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1

The only holomorphic functions that have this property are monomials $cz^n$. Indeed, the property implies (by continuity) that $|x|\le |y|\implies |f(x)|\le |f(y)|$, which in turn yields $|x|=|y|\implies |f(x)|=|f(y)|$. So, for every unimodular constant $\zeta $ the function $f(\zeta z)/f(z)$ has constant modulus, hence is constant. So, $f(\zeta z)\equiv ...


2

For the $s=0$ case, let $s_{n}=\sum_{k=1}^{n}c_{k}$ for $n \ge 1$ and let $s_{0}=0$. Then $s_{n}-s_{n-1}=c_{n}$ for all $n \ge 1$ and $\lim_{n} s_{n}=0$ by assumption. So the following are absolutely convergent for $0 \le r < 1$: $$ \begin{align} \sum_{n=1}^{\infty}r^{n}c_{n} & = \sum_{n=1}^{\infty}r^{n}(s_{n}-s_{n-1}) \\ & = ...


1

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3

hint: $$\frac1{z^2}\cos z=\frac1{z^2}\sum_{n=0}^\infty(-1)^n\frac{z^{2n}}{(2n)!}=\frac1{z^2}\left(1-\frac{z^2}2+\frac{z^4}{4!}-\ldots\right)$$


2

Write $log(2)=A+2\pi i n$ and $log(3)=B + 2\pi i m$ where $A$ and $B$ are the real logs of $2$ and $3$. Write $z=x+iy$, write out the real and imaginary parts of your equation and you'll get two equations in the four unknowns (two real, two integers) $x,y,m,n$, and it won't be hard to characterize the solutions.


7

To the best of my knowledge, you can't solve for $z$ analytically. As in my comment above, the expression simplifies to $-1=2^z + 3^z$, but there isn't a way to find $z$ explicitly. Even separating into real and imaginary parts yields a system of nonlinear equations. So to answer: is it solvable? No, it is not. So you'd have to use numerical methods to solve ...


1

Hint: There is a unique Möbius transformation $g$ which sends $0$ to $a$, $1$ to $b$, and $\infty$ to $c$ (provided $a$, $b$, and $c$ are distinct). Given $f \in M_{\{0, 1, \infty\}}$, can you see how to construct an element of $M_{\{a, b, c\}}$ using $g$?


0

Why are calculated The function {f\left( x \right)e^x } \begin{array}{l} {\mathop{\rm Re}\nolimits} s\left( {f\left( x \right); - 2i} \right) = \mathop {\lim }\limits_{z \to - 2i} \left( {z + 2i} \right)\left( {f(z)} \right) \\ = \mathop {\lim }\limits_{z \to - 2i} \frac{{z + 2 + 2\sqrt z }}{{z - 2i}} = \frac{{ - 2i + 2 + 2\sqrt { - 2i} }}{{ - 4i}} ...


0

Hint: show there exists a $B$ such that $e^{B\lambda_1} = a$ and $e^{B\lambda_2} = b$. Then consider the function $f(z)/e^{Bz}$, and use what you know about doubly periodic entire functions.


0

Hint: what are the residues of the function $$ f(z) = \frac \pi{\sin \pi z} $$?


1

For a complex variable $s$, whose real part is greater than zero, the Dirichlet eta function is defined by the series $$\eta(s) := -\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}.$$ In particular, one has that $$\eta(s) = \left(1-2^{1-s}\right)\zeta(s),$$ where $\zeta$ denotes the Riemann zeta function. With that in mind, one need only substitute $s=2$ into the ...


0

A path-connected topological space $X$ is simply connected if for any given point $x_0$ on $X$, any loop $\sigma$ based at $x_0$ is path-homotopic to the constant loop $e_{x_0}$ at $x_0$. The third space (subset of $\Bbb C$) is not connected, so there is no path between a point in the first component and a point in the other, say. Thus this space is not ...


1

Since $g(z)$ has a simple zero in $z_0$ you can freely add or subtract $g(z_0)$ in the denominator, because $g(z_0)=0$: $$ (z-z_0) \frac{f(z)}{g(z)}=(z-z_0)\frac{f(z)}{g(z)-g(z_0)}=\frac{f(z)}{\frac{g(z)-g(z_0)}{z-z_0}}\to_{z\to z_0} \frac{f(z_0)}{g'(z_0)}. $$


2

Hint:$$\lim_{z\to z_0}\frac{f(z)}{\frac{g(z) - g(z_0)}{(z-z_0)} + \frac{g(z_0)}{(z-z_0)}}$$


4

Even though I agree that the solution by @math110 is the best one, I want to add that this can be done using the theory of residues, using the common trick with the cotangent function. I leave it to you to fill in the details, if necessary. We let $$ f(z)=\frac{\pi\cot \pi z}{z^3(z+1)^3}. $$ This function $f$ has poles at all integers. For integers ...


1

You need stronger conditions than uniform convergence to ensure that $$f_n(x) \to f(x) \implies f_n'(x) \to f'(x).$$ Here is a standard theorem found in virtually all real analysis books. Suppose $(f_n)$ is a sequence of differentiable functions that converges pointwise at some point in $[a,b]$ and $(f_n')$ converges uniformly on $[a,b]$ to $g.$ Then ...


2

Uniform convergence does not preserve differentiability in the real case. For example, the functions $f_n(x)=\sqrt{x^2+1/n}$ converge to $f(x)=|x|$ uniformly. Even worse, every continuous function on a closed bounded interval, including the nowhere-differentiable Weierstrass function, is the uniform limit of polynomials. But in the complex case we have ...


5

If you're still interested in approaches that use contour integration, consider the function $$f(z) = \frac{\log(1+z) \log(-z)}{1+z^{2}}.$$ Using the principal branch of the logarithm, there is a branch cut along $[0,\infty)$ and a branch cut along $(-\infty, -1]$. Then integrating counterclockwise around a keyhole contour deformed around the branch cuts ...


1

My preferred way of avoiding losing track of notation is by using linear approximation to $u$ and $f$. That is: express the given derivatives as $$ f(a+h)=f(a)+hf'(a) + o(|h|) \tag{1} $$ and $$ u(x+h_1,y+h_2)=u(x,y)+h_1 \frac{\partial u}{\partial x}(x,y)+h_2 \frac{\partial u}{\partial y}(x,y)+o\left(\sqrt{h_1^2+h_2^2}\right) \tag{2} $$ Before even doing ...


7

Let $Q(z)=P(z)+3=z^{10}+2z^9+z^5+4$. We have: $$\lim_{R\to +\infty}\frac{1}{2\pi i}\oint_{|z|=R}(2z^2+z-1)\frac{Q'(z)}{Q(z)}\,dz=\sum_{\xi\in Z}(2\xi^2+\xi-1) $$ where $Z$ is the set of zeroes of $Q(z)$. Since, by Viète's theorem, $$\sum_{\xi\in Z}\xi = -2,\qquad \sum_{\xi\in Z}\xi^2 = (-2)^2-2\cdot 0 =4,$$ we simply have: $$\lim_{R\to +\infty}\frac{1}{2\pi ...


0

$$ \|u \circ \phi_z\| = \sup_{\zeta \in \mathbb{D}} |u(\phi_z(\zeta))| = \sup_{\zeta\in\mathbb{D}} \{ |u(w)| : w=\phi_z(\zeta) \} = \sup_{w\in \mathbb{D}} |u(w)| = \| u \| $$ where the third equality uses that $\phi_z(\mathbb{D}) = \mathbb{D}$.


1

Assuming "$\ln$" is the principal branch of the complex logarithm: $$ \operatorname{Res}\limits_{z=i}\left( \frac{e^{zt}\ln z}{z^2+1} \right) = \frac{e^{it}\ln i}{2i} = \frac{e^{it} i\pi}{4i} = \frac{\pi e^{it}}{4} $$ and $$ \operatorname{Res}\limits_{z=-i}\left( \frac{e^{zt}\ln z}{z^2+1} \right) = \frac{e^{-it}\ln(-i)}{-2i} = \frac{e^{-it} i\pi}{4i} = ...


5

Let $$ I(a)=\int_0^\infty\frac{\ln(1+at)}{1+t^2}dt. $$ Then $I(0)=0$ and \begin{eqnarray} I'(a)&=&\int_0^\infty\frac{t}{(1+at)(1+t^2)}dt\\ &=&\frac{1}{1+a^2}\int_0^\infty\left(\frac{a+t}{1+t^2}-\frac{a}{1+at}\right)dt\\ &=&\frac{1}{2(1+a^2)}(2a\arctan t-2\ln(1+at)+\ln(1+t^2))\bigg|_{0}^\infty\\ &=&\frac{a\pi-2\ln a}{2(1+a^2)}. ...


1

Define $$ h(z) = \begin{cases} f(z), & |z|\le 1 \\ g(z), & |z| > 1. \end{cases} $$ By assumption, $h$ is continous on $\mathbb{C}$, so by Morera's theorem, it suffices to check whether $$ \int_\gamma h(z)\,dz = 0 $$ for all closed curves $\gamma$. In fact, it is enough to check that the integral vanishes for all triangles. Let $\gamma$ be an ...


0

Following on from the comments we do not needed to use the bounded inverse theorem as I had originally thought. As $T$ is continuous and surjective by the open mapping theorem $T$ is an open map. This means that for some $r >0$ we have $r T(B_{l_\infty}) \subset T(B_{\mathbb{H}^{\infty}})$ where the $B$ are the open unit balls in the respective spaces. ...


2

As $$e^t=\sum_{n=0}^\infty\frac{t^n}{n!},$$ we have $$e^{-t^2}=\sum_{n=0}^\infty\frac{(-1)^nt^{2n}}{n!}.$$ Integrating term to term (why is possible?), $$\frac{\sqrt\pi}{2}\text{erf}(z)=\sum_{n=0}^\infty\frac{(-1)^nt^{2n+1}}{(2n+1)n!}.$$


0

For functions obeying just some very fairly modest restrictions, you differentiate under the integral sign by evaluating the function being integrated at the endpoint of the integral; this is the fundamental theorem of integral calculus. So $$ \frac{d}{dz} \frac{2}{\pi} \int_0^z e^{-t^2}dt =\frac{2}{\pi} e^{-z^2} $$ At $z = 0$, this is $\frac{2}{\pi}$ And ...


19

Hint $$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$$ and $$(a-b)^3=(a^3-b^3)-3ab(a-b)$$ so \begin{align*}\dfrac{1}{n^3(n+1)^3}&=\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right)-3\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\left(\dfrac{1}{n}-\dfrac{1}{(n+1)}\right)\\ ...


2

This is the transpose of the dual unit $\epsilon$. So if you represent your duals by $ \begin{pmatrix}0 & 1 \\0 & 0 \end{pmatrix} $ it will not correspond to any "real" number.


1

For a function $f(x)=\frac{1}{\pi}\frac{a}{a^2+x^2}$ (also kwnown as Lorentzian function) the Fourier transform is: $$ \int_{-\infty}^{\infty}\frac{1}{\pi}\frac{a}{a^2+x^2}e^{-itx}dx=\frac{a}{\pi}\int_{-\infty}^{\infty}\frac{1}{(a+ix)(a-ix)}e^{-itx}dx $$ Now we will use Cauchy's integral formula. Although we only need to solve our integral for the real axis, ...


7

This is identical to showing that the fractional parts of $\dfrac{n}{2\pi}$ are dense in $[0,1)$, which is true since $\dfrac{1}{2\pi}$ is irrational (see e.g. this question).


0

Lemma. If $a,b$ belong to the same connected component of $\mathbb C\smallsetminus\mathbb \Omega$, where $\Omega\subset\mathbb C$ open, then $$ f(z)=\frac{z-a}{z-b}, $$ possesses an analytic logarithm in $\Omega$ (i.e., there exists a $g\in\mathcal H(\Omega)$, such that $\exp(g(z))=f(z)$, for all $z\in\Omega$). Applying this lemma, we obtain an analytic ...


1

Since the square of your matrix is $0$, I would say that your matrix is $\epsilon$. Note that the representation given in the Wikipedia article you linked to is not the only representation using $2 \times 2$ real matrices.


1

In your Calculation, x shouldn't be the lower border in the integral. It must be H(x)*x. Because, if x < 0, the integral is cut, then the lower border is 0. And in the end it is (x-2H(x)*x) = -|x| ... :)


1

If $\Gamma$ is subgroup of $ \mathrm{SL}_2(\Bbb{Z})$ which $-\mathrm{id}\notin \Gamma$, then there is weight $1$ modular form such as $$ \mathbb{G}_1(z)=\frac{1}{2}L(0,\chi)+\sum^{\infty}_{n=1}\left(\sum_{d|n}\chi(d)d\right)e^{2\pi i n z}$$ over $ \Gamma(4)$. where $\chi:(\Bbb{Z}/n\Bbb{Z})^{\times}\to \Bbb{C}$ is Dirichlet character. but if ...


1

There are no nonzero modular forms of odd weight for $SL(2,\mathbb{Z})$. Because of the invariance of a modular form $f$ under the action of $−id$, where $id$ is the identity matrix in $SL(2,\mathbb{Z})$, it follows that $f$ of odd weight $m$ is zero: $f(z) =(−1)^mf(z)$. So we may assume that $m=2k$ is even. It is easy to see that for $k=0$ there are only ...


3

Consider the function $f(z)=\sum_{k=0}^n a_k z^{n-k}=z^n p(1/z)$. We have for $|z|=1$,$|f(z)| = |p(1/z)| \leq M$. By maximum modulus principle, we have $|f(z)| \leq M, \forall |z|\leq 1$. Thus $|z^np(1/z)| \leq M, \forall z| \leq 1$, i.e. $$|p(1/z)|\leq M|\frac{1}{z}|^n, \forall 0< \left|z\right| \leq 1$$ So $|p(z)| \leq Mz^n, \forall |z|\ge 1$


2

The integral is zero. Your trouble stems from the fact that you have a double pole. The residue at the pole is clearly zero, so the integral is zero.


3

The poles of $1/(z \sin{z})$ are at $0$, and $\pm \pi$ within $|z| \lt 5$. The residue at $0$ is $0$ - this should be apparent because $z \sin{z}$ is even. The residue at each of $\pm \pi$ is given by $(\pm 1/\pi)/(\cos{(\pm \pi)}) = \mp 1/\pi$. Thus, the integral is zero.


0

It could be useful, even if not proper/formal, to integrate the function over a small loop around the suspicious point (withtin the domain of analiticity of the function in order to take advantage of Cauchy theorems): this should result in an expression involving quantities path-related (eg. the radius of a small circle: try with $\sqrt{z}$ around the ...


3

This is globally correct, except that there are infinitely many roots $z$ to this equation, that differ by a multiple of $2i\pi$.


1

None of the four books in his "Princeton Lectures in Analysis" series have solutions but on occasion there are hints after a problem is stated.


0

By the identity theorem both claims follow trivially.


0

Note that $f$ is the composition of two functions: a Möbius transformation and the square root. So, the domain of the Möbius transformation is $\mathbb{C}\backslash\{1\}$ and its image is all the plane (Why?). On the other hand, the square root is only defined in a branch of the logarithm. Where does the Möbius transformation maps the interval $[-1,1]$? ...


3

Here's a hack at a proof-sketch. A bounded entire function is constant -- that's a standard theorem in complex variables. If your function is unbounded but entire, there must be a sequence $z_i$ with $\lim |f(z_i)| = \infty$, and since the function is bounded on compact sets, we know $\lim |z_i| = \infty$. I'm pretty sure that with a little fiddling, you ...


0

There is a lot nicer and more general solution. You could write the poles as $z=\sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}$ with $k \in \{ 0,1,2,3\}$. Then $$\text{res}(f,\sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}) = \lim_{z\to \sqrt{2}e^{i\left(\frac{\pi}{4}+k\frac{\pi}{2} \right)}} \left( ...


1

Hint: Use $$ \tan(z) = \frac {\sin z}{\cos z} = \frac {\frac {1}{2i}(e^{iz} - e^{-iz})}{\frac {1}{2}(e^{iz} + e^{-iz})} = \frac 1i \frac {e^{2iz} - 1}{e^{2iz} + 1} $$ and investigate the mappings $$ z\to u = 2iz, \quad u \to v = e^u, \quad v \to w = \frac 1i\frac{u-1}{u+1} $$


0

Let $$\Omega(z)=U\left((z-z_0)e^{-i\alpha}+\frac{\mu^2e^{i\alpha}}{(z-z_0)}\right)$$ where $z_0 = \epsilon +\delta i$, $\mu = \sqrt{(b-\epsilon)^2 + \delta^2}$ and b is the critical point of the Joukowski mapping. This represents a flow past an offset cylinder centred at $z_0$ and passes through the critical point $b$. Then consider Joukowski mapping ...


1

A curve which is contractible to the unit circle has necessarily winding number $1$, because curves with different winding numbers can not be continuously deformed into each other ("Are not homotopic", if you want to be more sophisticated). Therefore, the answer is "No" because your curve has (as can be easily shown) a winding number of $2$. Is this fine ...


0

Let $$G(s) = \frac{(s-z_1) \dots (s-z_m)}{(s-p_1) \dots (s-p_n)} = \frac{\prod_{i=1}^m |s-z_i|}{\prod_{i=1}^n |s-p_i|} \exp \left(i \left(\sum_{i=1}^m \angle(s-z_i)-\sum_{i=1}^n \angle(s-p_i)\right) \right)$$ Now when $\Gamma$ semi-encircles a pole (the small $r$) the phase difference would be $\pi$ radians. Multiplicity comes from the above summation. If ...



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