(\pi \left(z+\tfrac{1}{2}\right)\right),$$ and with $$s(z) = \frac{1}{z} + \sum_{n\neq 0} ...


1

Okay, I think I've got it. The solution to your integral is likely $$f(z)=\sqrt{\pi}e^{z^2}(\operatorname{sgn}(\operatorname{Re}(z)) - \operatorname{erf}(z) ).$$ To see this, let first $\operatorname{Im}(z)=:v> 0$ and denote $z=u+iv$. First, note that $$\int_{-\infty}^\infty \frac{e^{-t^2}dt}{t+z}= e^{-z^2} \int_{-\infty+iv}^{\infty+iv} ...


1

The last step (before the limit $r\to1$ is taken) of the proof shows $$ \sum_{n=0}^\infty nr^{2n}|a_n|^2\leq\frac{1}{r^2} $$ for each $r\in(0,1)$. Now for the left hand side we have $$ \sum_{n=0}^Nnr^{2n}|a_n|^2\leq\sum_{n=0}^\infty nr^{2n}|a_n|^2\leq\frac{1}{r^2} $$ for each $N\in\mathbb N$, since the coefficients of the sum are non-negative. For fixed ...


1

Use that $$ |\operatorname{Log}z|=|\log|z|+i\operatorname{arg}z|\le\sqrt{(\log|z|)^2+\pi^2}. $$ I am assuming that $-\pi<\operatorname{arg}z\le\pi$.


1

Note that $$ \sqrt{(b-x)(x-a)}=\sqrt{\left(\frac{b-a}{2}\right)^2-\left(x-\frac{b+a}{2}\right)^2}\tag{1} $$ Therefore, we can substitute $$ x=\frac{b+a}{2}-\frac{b-a}{2}\cos(\theta)\quad\text{so that}\quad\sqrt{(b-x)(x-a)}=\frac{b-a}{2}\sin(\theta)\tag{2} $$ Furthermore, by the definition of $y$, $$ ...


1

$P_N(z)$ converges uniformly towards $e^z$ over any compact subset of $\mathbb{C}$, and the Residue Theorem grants that: $$\oint_{|z|=2}\frac{dz}{P_N^3(z)-1}=2\pi i\cdot\sum_{\xi\in Z} \operatorname{Res}\left(\frac{1}{P_N^3(z)-1},z=\xi\right),$$ where $Z$ is the set of zeroes of $P_N^3(z)-1$ that lie inside the disk $|z|\leq 2$. Since the zeroes of ...



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