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3 votes

How is $|z|^2$ not analytic?

Let me summarize the points already made in comments: The definition of "analytic at $p$" is "differentiable in some neighborhood of $p$". I.e., it is not equivalent to "...
Paul Sinclair's user avatar
1 vote

Find the $\int_C \frac{e^{-2z^2}}{(z-2i)(z-5)} dz$

By Cauchy's Integral formula, $$ \int_C \frac{e^{-2 z^2}}{(z-2 i)(z-5)} d z= 2\pi i\left( \lim_{z\to 2i} \frac{e^{-2 z^2}}{(z-5)}\right)=2\pi i\left(\frac{e^{-2(2i)^2}}{2i -5}\right) =\frac{2\pi ie^8}...
random_0620's user avatar
  • 2,285
1 vote

Find the $\int_C \frac{e^{-2z^2}}{(z-2i)(z-5)} dz$

Using Cauchys integral formula its: $$2\pi i \bigg(\frac{e^{-2z^2}}{z-5}\bigg\vert_{z=2i}\bigg).$$ As the residue is zero for $z=5$, youre above answer is correct! Hope you can finish from here!! You ...
homosapien's user avatar
  • 4,177
1 vote

Proving an infinite series is $o(1)$ using asymptotics

A comment that got too long and gives a hint of how to prove this. First, you need a uniform asymptotic for $f_N(r)$ (in other words if $f_N(r)=\frac{(\log\log N)^{r-1}}{(r-1)!\log N}(1+\epsilon(r,N))$...
Conrad's user avatar
  • 27.5k
0 votes

Cauchy product of zeta function

The Euler product of $\zeta(s)$ is $$\zeta(s) = \prod_{p\in prime} \frac{1}{1-\frac1p}$$ $\zeta(s)^2$ is therefore $$\zeta(s)^2 = \prod_{p\in prime} \frac{1}{\left(1-\frac1p\right)^2}$$ $$ = \prod_{p\...
JMP's user avatar
  • 21.8k
1 vote

Classifying Singularities of a Rational Function

You can actually compute the Laurent series for the whole function: Note: $$\frac{1}{z-3} = -\frac{1}{3\left(1-\frac{z}{3}\right)} = -\frac{1}{3}\sum_{j=0}^{\infty}\frac{z^j}{3^j} \quad |z|<3 $$ $$\...
Bertrand87's user avatar
  • 2,181
1 vote
Accepted

Classifying Singularities of a Rational Function

As $$\lim_{z\to 0} \dfrac{z+2}{z-3}=-\dfrac{2}{3},$$ which is neither $0$ nor $\infty$, this factor does not affect the classification of $z=0$. And $\cos(1/z)$ has a essential singularity at $z=0$, ...
Julio Puerta's user avatar
  • 4,391
1 vote

Why is the Jacobi amplitude real when $k>1$?

Simply, because the substitution $1-k^2 \sin^2 \phi \mapsto 1 - \sin \theta$ for $k>1$ yields the elliptic $F(\theta, \frac{1}{k} )$. The inverse is an oscillating function as as solutions of $$\...
Roland F's user avatar
  • 2,212
1 vote

Complex Integral and Residue involving multiple Branch Points inside the contour, without deforming it

The strategy will depend a lot on $f$. I'll assume for simplicity that $f$ is meromorphic. Technically your integrand is ill defined since the square root is double valued: you need to specify the ...
LPZ's user avatar
  • 2,948
1 vote

Show that $f(\frac{1}{z})$ have a essential singularity at $0$.

Yes, that's enough. If $z=0$ is removable for $f(1/z)$, then $f(1/z)$ is bounded around $0$, therefore $f$ is bounded around $\infty$, hence bounded over the whole $\mathbb C$. To be a little bit more ...
Just a user's user avatar
  • 15.2k
1 vote

Show that $f(\frac{1}{z})$ have a essential singularity at $0$.

$f(\frac 1 z) \to f(0)$ as $ |z|\to \infty$. If $f(\frac 1 z)$ has a removable singularity at $0$, then it extends to a bounded entire function. Hence, it would be constant, by Liouville's Theorem.
geetha290krm's user avatar
  • 37.1k
1 vote

Show that $f(\frac{1}{z})$ have a essential singularity at $0$.

Assume that $f(1/z)$ has removable singularity at $z=0$. Then we know that this can happen only if $f(1/z)$ is bounded near $z=0$. But this means $f(z)$ is a bounded entire function, hence is constant ...
Tri's user avatar
  • 333
0 votes

Show that $f(\frac{1}{z})$ have a essential singularity at $0$.

Since $f(z)$ is entire, we may write it as $f(z)=\sum_{k=0}^\infty a_kz^k$. Since $f$ is non-constant, there is some $a_k\neq 0$ where $k>0$. We will show infinitely many of the $a_k$ are non-zero. ...
random_0620's user avatar
  • 2,285
1 vote
Accepted

Computing Singularities of a function

It doesn't matter if the limit of the mentioned expression tends to zero. The only thing that matters for a function like that is the existense of the limit.
Egor Larionov's user avatar
1 vote

Examples of non-elliptic Doubly Periodic Functions

Complex analysis is a very rigid subject: Given the values of functions on a countable complex set with a finite accumulation point, the function is fixed inside a circle of convergency around this ...
Roland F's user avatar
  • 2,212
0 votes

Prove that, except for the identity function, a holomorphic map of the open unit disk into itself has at most one fixed point in the disk.

Let $z_{1},z_{2}\in B(0,1)$ be distinct ,$f\in H({B(0,1)}) ,f(z_1)=z_1 $, $f(z_2)=z_2$. Def:$$\phi_{a}(z)=\frac{a-z}{1-\bar{a}z}$$ Notice that:$$\phi_{z_1}(f(z_2))=\phi_{z_1}(z_2)$$ Using Schwarz-...
Yuechuan Rom's user avatar
1 vote

Examples of non-elliptic Doubly Periodic Functions

Elliptic functions come from inverting elliptic integrals. In particular, when $0<k<1$ and $$ w=\int_0^z{\mathrm dt\over\sqrt{1-t^2}\sqrt{1-k^2t^2}}. $$ The inverse $z=\operatorname{sn}(w)$, ...
TravorLZH's user avatar
  • 6,788
3 votes

Prove that, except for the identity function, a holomorphic map of the open unit disk into itself has at most one fixed point in the disk.

Hints: Let $c,d$ be distinct fixed points of $f$. Define $g=\phi^{-1}\circ f\circ \phi$ where $\phi (z)=\frac {z+c} {1+\overline c z}$. Then $g(0)=\phi^{-1}c=0$. Since $\phi$ maps the unit disk to ...
geetha290krm's user avatar
  • 37.1k
1 vote
Accepted

exponential map property

The answer is yes. You can even take some larger domain for $f$: any interval $I$ of length $L<\pi$, to make sure that $$\frac{|f(s)-f(t)|}2\le a:=\frac{|e^{iL}-1|}2<1\quad(\forall s,t\in I).$$ ...
Anne Bauval's user avatar
  • 35.3k
3 votes

$\sum_{n=R}^{\infty}{\binom{n}{R}\frac{a^n}{(1+a)^{n+1}}} = a^R$

$$\sum_{n=R}^{\infty}{\binom{n}{R}\frac{a^n}{(1+a)^{n+1}}}=\frac 1{a+1}\sum_{n=R}^{\infty} {\binom{n}{R}}x^n \qquad \text{with} \qquad x=\frac a{1+a}$$ $$\sum_{n=R}^{\infty} {\binom{n}{R}}x^n= \frac ...
Claude Leibovici's user avatar
0 votes

Applying the Identity Theorem to Analytic Functions Agreeing on 1D Curves

By Identity theorem, we can see that $g,h$ has finite poles inside the unit disk. Let's denote these poles of $g$ and $h$ by $a_1,\ldots,a_n$ with order $m_1,\ldots,m_n$, respectively. Then $$g_1=\...
Tri's user avatar
  • 333
3 votes

$\sum_{n=R}^{\infty}{\binom{n}{R}\frac{a^n}{(1+a)^{n+1}}} = a^R$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} ...
Felix Marin's user avatar
  • 89.6k
3 votes

Sum of reciprocals of norms of Gaussian primes

If $p\in\Bbb{Z}$ is an integer prime with $p\equiv3\pmod{4}$, and $m+ni\in\Bbb{Z}[i]$ is such that $m^2+n^2=p^2$, then reducing mod $p$ shows that $$m^2+n^2\equiv0\pmod{p},\qquad\text{ and so }\qquad ...
Servaes's user avatar
  • 63.4k
2 votes

Rudin theorem $7.8$

Let $x_{0}$ be a Lebesgue point of $f$. If $r>0$, then \begin{align*} \left| \frac{1}{m(B_{r})}\int_{B(x_{0},r)} f \, {\rm d}m - f(x_{0}) \right| &= \left| \frac{1}{m(B_{r})}\int_{B(x_{0},r)} f ...
Dean Miller's user avatar
  • 1,147
0 votes

Sum of reciprocals of norms of Gaussian primes

If $m^2+n^2=p^2$ where $p$ is a prime congruent to $3$ modulo $4$ then $$\sum_{m+ni\in\mathbf{P}}\frac{1}{m^2+n^2}$$ converges because it is known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{...
Piquito's user avatar
  • 29.9k
0 votes
Accepted

Meromorphic continuation of Euler product

A comment that got too long. First I assume one wants $p \ge 3$ in the first product. Then I find it highly likely that the first product has the $\Re s =-1$ as natural boundary while the second has $\...
Conrad's user avatar
  • 27.5k
1 vote

Solve complex equation $(z-i)^4=(1+2i)^8.$

Since $(1+2i)^2= -3 + 4i$, the equation can be written as $(z-i)^4= (-3+4i)^4$. Note that for $a\in\mathbb{C}$, the solutions of $z^4= a^4$ are $z\in\{\pm a, \pm ai\}$. Hence the roots of the above ...
Etemon's user avatar
  • 6,515
2 votes
Accepted

Confusion in proof of $H^1(X,\mathcal{O})=0$ where $X$ is an open disk

Your shortened proof is true, but it's circular until you know the cohomology of the disk vanishes. The definition of Cech cohomology says we have to take a direct limit over all the opens; Leray's ...
hunter's user avatar
  • 30.3k
3 votes
Accepted

Calculate $\int_{-\infty}^{\infty}\frac{e^{itz}}{(z+i)^2}\,\mathrm{d}z$

The correct answer is $$ \int_{-\infty}^{\infty} \frac{e^{itz}}{(z+i)^2} \, \mathrm{d}z = \begin{cases} 0, & t \geq 0, \\[0.25em] 2\pi t e^t, & t < 0. \end{cases} $$ So, what have gone ...
Sangchul Lee's user avatar
0 votes

Calculate $\int_{-\infty}^{\infty}\frac{e^{itz}}{(z+i)^2}\,\mathrm{d}z$

Trying to calculate $$ \int_{-\infty}^{\infty}\frac{e^{itz}}{(z+i)^2}\,\mathrm{d}z $$ ...for $t > 0$, I use the lower semicircle that encapsulates the singularity at $z = -i$. ...$= 2\pi te^t$ ......
hft's user avatar
  • 556
0 votes

Branch points of $\sqrt{z-1}$

Consider first the circle $\gamma_1(\theta) = e^{i\theta}$ around $z=0$, $\theta = \Theta+2\pi n, \; n\in \mathbb{Z}, \;-\pi<\Theta\leq \pi$ We want to evaluate $f(e^{i\theta}) = \sqrt{e^{i\theta} -...
Bertrand87's user avatar
  • 2,181
1 vote
Accepted

Differentiability of function $f:\mathbb{C} \to \mathbb{C}; f(z)=z\exp(\bar{z})$

Yes your argument is correct. More directly: for any $z\in\Bbb C$, $$\frac{f(z+h)-f(z)}h=\frac{(z+h)e^{\overline{z+h}}-ze^{\overline z}}h=ze^{\overline z}\frac{e^{\overline h}-1}h+e^{\overline{z+h}}$$ ...
Anne Bauval's user avatar
  • 35.3k
0 votes

Differentiability of function $f:\mathbb{C} \to \mathbb{C}; f(z)=z\exp(\bar{z})$

Yes, this argument is correct. As pointed out in the comments, this argument implicitly uses that $\mathbb{C}\setminus\{0\}$ is open as in the definition of complex differentiability. The fact that $\...
Dowdow's user avatar
  • 56
2 votes
Accepted

Non-constant holomorphic function definied on open disk $D(0,1)$

Take $r_1,r_2\in[0,1]$ such that $r_1<r_2$. We know that $h(r_2)$ is attained at the circle of radius $r_2$. If $h(r_1)=h(r_2)$ then there would exist some $x_1$ in the circle of radius $r_1$, and ...
Julio Puerta's user avatar
  • 4,391
1 vote

Non-constant holomorphic function definied on open disk $D(0,1)$

The maximum of $|f|$ on $\overline {D(0,r_2)}$ is attained at some point on its boundary. If $h(r_1)=h(r_2)$ the maximum would also be attined at some point with $|z|=r_1$. By Maximum Modulus ...
geetha290krm's user avatar
  • 37.1k
3 votes
Accepted

$f$ with an essential singularity at $0$ intersects $1/z^4$ in every neighborhood of $0$.

$z$ is an element of the image of $g$, it may not even be near $0$. Nonetheless, the proof is still not very difficult. As $\dfrac{1}{z^4}$ has a pole at $0$, $f(z)-\dfrac{1}{z^4}$ has an essential ...
Julio Puerta's user avatar
  • 4,391
1 vote
Accepted

Uniform convergence of $\sum_{n=1}^{\infty} \frac{\exp(-nz)}{\sqrt{n(n+1)}}$

Another way is to use Cauchy's criterion for uniform convergence, and show it fails. Since the series $\sum\limits_{n=1}^{\infty}\frac{1}{\sqrt{n(n+1)}}$ diverges, there is some $\epsilon>0$ such ...
Mark's user avatar
  • 40.2k
0 votes

Finding linear imaginary and real factors of cubic P(z).

$$P(z)=z^3+az^2+3z+9$$ Let zeroes be represented as $p,iq,-iq$ where $p,q \in R$, because complex roots occur in conjugates, and here they are purely imaginary. Sum of roots $=-a=p+iq-iq=p$ $p$ ...
Gwen's user avatar
  • 1,135
1 vote

Uniform convergence of $\sum_{n=1}^{\infty} \frac{\exp(-nz)}{\sqrt{n(n+1)}}$

Your justification is sufficient. Actually, we have the following result: Suppose $z_0$ is a limit point of $E\subset\mathbb C$, the series $\sum\limits_{n=1}^{\infty}u_n(z)$ is uniformly convergent ...
Riemann's user avatar
  • 7,265
5 votes
Accepted

Question on determining $f(1-i)$ for an holomorphic function $f:\mathbb{C} \mapsto \mathbb{C}$

It's a simple exercise that if $f$ is holomorphic in $\mathbb{C}$ then the function $g:\mathbb{C}\to\mathbb{C}$ given by $g(z)=\overline{f(\overline{z})}$ is holomorphic as well. Note that for all $x\...
Mark's user avatar
  • 40.2k
1 vote

a small doubt in the proof of the quantitative form of the prime number theorem

Let's clarify what we need to prove: it's asserted that if $0<c<1$ then $$ x(\log x)^2 \exp(-\sqrt{c\log x}) \ll x \exp (-c\sqrt{\log x}), $$ which is equivalent to each of \begin{align*} (\log ...
Greg Martin's user avatar
  • 78.9k
2 votes
Accepted

How many roots has the equation $z = \varphi(z)$ in $|z| < 1$ if for $|z| \leq 1$, $\varphi(z)$ is analytic and satisfies $|\varphi(z)| < 1$?

This is indeed an application of Rouche's Theorem. Notice that your aim is to find the roots of $z-\varphi(z)$. We want to use the hypothesis that $|\varphi(z)|<1$. Therefore, using the same ...
Julio Puerta's user avatar
  • 4,391
1 vote

Show that $\int_{-\infty}^{\infty} \frac{x^2}{\left(x^2+a^2\right)^2} dx = \frac{\pi}{2a}$

If you like, you can use IBP to get the answer just in a single step. In fact $$\int_{-\infty}^\infty \frac{x^2}{\left(x^2 + a^2\right)^2} dx =-\frac12 \int_{-\infty}^\infty xd\bigg(\frac{1}{x^2+a^2}\...
xpaul's user avatar
  • 44.1k
1 vote
Accepted

Complex contour integral $\int\limits_{\partial \Omega}e^{\frac{x}{2}(z+\frac{1}{z})}dz$

We have $$e^{i\phi+\cos^2(\phi)}=e^{i\phi}e^{\cos^2(\phi)}=\cos(\phi)e^{\cos^2(\phi)}+i\sin(\phi)e^{\cos^2(\phi)}$$ Now, $f(\phi)=\cos(\phi)e^{\cos^2(\phi)}$ is even, so its integral on $[-\pi,\pi]$ ...
Julio Puerta's user avatar
  • 4,391
0 votes
Accepted

Locally uniform convergence of given series

Hints: STEP I It is enough to show that $\sum [\frac {(-1)^{n}} {n+z}-\frac {(-1)^{n}} n]$ is locally uniformly convergent for which it is enough to show that $\sum \frac {(-1)^{n}} {n(n+z)}$ is ...
geetha290krm's user avatar
  • 37.1k
1 vote

Holomorphic mappings of Riemann Sphere that preserves each hemisphere

A special easy Moebius transformation mapping the unit circle to itself is $$w=\frac{z-z_0}{\overline{z_0} z +1 }$$ It can be combined with rigid rotations $z=e^{i \alpha }\ z', w=w'\ e^{i\beta}$ ...
Roland F's user avatar
  • 2,212
1 vote
Accepted

Abel test for uniform convergence

Let $F_N (z) = \sum_{n=N}^{+\infty} f_n (z)$, and $f_n (z) = F_n (z) - F_{n+1} (z)$. (This is different from yours. As $N \to +\infty$, my $F_N (z) \to 0$ uniformly, therefore, it's easier to be used.)...
Y.D.X.'s user avatar
  • 364
4 votes
Accepted

Show that $\int_{-\infty}^{\infty} \frac{x^2}{\left(x^2+a^2\right)^2} dx = \frac{\pi}{2a}$

To compute the residue of a double pole, we need to compute the first derivative of the relevant factor of the integrand, not the second derivative. Here's another method that builds on your approach: ...
Travis Willse's user avatar
3 votes

Show that $\int_{-\infty}^{\infty} \frac{x^2}{\left(x^2+a^2\right)^2} dx = \frac{\pi}{2a}$

$$I:=\int_{-\infty}^ {\infty} \frac{x^2}{(x^2+a^2)^2}dx =\frac{1}{a}\int_{0}^ {\infty} \frac{t^{1.5-1}}{(t+1)^{1.5+0.5}}dx \ \ \ \ \ \ \ \ \ \ ;x= a\sqrt{t} $$ $$B(a,b)= \int_0^1 t^{a-1}(1-t)^{b-1} ...
pie's user avatar
  • 4,416
3 votes

Show that $\int_{-\infty}^{\infty} \frac{x^2}{\left(x^2+a^2\right)^2} dx = \frac{\pi}{2a}$

This integral can be easily evaluated by substituting $x=a\tan(\theta)$, $$I=\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)^2}dx$$ $$=\int_{-\pi/2}^{\pi/2}\frac{a^2\tan^2(\theta)}{a^4\sec^4(\theta)}a\sec^...
Sparsh Gupta's user avatar

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