New answers tagged

0

HINT.-$$\frac{c_kx}{(x_k-ix)^2}=\frac{c_k}{x_k-ix}\cdot \frac{x}{x_k-ix}$$ $$\left|\frac{c_kx}{(x_k-ix)^2}\right|=\left|\frac{c_k}{x_k-ix}\right|\cdot \left|\frac{x}{x_k-ix}\right|\le\left|\frac{c_k}{x_k-ix}\right|\cdot\left|\frac{x^2}{x_k^2+x^2}\right|\le \left|\frac{c_k}{x_k-ix}\right|$$


1

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1

Alright, let's take $e^{1/z}=-7$ as an example. Let $w=1/z$; certainly solving $e^w=-7$ is easy enough: $w=\ln(7)+(2k+1)\pi i, \, k \in \mathbb{Z}$. Now, all we do is go back to $z=1/w$. $$z=\frac{\overline{w}}{{|w|}^2}=\frac{\ln(7)-(2k+1)\pi i}{{\ln(7)}^2+{(2k+1)}^2{\pi}^2}, \, k \in \mathbb{Z}$$ Of course, the situation with $e^{1/z}=-1$ is similar and ...


0

The def'n of rectifiable path, whatever it is in your source, implies that $\{\gamma\}$ is compact. Let $A$ be a non-empty compact subset of the open set $B. $ For every $x\in A$ there is an open ball $B(x,r_x)$ of radius $r_x>0$ such that $B\supset B(x,r_x).$ Now $\{B(x,r_x/2):x\in A\}$ is an open cover of $A, $ so there exists a non-empty finite $A^*\...


1

To show this function is analytic, you need to show that the Cauchy Riemann equations are valid; equivalently, the differential operator below satisfies $$ \frac{\partial g}{\partial \bar z} = \frac{1}{2}\bigg( \frac{\partial g}{\partial x} - \frac{1}{i}\frac{\partial g}{\partial y} \bigg) = 0.$$ If you are unfamiliar with this very useful formulation of ...


1

Since $b_m\neq 0$, the fraction $$\frac{a_0 z^{m-2}+a_1 z^{m-3}+a_2 z^{m-4}+\cdots +z_n z^{m-n-2}}{b_0 z^m+b_1 z^{m-1}+b_2 z^{m-2}+\cdots +b_m}$$ is defined and continuous at $z=0$ (the denominator does not vanish). This means that the singularity of $$g(z)=\frac{1}{z^2} \cdot \frac{P(1/z)}{Q(1/z)}$$ at $z=0$ is removable, so $g(z)$ can be extended to be ...


0

The intersection of $\{\gamma\}$ and $\partial G$ is trivial, since the image of $\gamma$ is contained in $G$. There exists $(x_n), x_n\in \{\gamma\}$ $(y_n\in \partial G)$ such that $lim_nd(x_n,y_n)=d(\{\gamma\},\partial G)$. Since $\{\gamma\}$ is compact, there exists a subsequence $(x_{f(n)}$ of $(x_n)$ which converges towards $x\in\{\gamma\}$, this ...


0

For a fixed set, the distance from it is a continuous function. If the set is closed, the distance is positive for each point not contained in it. A continuous function on a compact set has a minimum.


1

Because the distance between disjoint sets, one of which is closed and the other compact, is positive. Since the boundary of $G$ is closed, your statement follows.


2

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4

We can use complex analysis that begins by enforcing the substitutions that were used in the solution posted by @jackd'aurizio . There, we have $$\int_0^\infty \frac{x}{\sinh(x)}\,dx=\int_0^\infty \frac{\log(x)}{x^2-1}\,dx \tag 1$$ Now, we analyze the contour integral $$\begin{align} I&=\oint_C \frac{\log^2(z)}{z^2-1}\,dz\\\\ &=\int_{0^+}^R \...


1

Suppose we seek to evaluate $$K = \int_0^\infty \frac{\log x}{(x+a)^3} \; dx$$ with $a$ positive. We use $$f(z) = \frac{(\log z)^2}{(z+a)^3}$$ and integrate around a keyhole contour with the branch cut of the logarithm on the positive real axis and the argument between zero and $2\pi.$ We obtain $$2\pi i \mathrm{Res}_{z=-a} f(z) \\ = \int_0^\...


0

For 3., note that $f(z) - if(z)$ is constant by 1. It follows that $f'(0) = if'(0), f''(0) = if''(0), \dots.$ Thus all derivatives of $f$ at $0$ vanish, which implies $f$ is constant.


0

Princeton Lectures in Analysis is a good choice for undergrad level analysis. Here is the link for book 3: Real Analysis: https://www.amazon.com/Real-Analysis-Integration-Princeton-Lectures/dp/0691113866


8

A simpler way to use complex analysis is to consider the integral $$\oint_C dz \frac{z}{\sinh{z}} $$ where $C$ is the rectangle with vertices $-R$, $R$, $R+i \pi$, $-R+i \pi$, with a semicircular detour of radius $\epsilon$ at $z=i \pi$ into the rectangle. Thus, the contour integral is equal to $$\int_{-R}^R dx \frac{x}{\sinh{x}} + i \int_0^{\pi} dy \...


1

Let $i:G\hookrightarrow \mathbb{C}$ denote the canonical inclusion. Then $i \circ g^{-1}:\mathbb{C}\to\mathbb{C}$ is an injection, and therefore, $$i \circ g^{-1}(z)=az+b,~~~~a\in\mathbb{C_*},b\in\mathbb{C}$$ It follows that $i=ag+b$, and since $g$ is a surjection, so is $i$. That the inclusion map is surjective implies that $G=\mathbb{C}$.


5

Through the substitutions $x=\log t$, then $t=\frac{1}{v}$, we have: $$ I=\int_{0}^{+\infty}\frac{x\,dx}{\sinh{x}}=\int_{1}^{+\infty}\frac{2\log t}{t^2-1}\,dt = 2\int_{0}^{1}\frac{-\log v}{1-v^2}\,dv\tag{1}$$ and now a simple Taylor series expansion is enough, since: $$ \int_{0}^{1}(-\log v)v^{2k}\,dv = \frac{1}{(2k+1)^2} \tag{2}$$ leads to: $$ I = 2\sum_{k\...


2

All four follow easily from Liouville's theorem: Theorem If $f$ is an entire function and $f(\Bbb C)$ is not dense in the plane then $f$ is constant. Proof: Since $f(\Bbb C)$ is not dense there exist $p\in \Bbb C$ and $r>0$ so $$f(\Bbb C)\cap D(p,r)=\emptyset.$$This says that $|f(z)-p|\ge r$ for every $z$. So the function $g=1/(f-p)$ is bounded; ...


0

For Complex Analysis my favorite book is "A Course in Complex Analysis"- Fischer and Lieb. For Real Analysis I suggest you the Baby-Rudin or Real Analysis-Sullivan and Haaser.


0

We have the function $\cosh:\mathbb{C}\rightarrow \mathbb{C}$. We can show that it isn't injective if we can find $x \neq y$ such that $\cosh{x} = \cosh{y}$. (Relatedly, we can show that it isn't surjective if we can find $y$ such that there is no $x$ for which $\cosh{x} = y$.) Because the function is even, it is easy to show that the function isn't ...


1

METHODOLOGY $1$: If $f(z)=u(x,y)+iv(x,y)$ is analytic inside the unit disk, then we have $\nabla^2 u(x,y)=\nabla^2 v(x,y)=0$ for $x^2+y^2<1$. Note that if in addition, $f(z)=i$ on the unit disk, then the problem is equivalent to the pair of interior Dirichlet Problems $$\begin{align} \nabla^2 u(x,y)&=0, \,\,x^2+y^2<1\\\\ u(x,y)&=0,\,\,x^2+...


1

If $f$ has a single zero at $z_0$ then the function $g(z):={f(z)-f(z_0)\over z-z_0}$ has a removable singularity at $z_0$. In fact, $g$ is analytic in a neighborhood $U$ of $z_0$, and $g(z_0)=f'(z_0)\ne0$. It follows that $h:={1\over g}$ is analytic in $U$, and that $${1\over f(z)}={h(z)\over z-z_0}\qquad(z\in\dot U)$$ has a first order pole at $z_0$ with ...


1

Here are some answers to your questions: "The reason $\mathbf{g(z_0)\not=0}$": The fact that $g(z_0) = a_1 \not= 0$ is due to the assumption that $f(z)$ has a zero of order $1$ at $z=z_0$. If $a_1 = 0$ then we would have $f(z) = (z-z_0)^2(a_2 + a_3(z-z_0)+\ldots)$ and the zero would be (at least) of order $2$ contradicting this assumption. "Where is $\...


1

Another solution is this... We have $$I:=res\left(\dfrac{1}{f},z_0\right)=\dfrac{1}{2\pi i}\int_{|z-z_0|=\epsilon}\dfrac{dz}{f(z)},$$ where $\epsilon>0$ is very small, such that $f$ admits power series expansion in $U=\lbrace z\in \mathbb{C}:|z-z_0|<\epsilon\rbrace$, and in this set the unique zero of $f$ is $z_0$. Now, since $f$ has a zero of order $...


1

If $\cos(iz)=\cos(-iz)$ then by definition its not injective from $C$ to $C$.


0

Here is an alternative way to show a more general version. Claim. $$\operatorname{Res}_{z_0}\left(\frac{f}{g}\right)=\frac{f(z_0)}{g'(z_0)}$$ Proof. If $f$ has a zero of any order in $z_0$ it can be removed (since $g$ has a zero of order 1) thus there is no principal part of the Laurent expansion and the residue is $0$. We can assume $f$ has no zero so we ...


1

This is quite non-trivial. Suppose $z_1,z_2\in B(0,1)$ are such that $f(z_1)=f(z_2)=z_0$. Take a circle $\gamma\in A$ such that $z_1,z_2\in Int(\gamma)$. Then observe that $f(\gamma)$ is a Jordan curve (this is where you use the injectivity of $f$ in $A$) and thus by Jordan Curve Theorem, $\text{Ind}(f(\gamma),z_0)=1$, ignoring orientation. However by the ...


3

A example of function that have a pole of order $p$ at $z_0$ and residue $r$ is given by $$f(z) = \frac{rz^{p-1}}{(z-z_0)^p}.$$ Indeed, to compute the residue, we can use the formula $$Res(f,z_0) = \lim_{z\to z_0}\frac{1}{(p-1)!}((z-z_0)^pf(z))^{(p-1)}=r.$$ Hence a example of function with pole of order $2$ with residue $1$ at $i$ and simple pole with ...


4

Just adding a graphical figure to Martin R's answer. The transformation used in the figure differs slightly from Martin's in that $$f(z)=4+\frac{16}{z-4}=\frac{4z}{z-4}=2\left(1+T(z)\right)$$ but both are equally usable. The important point is that the inversion circle has its center at $c=4$. Well, $c=-4$ would have worked as well, both possibilities follow ...


5

As already said in the comments, it is useful to map $\Omega$ with Möbius transformation $T$ to an annulus between two concentric circles, and we can assume that annulus is centered around $w=0$. Some geometric observations help to find such a transformation $T$. Let $a$ and $b$ be the pre-images of $\infty$ and $0$, respectively. From the symmetry of $\...


1

As already answered, the root test is the easiest. We can work the ratio test in a slightly different manner (compared to Soke's answer). Consider $$u_n=\frac{\left(1+\frac{1}{n}\right)^{n^{2}}}{n^{3}}$$ and take logarithms $$\log(u_n)=n^2\log\left(1+\frac{1}{n}\right)-3 \log(n)$$ $$\log\left(\frac{u_{n+1}}{u_n}\right)=\log(u_{n+1})-\log(u_n)$$ $$\log\left(\...


0

How I would do it: In $\sum\limits_{n = 0}^\infty \frac{(1+\frac{1}{n})^{n^{2}}}{n^{3}}z^{n} $, the $n^3$ doesn't matter (except, possibly, at the edge). Also, $(1+\frac{1}{n})^{n^{2}} =((1+\frac{1}{n})^{n})^n \approx e^n $. So the terms are about $e^nz^n =(ez)^n $, and the sum of these converges for $|ez| < 1$ or $|z| < \dfrac1{e}$. To see what ...


1

As carmichael561 showed, the root test is a lot easier here. Also, your textbook answer is wrong. Since you asked, here's how you can do it using the ratio test: let $a_n = \dfrac{\left(1 + \frac{1}{n} \right)^{n^2}}{n^3}$. Then $$ \begin{align}\lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^{n^2} \left( \frac{...


2

Let $a_n=\frac{(1+\frac{1}{n})^{n^{2}}}{n^{3}}$, then $$ \lim_{n\to\infty}a_n^{\frac{1}{n}}=\lim_{n\to\infty}\frac{(1+\frac{1}{n})^n}{n^{\frac{3}{n}}}=e$$ Therefore the radius of convergence is $\frac{1}{e}$ by the Cauchy-Hadamard theorem.


2

$e^{-i7\pi/4}$ is $e^{i\pi/4}$.


2

No, it's not right. To do relative deRham cohomology, in your relative cochain complex $C^k(X,A)$ consists of $k$-forms on $X$ whose pullback to $A$ vanishes. This is a bit of a subtle game. But, in particular, when $A$ is a point, any $k$-form with $k\ge 1$ will pull back to be $0$. It's hard to find references for this, but I know of a book (in French) by ...


5

Ok, so lets start by identifying what group we're actually working with. We're acting with the action of a fractional linear transformation right? So we are looking at subgroups of $PSL(2, \mathbb{C})$. Ok, so with the restrictions given we know that we have to be in the matrix group generated by $$ \begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} $$ ...


1

If a function from the $2$-sphere to itself has no fixed points, then it is homotopic to the antipodal map. In particular it has degree $-1$.


2

Hint : $$z^2+2z+2=(z+1-i)(z+1+i)$$


2

Suppose $f$ has a pole at $0$. For any $\epsilon > 0$, let $U_\epsilon = \{z: 0 < |z| < \epsilon\}$. By the Open Mapping Theorem, $(1/f)(U_\epsilon) = \left\{\dfrac{1}{f(z)}: z \in U_\epsilon\right\}$ contains a deleted neighbourhood of $0$, so $f(U_\epsilon)$ contains a deleted neighbourhood of $\infty$, i.e. $\{w: |w|>R\}$ for some $R > 0$...


1

Put $g(z)=\exp(f(z))$, and suppose that $g$ has a removable singularity at $0$. Then you can write $g(z)=z^m h(z)$, with $m$ integer $m\geq 0$ ($m=0$ is the case you have done) and $h$ analytic at $0$, $h(0)\not =0$. Then $$ \frac{g^{\prime}(z)}{g(z)}=f^{\prime}(z)=\frac{m}{z}+\frac{h^{\prime}(z)}{h(z)}$$ show that if $m=0$, $f^{\prime}$ has no pole at zero,...


2

Yes, this is correct and your argument is correct. (One can also think about this in terms of linear systems. If $D$ is a divisor of degree $0$, then $h^0(D)\le 1$, with equality holding precisely when $D$ is the trivial divisor.)


0

One way to define $x(z)^{y(z)}$ is to use the universal cover of $\mathbb{C}^\times = \mathbb{C} \setminus \{0\} $, which can be defined by the map : $$ \mathbb{C} \rightarrow \mathbb{C}^\times, t \mapsto z=\exp(t).$$ When $\mathrm{Re}(t) \to -\infty$ then $z \to 0$. Hence the two functions $z \mapsto x(z)$ and $z \mapsto y(z)$ on $\mathbb{C}^\times$ define ...


1

Some hints/steps : If an entire function has a non-essential singularity at $\infty$, then it is a polynomial. In order to prove this, consider $g(z)=f(\dfrac{1}{z})$, and compare power series expansions. From (1), conclude that $\infty$ is an essential singularity of $f$. Casorati-Weierstrass tells you that the image of a deleted neighbourhood of a ...


2

We define complex power functions using the exponential and logarithm functions: $$ z^\alpha = e^{\alpha\log z}. $$ The problem with this definition is that the complex logarithm function cannot be defined continuously on the entire punctured complex plane. (The punctured plane is the plane with zero removed. Of course you cannot define the logarithm to be ...


1

The law of exponents $$ x^ax^b=x^{a+b} $$ Remains valid. However, as you note, exponentiation is a bit more complicated and is defined as follows, for a complex number $s$ and a real base $a$: $$ a^s=e^{s\log n} $$ As is seen in the (super important) class of series called Dirichlet series, for $a$ natural, a special case of this being the Riemann Zeta ...


1

Integration is path independent as integrand is entire function so$$\int_{c}\frac{z}{i}dz=\int_{1}^{i}\frac{z}{i}dz=\frac{z^{2}}{2i}|_{1}^{i} =i$$


1

$$z=e^{i\theta}\implies dz=ie^{i\theta}d\theta\implies {z\over i}dz=(e^{i\theta})^2d\theta\implies\int_C{z\over i}dz=\int_0^{\pi\over2}e^{2i\theta}d\theta={e^{2i\theta}\over2i}\Big{|}_{0}^{\pi\over2}=i$$


1

You want to find a sequence for $z \to -1$ such that $2πk = \frac{z}{z+1} = 1 - \frac{1}{z+1}$. Can you now? You basically have to choose a sequence for $k$ so that the required $z$ tends to $-1$. So the original expression is zero on this sequence for $z \to -1$. You also want another sequence for $z \to -1$ such that the original expression does not tend ...


0

From your assumptions it only follows that $\int_\gamma f(z)\>dz=0$ for tiny closed curves $\gamma$, where "tiny" means that such a $\gamma$ has to lie in the domain of convergence of a single power series locally representing $f$. For a full proof you would have to show that the interior of any (large) Jordan polygon $\pi$ can be triangulated into tiny ...



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