New answers tagged

2

Your answer is correct, and here is another way to try, in particular when taking derivatives or limits can be a little troublesome, and using the fact that we're only interested in low powers of $\;z\;$ in power or Laurent series since we want to find out something at $\;z=0\;$: $$\frac{\cos z}{\sin ...


1

Yes your answer is right. In general, assume to have two holomorphic functions $f$, $g$ with a zero at $z_0$. Assume $z_0$ is a zero of multiplicity $p$ of $f$ and multiplicity $q$ of $g$, then $z_0$ is a pole of order $q-p$ of $f/g$. If $q-p \leq 0$, the singularity is removable. Here $0$ is a zero of multiplicity $1$ of $\sin(z)$ (since it doesn't ...


0

There should be $d_∞$ in the proof. All the norms are equivalent and all the properties being proved are topological, so it really holds for balls of any norm. But the proof should use the same norm as the statement. Regarding your points a), b), they actually hold since $d_∞ ≤ d_2$ and $\lVert·\rVert_∞ ≤ \lVert·\rVert_2$. But there is a problem with the ...


1

In the last calculation, rememver that $\frac{1}{i}=-i$, so you need to add a minus in the exponent. Other than that, you can now use Euler's formula: $$e^{x+iy}=e^x\cdot\left(\cos y+i\sin y\right)$$


1

Necessary and sufficient condition for existence of an entire function $g$ extending $f$ to the whole complex plane: $f$ is infinitely differentiable at $0$, and the power series for $f$ at the origin converges to $f$ on the real line. A counterexample for something more is (as noted in a comment) $f(x) = 1/(1+x^2)$. It is real analytic on the real line, ...


3

Approach 1. Suggested by Lost in a Maze. Integrate $\frac{e^{az}}{1+e^z}$ along the rectangle with vertices $\pm R,$ $\pm R +2\pi i$ and let $R$ tend to $\infty$, we would have \begin{align} \left(1-e^{2a\pi i}\right)\int_{-\infty}^\infty \frac{e^{ax}}{1+e^x} dx&=2\pi i\,\operatorname{Res} \left(\frac{e^{az}}{1+e^z}; \pi i\right)\\ &=-2\pi ...


5

In my view it is not really the definition of the radius of convergence. Rather the definition of the radius of convergence of $$\sum_{n = 0}^{\infty} a_n(z - c)^n $$ is $R=\sup\{|z_0 - c | \colon \sum_{n = 0}^{\infty} a_n(z_0 - c)^n \text{ converges}\}$. With this definition, it is clear that this is the largest disk where the convergence could ...


1

The function is holomorphic at each point where $\sin z\ne0$, because a quotient of holomorphic function in a neighborhood of that point. For the residue at $k\pi$, just consider that $\cot(z+k\pi)=\cot z$, so the residue is the same as the residue at $0$; since $$ \lim_{z\to0}z\frac{\cos z}{\sin z}=1 $$ the residue is $1$.


0

This "answer" is rather far from a complete answer. It only transforms the problem into a kind of conformal representation problem and makes remarks about the tightness of the minoration. These indications and remarks can, hopefully, help other people as a kind of base camp before undertaking ascension to the summit. $$|z|>1 \Rightarrow \left( ...


2

Yes. This is even true for a real analytic section of a real analytic bundle over a real analytic manifold. To prove this, suppose $f=g$ on a set $U$ with nonempty interior (which we will henceforth also call $U$); if $U$ is not everything, then there is a point $x \in \partial U$. Work instead with $f-g$; we wish to show it vanishes in a neighborhood of ...


1

Take $u = ( \frac{1}{2}, \frac{1}{2})$ and $v = ( \frac{1}{2}, \frac{1}{3})$, then $|u|^2 = \frac{1}{2}$, $|v|^2 = \frac{13}{36}$ and $|u - v|^2 = \frac{1}{36}$


2

It is not true. Take $u = \frac{1}{3}, v= \frac{1}{2}$ and n = 2. Now your inequality tells $\frac{5}{36} = |\frac{1}{9}-\frac{1}{4}| \leq |\frac{1}{3}-\frac{1}{2}|^2$ or $\frac{5}{36} \leq \frac{1}{36}$ which is not.


4

Consider the branch $f(z) = \frac{\ln z}{z^2 +1}$ where $|z| > 0 , -\frac{\pi}{2}< \arg z < \frac{3\pi}{2}$. Take the path $C = L_2 + L_1 + C_{\rho} + C_R$ where $\rho < 1 < R$ and $C_R$ and $C_{\rho}$ are the semi-circles with radius $R$ and $\rho$ respectively. See the figure below. $\hskip.75in$ By Cauchy's Theorem we have $$\int_{L_1} ...


3

To first answer the question in the OP regarding the applicability of a branch cut along the negative imaginary axis, we note that we can certainly evaluate the integral of $\frac{z^a \log(z)}{z^2+1}$ over the proposed contour as $$\begin{align}\oint_C \frac{z^a \log(z)}{z^2+1} &=\int_r^R \frac{x^a\log(x)}{x^2+1}\,dx-\int_{r}^{R}\frac{e^{i\pi ...


0

With $\;z=x+iy\;,\;\;x,y\in\Bbb R\;$: $$\left|\int_\gamma e^{\sin z}dz\right|\le\int_\gamma\left|e^{\sin z}\right|dz\le e^{\sin x\cosh y}\cdot1= e^0=1$$ since on the real line between zero and one we have $$\;x=0\implies \sin0=0 $$


1

Note: If you proceed with the geometric series approach, you find (formally) that: $$\frac{1}{e^z-1}=\sum_{n\ge1}e^{-nz}=\sum_{n\ge1}\sum_{k\ge0}\frac{(-1)^kn^kz^k}{k!}=\sum_{k\ge0}\frac{(-1)^kz^k}{k!}\sum_{n\ge1}n^k$$ The term at the end, $\sum_{n\ge1}n^k$, is obviously infinite and divergent. That being said, some treatments allow us to assign value to ...


0

We can check your answer with another method, with residues $$\int_{|z|=1}\left(\frac1z+\frac2{z^2}\right)dz=2\pi i\ \text{Res}\left[\frac1z+\frac1{z^2};0\right]=2\pi i.$$ Seems like your answer is correct.


2

Your function's only possible pole is at $-1$, and it has to be a simple pole: a pole of order $p$ would have $|f(z)| \sim |z+1|^{-p}$ as $z \to -1$, and this is greater than your right side if $p > 3/2$. Your function must have a removable singularity at $\infty$, because it is bounded as $z \to \infty$. After removing the pole you'll be left with a ...


0

Since $3/2<2$, $C_{-2}$ must be $0$. Except for that, your work is correct. Also, there is no need to write $C_1(z+1)$; you can write $C_1\,z$.


0

In the context of topological spaces and continuous maps this is a standard fact about covering spaces. Locally you have, speaking very informally, $f=f_2^{-1}\circ f_1$. Of course $f_2$ is not invertible, but it has local inverses near each point. The fact that $f$ is analytic now follows for free.


0

That depends on the branch of $\sqrt z$. Usually, $\sqrt z$ is taken to be the branch defined on $\mathbb C\setminus(-\infty,0]$ and $\sqrt 1=1$. In that case, $f$ does not have isolated singularities.


2

Consider the letter "X" (i.e., two crossing line-segments). It's starlit from its center, but every point except the crossing-point is not a star center.


1

Split them into partial fraction and upon solving you will get, $ \frac {2(z+2)}{(z-1)(z-3)} = \frac {3} {(z-(-1))} + \frac {5} {z-3} $ taylor's series expansion is $ \frac {1}{1-x} = \frac{1}{1-a} + \frac {x-a}{(1-a)^2} + \frac {(x-a)^2}{(1-a)^3} + ...$ so for $\frac {3}{(z-(-1)} $ , value of a=-1 i.e., $ \mathcal\ z ^\left(-1 \right) $ $\left[ ...


4

One may observe that $\displaystyle z \to \frac{z}{e^z-1}$ is analytic in $0<|z|<2\pi$, then it admits a power series expansion $$ \frac{z}{e^z-1}=\sum\limits_{n=0}^{\infty}b_n\frac{z^n}{n!}, \quad 0<|z|<2\pi, \tag1 $$ with $b_0=1$, $b_1=-1/2$. Then, multiplying $(1)$ by $\displaystyle e^z$, $$ ...


1

You only have to compute the modules and see if they are less than $2$. For exemple $|\pm \frac{\pi}{2}|<2$ since it is well known that $3<\pi<3,5$. However the other singularities are not in the disk because the module is greater than $2$.


3

$$\left(1+\frac1z+\frac1{2z^2}+\cdots\right)\left(\frac1{z^2}+\frac1{z^4}+\cdots\right)= \cdots+\left(1\frac1{z^4} + \frac1{2z^2}\frac1{z^2}\right)+\cdots$$


3

Hint: Note that $4^n+3^n=4^n\cdot(1+(3/4)^n)$.


2

Hint. Starting with a standard partial fraction decomposition $$ \frac{2(z+2)}{z^2-4z+3}=\frac{3}{1-z}-\frac{5}{3-z} $$ you may then set $u:=z-2$, that is $z=u+2$, obtaining $$ \frac{3}{1-z}-\frac{5}{3-z}=-\frac{3}{1+u}-\frac{5}{1-u} $$ with $u \to 0$ as $z \to 2$. Can you take it from here?


0

I figured it out. Thanks to @DanielFischer for the comment. I still don't know how to proceed without exponential form though. Let $a,b$ be real positive numbers. $$a+i b=\sqrt{a^2+b^2} \exp \left(i \arctan \frac{b}{a} \right)$$ $$\sqrt{a+i b}=\sqrt[4]{a^2+b^2} \exp \left(\frac{i}{2} \arctan \frac{b}{a} \right)=$$ I use the principal value of the root. ...


0

Consider a neighborhood of radius $R$ around $i.$ Then $$|z^2+2|=|z+i||z-i|\le (|z-i|+|2i|)|z-i)\le (R+2)|z-i|.$$


2

By taking $0<\delta<\min\left\{1,\frac{\epsilon}3\right\}$ we get \begin{align*} |z^2+1|&=|z+i||z-i|\\ &\le\left(|z-i|+|2i|\right)|z-i|\qquad\text{from the triangular inequality}\\ &<(\delta +2)\delta\\ &<3\cdot\frac{\epsilon}3 \end{align*} provided $|z-i|<\delta$.


1

Let $\delta = \min\{1,\frac{\epsilon}{3}\}$, then $$ 0<|z-i|<1 $$ and so $$ |z|-1<1 $$ by triangle inequality. Thus, $$ |z+i|\le |z|+1< 3 $$ by triangle inequality again.


0

I think most of the time, only the topology of the Riemann surface is of concern... so I wouldn't think there is an equation that could describe the surface in general.


0

Like Cameron Williams says, you can perform these by contour integration. Write $$ \int^{2 \pi}_0 \frac{1}{ \sqrt{5}+\cos t}\mathrm{d}t = \int^{2\pi}_0 \frac{\mathrm{d}t}{\sqrt{5} + \frac{1}{2} e^{it} + \frac{1}{2}e^{-it}} = \oint_{|z|=1} \frac{\mathrm{d}z}{iz \left (\sqrt{5} + \frac{z}{2} + \frac{1}{2z} \right )}, $$ where we have substituted $z = e^{it}$ ...


1

I'm going to assume that in your question the indeces in the summation are from $n=0$ to $\infty$ instead of $j=0$ to $\infty$; if they are not then both series diverge, as $$ \sum_{j=0}^{\infty} z^n = \lim_{m \to \infty} \sum_{j=0}^{m} z^n = \lim_{m \to \infty} mz^n = \infty. $$ The basic idea for the Weierstrass $M$-test is that you bound the norm of your ...


1

With respect to the new bounty: Here's a way to see this. Suppose $u$ is a compactly supported solution to $\bar{\partial}u=f$. Then we have, for large enough $R>0$ $$0=\int_{\partial D(0,R)} u(z)dz = 2i\int_{D(0,R)} \bar{\partial}u(z)\: d\bar{z}\wedge dz=2i\int_{D(0,R)} f(z)\:d\bar{z}\wedge dz$$ Taking $R\to\infty$, this implies that if a compactly ...


0

The expression $f(z)=\frac1{z-3}$ is already the Laurent series around $|z-3|>0$.


1

Since you want powers of $z+1$ anyway, I find it much easier, even if it’s a purely mechanical device, to make the substitution $\zeta=z+1$ and $z=\zeta-1$. This gives you $$ \frac z{(z+1)(z-2)}=\frac{\zeta-1}{\zeta(\zeta-3)}=\frac1{\zeta-3}-\frac1{\zeta(\zeta-3)}\,, $$ at which point the problem just becomes writing out the power series for $1/(\zeta-3)$. ...


1

Sketch of a possible argument. Use the branch with the argument from zero to $2\pi$ of the logarithm, the function $$f(z) = \frac{\exp(1/3\log z)\log z}{z^2+1}$$ and a keyhole contour with the slot aligned with the positive real axis. The sum of the residues at $z=\pm i$ is (take care to use the chosen branch of the logarithm in the logarithm as ...


3

We use the fact that $$\left(\cot x\right)\left(\cot\left(\frac{\pi}{2}-x\right)\right)=1.$$ The product of the entry that is $k$ from the beginning and the entry that is $k$ from the end is $1$. (If $m$ is even, there is a "middle" term, but it is $1$.)


2

Here is a useful finite evaluation: $$ 1+z+z^2+...+z^n=\frac{1-z^{n+1}}{1-z}, \quad |z|<1. \tag1 $$ Then by differentiating $(1)$ you get $$ 1+2z+3z^2+...+nz^{n-1}=\frac{1-z^{n+1}}{(1-z)^2}+\frac{-(n+1)z^{n}}{1-z}, \quad |z|<1, \tag2 $$ and by making $n \to +\infty$ in $(2)$, using $|z|<1$, and multiplying by $z$ gives $$ \sum_{n=1}^\infty ...


0

$$\sum_ {n=1} ^{\infty} {nz^n} =\left(\sum_ {n=1} ^{\infty} {z^n} \right)^2$$


2

the functional equation is proven first for $Re(s) \in \ ]0,1[$, but then by analytic continuation it is valid everywhere (except at $s=1$ or $s=0$ of course). you can also analytically continuate $\zeta(s)$ from $$\Gamma(s) \zeta(s) = \int_0^1 \left(\frac{x}{e^x-1}-\sum_{k=0}^K \frac{B_k}{k!} x^k\right)x^{s-2}dx + \int_1^\infty \frac{x^{s-1}}{e^x-1}dx + ...


0

$h$ is a composition of products and sum of continuous functions, so it is continuous.


1

Clearly the function $$ f(z)=\frac{\sin(z)}{z} $$ is holomorphic on $\mathbb C \backslash \{0\}$. Therefore $f$ has an isolated singularity at $z=0$ which is removable since $$ \lim_{z\to 0}zf(z)=\lim_{z\to 0}z\frac{\sin(z)}{z}=\lim_{z\to 0}\sin(z)=0 $$ So we can remove this singularity. Now we just need to extend $f$ in $z=0$ continuously which we can do ...


1

It has an isolated singularity at $0$ because dividing by $0$ is undefined and thus $f(z)$ is not defined at $0$. Second, there are two ways to think about this. 1) Since we already accept that this singularity is removable, the only way to define $f$ at $0$ is by its limit for else it would not be continuous and thus we would not have removed the ...


1

The substitution $a=\cosh \tau$ is convenient here, since then $$\sqrt{a^2-1}-a=\sinh \tau-\cosh \tau = -e^{-\tau }.$$ Hence the first term of the expression in the question becomes $$\frac{(e^{-2\tau} - 1)^2}{e^{-2\tau} (e^\tau-e^{-\tau})}=\frac{(e^{-\tau} - e^{\tau})^2}{(e^\tau-e^{-\tau})}=e^{\tau}-e^{-\tau}=2\sinh\tau=2\sqrt{a^2-1}$$ in agreement with ...


2

A connected set is a set that cannot be divided into two disjoint nonempty open (or closed) sets. Intuitively, it means a set is 'can be travelled' (not to be confused with path connected, which is a stronger property of a topological space - every two points are connected by a curve). A simply connected set (let me short it to SC for now) is path-connected ...


1

A connected set is a set that cannot be split up into two disjoint open subsets (this of course depends on the topology the set has; for the case of $\mathbb{C}$, this is the same as the Euclidean topology on $\mathbb{R}^2$). Now, a simply connected set is a path-connected set (any two point can be joined by a continuous curve) where any closed path (a ...


1

A connected set is a set which cannot be written as the union of two non-empty separated sets. This is when the set is made only of one-part, if one wants to think of it intuitively. However, simple-connectedness is a stronger condition. It requires that every closed path be able to get shrunk into a single point (continuously) and that the set be ...



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