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0

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} ...


0

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 $$ 2 \pi i \sum Res_{z=z_n}f(z) $$


0

Since $f$ has a simple pole in $z_0$, $$\lim_{z\to z_0}(z-z_0)\,f(z) = C, $$ hence by replacing $f$ with its Taylor series in zero, $$\lim_{z\to z_0}\left(-a_n z_0 + \sum_{n=0}^{+\infty} (a_n-z_0 a_{n+1})\, z^{n+1}\right)=C.$$ This gives that $\{(a_n-z_0 a_{n+1})\,z_0^{n+1}\}_{n\in\mathbb{N}}$ is a Cauchy sequence, from which: $$\forall ...


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 ...


1

Observe that $$ H_{n}-\ln n-\gamma -\frac{1}{2n} = \psi (n) - \ln n + \frac{1}{2n} $$ where $\psi := \Gamma'/\Gamma$ is the digamma function, using $\displaystyle \psi (n)= H_{n-1}-\gamma = H_n-\gamma- \frac{1}{n}$, $n\geq 1$. Our initial series thus rewrite $$ \sum_{n=1}^{\infty} \left( \psi(n )- \log n + \frac{1}{2n}\right) = \frac{\gamma}{2} - ...


-3

Please try value as 4/[sqrt(φ)]= [3.14460551..]. Attach links of my proof, based on a novel method based on the "quadrature triangle": - http://www.stefanides.gr/Html/QuadCirc.htm http://www.stefanides.gr/pdf/BOOK%20_GRSOGF.pdf Regards from Athens, Panagiotis Stefanides


0

Assume that such an $f$ exists, and consider the function $$g(z):=\left({f(z)\over z}\right)^2\qquad(z\in\dot U)\ ,$$ where $U$ is a sufficiently small disk with center $0\in{\mathbb C}$. Since $\lim_{z\to0}{g(z)\over z}=1$ it follows that $g$ is in fact analytic in all of $U$, and that there is an analytic function $h:\ U\to{\mathbb C}$ with ...


1

If the order of the zero of f at $0$ is two, then you can write a Taylor expansion at zero, $$ f= \sum _{n=0}^{\infty} {a_n z^{n+2}}$$ and factor out $z^2$ so that $$f=z^2(a_0+a_1z+...)$$ and thus $$f^2=z^4(a_0+a_1z+...)^2$$ and then the limit divided by $z^3$ cant be one. Same goes if the order of zero is higher than two.


1

By Taylor expansion about $z_0$, you can formally compute the limit, but it is far from elegant. Write \begin{align} f(z) = \frac{W'(z)W'(z_0) - \left[\frac{W(z)-W(z_0)}{(z-z_0)}\right]^2}{[W(z)-W(z_0)]^2}. \end{align} The Taylor expansion \begin{equation} W(z)-W(z_0) = W'(z_0)(z-z_0) + \frac{1}{2}W''(z_0)(z-z_0)^2 + \ldots \end{equation} gives ...


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 $f'(0) = b$. ...


0

Hint: $$\lim_{z\to0}[z^{-3}f^2(z)]=1 \implies f(z)=\frac{1}{z^{3/2}}\frac{1+a_1z+a_2z^2+\cdots}{1+b_1z+b_2z^2+\cdots}$$


0

Harmonic functions share many of the properties you also have for analytic functions (see section properties of harmonic functions of the Wikipedia article for harmonic functions). Because any function fulfilling the Cauchy-Riemann equations is harmonic, you can have a look at the proofs for the properties of harmonic functions to have an alternative way to ...


1

I've recently come across this integral, and I think that your solution is wrong. If I am right, it holds that $f(z)=-\sqrt{\pi} e^{z^2} \text{erf}(z)$. To see this, note that $$f(z)= -\frac{i}{\sqrt{\pi}} e^{z^2} \int_{-\infty}^\infty \frac{e^{-t^2-2izt}dt}{t}=-\frac{i}{\sqrt{\pi}} e^{z^2} I(z)$$ via substitution. Using the fact that $\int_{-\infty}^\infty ...


6

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


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

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 $\frac18$.


2

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


1

The exponential factor is a scalar valued function. When we have a scalar valued function $\phi$ and a vector valued function $F$, the divergence satisfies the following property: $$\nabla\cdot(\phi \vec{F}) = \nabla(\phi)\cdot \vec{F} + \phi (\nabla \cdot \vec{F})$$ In our case, we have $$\nabla \cdot (e^{i(\vec{k} \cdot \vec{x}-wt)}\vec{E}) = ...


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< ...


1

Using the Taylor series of the $\sin$ function we see that $$f(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n},\quad \forall x\in\Bbb R$$ so the option (d) is correct.


0

Hint: $$\int_{a}^{b}\frac{f'(\theta)}{f(\theta)} \, d\theta = \ln(f(\theta))\Big|_{\theta = a}^{\theta=b} .$$


0

I want to show $\lim_{r \rightarrow 1^-}-\pi \sum_{-1}^\infty n r^{2n} |a_n|^2 = -\pi \sum_{-1}^\infty n |a_n|^2$ Either $-\pi \sum_{-1}^\infty n |a_n|^2$ is finite, or it is equal to $-\infty$. In the first case, we are done, by Abel's theorem. In the second case, there is a more general version of Abel's theorem that says $\lim_{r \rightarrow 1^-}-\pi ...


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 ...


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 ...


1

Hint:$$x^2+x+1=(x+1/2)^2+3/4$$


1

Yes, if so, you can write this as: $$|x|\ge c|y|$$ in particular, in the first quadrant you miss everything from that line over to the imaginary axis, so you can do $${1\over f(z)-i}$$ which is also entire since $f(z)\ne i$ for any $z\in\Bbb C$. But then this is a bounded entire function whether $f$ was or not, and by Liouville it is constant. ...


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} ...


0

It's more natural to think of these functions as defined on the unit circle in the complex plane. Then we look at the completion of the linear span of the functions $z^n$, $n=0,1,2,3\dots$. Of course the completion depends on the norm we use. If the uniform norm is used, we get the Disk Algebra. If $L^p$ norm is used, we get the Hardy space $H^p$. If by ...


1

You pretty much solved it already: $g_n(z)=z+na_2z^2 + \ldots$, so $g''_n(0)=2na_2$, but by Cauchy's formula this quantity is bounded. More explicitly, if $\overline{D}_R(0)\subset U$, then $$ g''_n(0) = \frac{1}{i\pi} \int_{\partial D_R} \frac{g_n(w)}{w^3}\, dw , $$ and this is $\lesssim 1/R^2$ since, by assumption, $g_n(w)$ takes values in the bounded set ...


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 ...


0

Yes, thats true, because you are able to differentiate (inside the convergence-disc) any power-series "term by term" i.e. $$f'(z)=\left(\sum_{k=0}^{\infty}a_kz^k \right)^{'}=\left(\sum_{k=0}^{\infty}ka_kz^{k-1} \right) $$


1

It is essentially because the part $e^{\frac{1}{z}}=\sum_{n=0}^{\infty}\frac{\frac{1}{z^n}}{n!}=\sum_{n=0}^{\infty}\frac{1}{z^n\cdot n!}$ attains infinity times often addends in the form $\frac{a_k}{z^k}$ (an the other parts are holomorphic or liftable in $0$, in a neighborhood of $0$) so the principal part has infinity many addeds. Thus $z_0=0$ as an ...


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 ...


1

He mentions that $Z(t) = e^{it}.$ You have $$|Z(t)| = |e^{it}| = \sqrt{\cos^2(t) + \sin^2(t)} = 1.$$


2

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


-1

Using the Hint from above: \begin{align} d\left(\tan^{-1}\left(\frac{x}{y}\right)\right) &= \frac{\partial \tan^{-1}\left(\frac{x}{y}\right)}{\partial\left(\frac{x}{y}\right)}\frac{\partial\left(\frac{x}{y}\right)}{\partial x} dx + \frac{\partial ...


2

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


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) = ...


5

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


0

We assume that $|f(r)|>0$ otherwise $f=0$ and the rest is trivial. Define $g(z)=f(z)/f(r)$. Therefore $g:D\to D$ is analytic, so we can apply Schwartz's lemma. This tells us that $|g(z)|\leq |z|$ on $D$ and we have $|g(r)|=1$. This can only hold if $|r|=1$. Let us write $r=\mathrm e^{\mathrm i\rho}$. As we have the equality $|g(z)|=|z|$ for $z=r$, the ...


0

If the derivative were ever negative, we would contradict the maximum principle: increase r slightly and you get a boundary with smaller modulus than in the interior. If it were ever zero, then by Rolle's theorem (the one from calculus) there are distinct $r_1,r_2$ with $f(r_1)=f(r_2)$. This implies $f$ is constant by the maximum modulus principle. Ask ...


0

A non-essential singularity is either a pole or removable. In either case, we can write $$f(z) = (z-a)^m\cdot g(z)$$ with a holomorphic function $g$ in a neighbourhood of $a$ with $g(a) \neq 0$ and a uniquely determined $m\in\mathbb{Z}$, unless $f \equiv 0$. You can read that off for example from the Laurent series of $f$. Then for any $k > -m$ we have ...


0

Use that $${{\rm Log}\,z\over z^2}=-{d\over dz}{{\rm Log}\,z+1\over z}\qquad\bigl(|{\rm Arg}\,z|<\pi\bigr)\ .$$ Cutting $|z|=R$ up at the point $-R\in{\mathbb C}$ and passing to the limit this allows to represent the exact value of the integral as a difference: $$\int_{\partial D_R}{{\rm Log}\,z\over z^2}\>dz=-{{\rm Log}\,z+1\over ...


0

Just show that any point on the line of reflection $r$ is mapped to a different point on $r$. Now, any other line $l$ intersecting $r$ can't be mapped to itself, since the image intersection point $A$ is on $r$ bot not equal to $A$, so is not on $l$. Thus, any line that is mapped to itself must be parallel to $r$. Any such line not equal to $r$ is not mapped ...


0

Note that $ \begin{align} \biggr|\frac{\log z}{z^{2}}\biggr| = \biggr|\frac{\log|z| + i\operatorname{Arg}(z)}{z^{2}}\biggr| &\leqslant\frac{\bigr|\log |z| + i\pi\bigr|}{|z^{2}|} \\ &= \frac{|\log R + i\pi|}{R^{2}} \\ &\leqslant \frac{\sqrt{2\log^{2} R}}{R^{2}} \\ &= \sqrt{2}\frac{\log R}{R^{2}} \end{align} $ Thus we have $ ...


0

For $m\geqslant 0, f(z) = (z - z_{0})^{m}$ is analytic on $\mathbb{C}$. Thus the line integral is 0 ( by FTOC ). For $m = -1, f(z) = 1$ is analytic on the closed disc $\rvert z - z_{0}\rvert\leqslant R$, so by Cauchy's integral formula this implies that $$1 = \frac{1}{2\pi i}\int_{\rvert z - z_{0}\rvert = R}\frac{1}{z - z_{0}}\,\mathrm{d}z.$$ Therefore ...


0

For the image of the imaginary axis under the Cayley's transformation: $$0 + yi\mapsto\frac{yi - i}{yi + i} = \frac{y - 1}{y + 1}\:\text{ ( the real-axis ) }$$ The map for the real-axis: $$x + 0i\mapsto\frac{x - i}{x + i}\:\text{ ( Note $\rvert\frac{x - i}{x + i}\rvert = \rvert\frac{\overline{x + i}}{x + i}\rvert = 1$, so this is the unit circle centered ...


1

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


0

Expand $\exp(itx)$ using Euler's and note the symmetry of the bounds. Explicitly we have $ \begin{align} I &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(\frac{-x^{2} - 2itx}{2}\right)\,\mathrm{d}x \\ &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left(-\frac{x^{2}}{2}\right)\big(\cos(tx) - i\sin(tx)\big)\,\mathrm{d}x \\ &= ...


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{*} ...



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