New answers tagged

0

Your interpretation of removable singularity is not the accepted one. You are not required to prove that $f$ is not analytic at $0$. For example $f(z)=0$ for all $z \neq 0$ has a removable singularity at $0$.


0

You are right: we have that $b_n=0$ for all $n \ge 1$ and therefore $f(z)=\sum_{n=0}^{\infty}{a_n}{z^n}$ on $A$. Now define $g(z):= \sum_{n=0}^{\infty}{a_n}{z^n}$ for $|z|<2.$ Then $g$ is holomorphic and $g$ is a holomorphic continuation of $f$ on the disc $|z|<2.$ Conclusion: $f$ has a removable singularity at $z=0.$


0

No. Let $f_n(z):=2^nz^n-1$ and $K:= \{1/2\}$. Then $f_n(1/2)=0 \to 0$ and $ f_n'(1/2) \to \infty.$)


0

@Arthur, I like your solution very much! Then apparently x = 0, x = sqrt(3)y and x = - sqrt(3)y are the three sought for three lines. There is really nothing new in this comment.


3

Let $m=\inf_{z\in\partial\Omega} |z-a|$ and $M=\sup_{z\in\partial\Omega} |z-a|$. Let $h_1(z)=f(mz+a)$ for $|z|<1$. It's easy to see that $h_1$ is well-defined($mz+a\in \Omega$) and $|h_1(z)|<1$ for all $|z|<1$ and $h_1(0)=f(a)=0$. Schwarz’s lemma implies that $|h_1’(0)|\leq 1$, which is exactly the left part of the desired inequality. For the ...


1

The residue at $z=1$ is $\lim_{z\to1}(z-1)\dfrac {2z-1}{z^2-1}=\dfrac 12$. Thus by the residue theorem, we get $2\pi i\cdot\dfrac 12=\pi \cdot i$.


2

You are integrating $\frac {(z-a)(\frac 1 z - \overline {a})} {(z-b)(\frac 1 z - \overline {b})z}$. Poles inside the contour are at $0$ and $b$. the residues are $\frac a b$ and $\frac {(b-a) (1-\overline {a} b)} {b(1-|b|^{2})}$. So the answer is $\frac 1 b + \frac {(b-a) (1-\overline {a} b)} {b(1-|b|^{2})}$.


1

If all else fails, you can always brute-force it. Let $c:[0,1)\to\Bbb{C}$ be a function mapping the interval $[0,1)$ to your circle. The usual formula for the contour integral applies $$\int_c\frac{2z-1}{z^2-1}\ dz=\int_0^1\left(\frac{2c(t)-1}{c(t)^2-1}\cdot\frac{d}{dt}c(t)\right)\ dt$$


3

Hint: From Cauchy's integral formula (CIF), since $1 \in D=\{ z : | z − 1| < 1\}$ and $-1 \notin D$ $$g(1)=\frac{1}{2\pi i}\int\limits_{|z−1|=1} \frac{g(z)}{z-1}dz$$ where $g(z)=\frac{2z-1}{z+1}$ (check that $g(z)$ satisfies the criteria required by CIF). Or $$\int\limits_{|z−1|=1} \frac{g(z)}{z-1}dz=\int\limits_{|z−1|=1} \frac{2z-1}{z^2-1}dz=2\pi \cdot i ...


1

You think of $a\in\mathbb{C}$ like a vector in $\mathbb{R}^2$ $$a = u + iv \sim \begin{bmatrix} u \\ v\end{bmatrix}, \ b = w + iz \sim \begin{bmatrix} w \\z\end{bmatrix}$$ Then $<a,b> := uw + vz$ and $a^*b = uw + vz + i(uz - vw)$; therefore $$<a,b> = 0 \iff Re(a^*b) = 0$$ Similarly if $a,b\in \mathbb{C}^n$, you may take $u,v,w,z\in \mathbb{R}^{...


2

A subgroup $\Gamma$ of $PSL_2(\Bbb R)$ acts properly discontinuously on the upper half plane if and only if it is discrete. So take any discrete abelian subgroup of $PSL_2(\Bbb R)$, i.e., an abelian Fuchsian group. Note that every abelian Fuchsian group is cyclic. Edit: Since the action also should be free, finite cyclic groups are ruled out, too.


0

You can use the Cauchy-Riemann equations. If you write $f(x+yi)=u(x,y)+v(x,y)i$, with $u(x,y),v(x,y)\in\mathbb R$, then $u(x,u)=\sqrt{\lvert xy\rvert}$ and $v(x,y)=0$. But then$$\left\{\begin{array}{l}\frac{\partial u}{\partial x}(x,y)=\frac{\partial v}{\partial y}(x,y)\\\frac{\partial u}{\partial y}(x,y)=-\frac{\partial v}{\partial x}(x,y)\end{array}\right.\...


1

A basic fact about differentiation in the complex analytic sense is that C-R equations are necessary for differentiation. In this case C-R equation show that $\Re z =\Im z$ and $\Re z =-\Im z$ so the only we only have to consider the the point $z=0$.


1

One option is to write $g$ as a composition of “simple” transformations: $$ g(z) = 2i \left( 1 + \frac{1}{z-1}\right) $$ and then successively determine the image of the right half-plane. Another option is to use that fact that $g$ is a Möbius transformation: It maps the imaginary axis to a circle (or line), and that image is uniquely determined by the ...


0

Hint The function $f(x)$ diverges only in its singularities. So do any of those singularities belong to the inside or on the boundary of $D$?


2

For any Hermitian matrix $H\in M_n(\mathbb C)$ and positive semidefinite matrix $P\in M_n(\mathbb C)$, by unitarily diagonalising $H$, we see that $$ \lambda_n(H)\operatorname{Tr}(P)\le\operatorname{Tr}(HP)\le\lambda_1(H)\operatorname{Tr}(P).\tag{1} $$ Therefore, for any complex square matrix $A$ and Hermitian matrix $B$, if $\bar{A}$ denotes the Hermitian ...


5

I would suggest that you think geometrically from the start. That's a lot easier. What kind of point lands on the imaginary axis when you cube it? But now that we're here: $$ x^3-3xy^2=x(x-\sqrt3y)(x+\sqrt3y) $$ is $0$ iff one (or more) of these three factors is zero. Each of the factors are zero along a straight line in the plane ($x=0$, for instance, is ...


0

Let $K$ be the kernel of the measure $\omega$: $$K = \{g\in L^2: ||g||_\omega=0\}. $$ When there are points on $T$ where $\omega =0$, the set $K$ is non-empty. If $g\in K$, then $\bar{g} \in K$ as well, so there exists $h\in K$ such that $h\not\in H^2$. Take any $f\in H^2$, then a sequence $f_k$ that converges to $f$ also converges to $f+h$ almost ...


1

This is too complicated. There is a much simpler proof of the distributional Fourier transform with $0<\Re(a)<d$ in this previous answer I gave: How to calculate $c_a$ where $\left(f\mapsto\int_{\mathbb{R}}\frac{f(t)-f(0)}{|t|^{a}}dt\right)=c_a\mathcal{F}_x(|x|^{-1+a})$ It is based on representing $\frac{1}{|x|^a}$ as a continuous superposition of ...


0

You know that $f'(z)=$ $\left[\begin{array}{cc} u_x & -v_x \\ v_x & u_x \end{array}\right] $ then $f'(a,b)$ equal $\left[\begin{array}{cc} u_x & -v_x \\ v_x & u_x \end{array}\right] \left[\begin{array}{c} a\\ b \end{array}\right]=\left[\begin{array}{c} au_x -bv_x\\ av_x+bu_x \end{array}\right]$ The book consider $a+ib = \left[\begin{...


2

I think you can find all ingredients in Ullrich's book on pages 4-6. Ullrich explains that $f$ is complex differentiable at $z$ iff there exists $a \in \mathbb C$ such that $f(z+h) = f(z) +a \cdot h + o(h)$. Then $f'(z) = a$. The map $L : \mathbb C \to \mathbb C, L(h) = a \cdot h = f'(z)\cdot h,$ is $\mathbb C$-linear and thus trivially also $\mathbb R$-...


3

Writing your limits with an $h$ makes both equations look like the same limiting procedure. But there is a subtle difference: in your first equation we have $h \in \mathbb{C}$, whereas the second equation is real and hence $h \in \mathbb{R}$. Consequently the first equation is a lot stronger than the second. Instead of approaching $0$ only from the left or ...


1

People sometimes say things like "$z^2$ is holomorphic because it depends only on $z$, while $|z|^2=z\overline z$ is not holomorphic because it depends on $z$ and $\overline z$." That's a very useful "intuitive" point of view, giving insight into what is and what is not holomorphic, except that of course taken literally it's nonsense, since $\overline z$ ...


1

If $f = \sum_n c_n e^{2i\pi n x}$ is periodic and $L^2[0,1]$ then $|c_n|$ is bounded so that $F_k(z)= \sum_n c_n z^n e^{-\pi n^2/k^2}$ is analytic on $\Bbb{C}^*$. Moreover $$\lim_{k \to \infty}\|F_k(e^{2i\pi x})- f\|_{L^2[0,1]}^2 =\lim_{k \to \infty} \sum_n |c_n|^2 (1-e^{-\pi n^2/k^2})^2 = 0$$ $H^2$ is the subspace with $c_n = 0$ for $n < 0$, the ...


1

Hint: to apply Fubini's Theorem it is enough to check integrability on the product space. Here $|t^{z-1}|=t^{\Re z -1}$. If $C$ is a closed path contained in $\{z: Re z >0\}$ the there exists $r>0$ such that $\Re z >r$ for all $z$ on $C$. Can now check integrability? Suppose $C$ is given by $\gamma:[0,1] \to \mathbb C$. Then $\int_0^{1}\int_1^{\...


6

Don't blame yourself for not understanding it. It is not clear at all. It seems to me that whoever wrote that was aiming at the concept of Wirtinger derivatives: if $U\subset\mathbb C$ and $f$ is a map from $U$ into $\mathbb C$, then, if you see $f(z)$ as $f(x+yi)$ (with $x,y\in\mathbb R$, we define$$\frac{\partial f}{\partial z}=\frac12\left(\frac{\partial ...


4

$\partial U$ is the boundary of $U$, and $d(z, \partial U)$ denotes the distance from $z$ to the boundary of $U$: $$ d(z, \partial U) = \inf \{ |z - w| : w \in \partial U \} \, . $$ If $0 \le \rho \le r < d(z, \partial U)$ then the disk $B(z, r)$ is contained in $U$, and an application of Cauchy's integral formula gives $$ |f(z)| \le \frac{1}{2 \pi} \...


2

If you draw a diagram, then the equation describes a circle of radius $6$ centred at $10$ units above the origin. Thus, we have that $r\cos\theta$ ranges from $-6$ to $6$ and that $r\sin\theta$ ranges between the extremes of $4$ and $16,$ with $r=|z|$ ranging between $4$ and $16.$ Now if we set $\cos\theta=x$ and $\sin\theta=y,$ then it follows that the ...


3

Let $z=r(\cos\theta+i\sin\theta).$ Thus, $$r^2\cos^2\theta+(r\sin\theta-10)^2=36$$ or $$\sin\theta=\frac{r^2+64}{20r}\geq\frac{2\sqrt{64r^2}}{20r}=\frac{4}{5},$$ which gives $$\arcsin\frac{4}{5}\leq\theta\leq\pi-\arcsin\frac{4}{5}.$$ Also, write $$8\sin\theta+6\cos\theta=10\sin\left(\theta+\arccos\frac{4}{5}\right).$$ Can you end it now? Actually, I ...


1

This problem exemplifies the necessity of taking a branch cut of $\mathbb{C}$ in order to define $\log z$. We begin with the case $\theta = 0$, which leads to the power series expansion $\log z = (z-1) - \frac{(z-1)^2}{2} + \dots$. Here, we consider $\log z$ under the principal branch cut $-\pi < \arg(z) < \pi$. This expansion only converges inside ...


0

Take a look at the following picture : and see the connection with the issue in the so-called Argand plane (knowing that what is represented is a partial sum of the series). It remains to prove that this "inward spiraling" movement terminates as a limit point... therefore, I don't say that this graphics constitue a "proof without words" ; it just ...


4

The key here is to understand that $\partial_z\frac{1}{z^*}=0$ everywhere but at the origin. This is how a delta-singularity can arise in this context. Once one is aware of this, integrating this function on a square shouldn't be too difficult, and can be done without use of multivariable tools like Stokes' theorem. To wit, using the fundamental theorem of ...


1

If $\tilde{F}(., t) = \tilde{f}_t$ denotes the lifting then it suffices to check that $\tilde{F}$ is continuous into $\tilde{X}$. Do it "horizontally" locally instead of "vertically" as you've been trying to do. That is, pick a point $(0, t_0) \in I \times I$ and show that there exists an open set $W \subseteq I \times I$ such that $I \times \{ t_0 \} \...


2

It converges by Dirichlet's test: the partial sumes of the series $\sum_{n=1}^\infty i^n$ are bounded, the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ is monotonic and converges to $0$.


8

Hint $$\sum_{n=1}^\infty \frac{i^n}{n}=\sum_{n=1}^\infty \frac{(-1)^{n}}{2n}+i\sum_{n=0}^\infty \frac{(-1)^{n}}{2n+1}.$$


1

This can be solved with elementary calculations, and without using Rouché's theorem. To simplify the notation we substitute $w = z-1$, so that the task is to find a solution $w$ of the equation $$ \tag{*} a(w+1)^2 - (w+1) + 1 = aw^2 + (2a-1)w + a = 0 \, . $$ satisfying $|w| \le 1$. If $a=0$ then $w=0$ is such a solution. For $a \ne 0$, $(*)$ is ...


1

First, convert what you are given on the LHS to polar coordinates. We have that $$e^{-i\theta}=\cos{\theta}-i\sin{\theta}$$ therefore \begin{align}e^{-i(0.4)\pi}=\cos(0.4\pi)-i\sin(0.4\pi)&=\cos(72^{\circ})-i\sin(72^{\circ})\\&=\frac{1}{4} (\sqrt{5} - 1)-i\Big(\sqrt{\frac{5}{8} + \frac{\sqrt{5}}{8}}\Big)\\&\approx0.309017-0.951057i\end{align} ...


2

Consider $1/f(z)=(z-i)(e^{\pi z}+1)=(z-i)g(z)$ say. Obviously $z-i$ has a simple zero at $i$. Now consider $g(z)=e^{\pi z}+1$. Then $g(i) =e^{\pi i}+1=0$ and $g'(i)=\pi e^{\pi i}=-\pi\ne0$. Therefore $g(z)$ also has a simple zero at $z=i$. Therefore $1/f(z)$ has a double zero at $z=i$. Thus $f(z)$ has a double pole at $z=i$.


4

Hint: Let $w=f+ig.$ Then write it as $$ \frac{w''}{w'}=\frac{3w'}{w}\\ \frac{(w')'}{w'}=\frac{3w'}{w}.$$


1

Hint: Note that $$ \sum_{n=1}^\infty \frac{z^n}{1 + z^{2n}} \leq \sum_{n=1}^\infty \frac{1}{z^n} $$


2

$$I = \int_{|z|=1}^{} \frac{(z+1/z)\frac{1}{2}}{(2+\frac{z+1/z}{2})} \frac{dz}{iz} = \frac{1}{i} \int_{|z|=1}^{} \frac{z^2+1}{z(z^2+4z+1)}dz = \frac{1}{i} \int_{|z|=1}^{} \frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}dz = 2\pi \color{red}{i} \ \sum_{j=1}^{2}\text{Res}\left[\frac{z^2+1}{z(z+2-\sqrt{3})(z+2+\sqrt{3})}, z_j\right]$$ Note that you have an extra ...


0

Express the exponential on LHS in rectangular form. Perform the addition and change the sum again to polar form.


2

Your calculation for the residue at $z=-2+\sqrt3$ is wrong. In fact $$ \text{Res}f(z)\bigg|_{z=-2 + \sqrt3}=\frac{z^2+1}{z(z+2+\sqrt3)}\bigg|_{z=-2 + \sqrt3}=-\frac{2}{\sqrt3}.$$


0

Wilf's "generatingfunctionology" (Academic Press, 1994) states (and proves) the formula as valid for formal power series (theorem 5.1.1, page 172).


1

Hint: if $a\neq 1$ then $g(z):=f\big(z+b/(1-a)\big)$ satisfies $g(az)=g(z)$.


1

From the very definition of Grothendieck Residue: Let $\omega = \sum_{j=1}^n f_jdz_j$ be a a local one form defining the distribution around a point $p$ (which we choose to be the origin in the coordinates $z$). Then, for a small polydisc $P$ with essential boundary $\Gamma$ we have $$ \text{Res}_{p} \left[ \begin{array}{cccc} df_{1}^{(j)} \wedge & \...


0

Reversing the direction of integration along the unit circle is done via the substitution $\theta = 2\pi - \phi$: $$ \int_0^{2 \pi} f(e^{i\theta}) \, d\theta = \int_{2 \pi}^0 f(e^{-i\phi}) \, (-1)d\phi = \int_0^{2 \pi}f(e^{-i\phi}) \, d\phi \, . $$ Both the substitution and the swapping of the integral bounds introduce a factor $(-1)$. In your case: $$ \...


0

$f(z)= z+1 + \frac{\sin (z-1)}{z-1} \to 1+1+1=3$ as $z \to 1.$ By Riemann, $z=1$ is a removable singularity of $f$. Consequence ?


1

Actually, one can characterize all functions $f$ holomorphic on the unit disc $D$ with $f(0)=0, f'(0)=1, f(D)=U$ starlike (at $0$) as the set of all compositions $f(z)=\frac{1}{B'(0)}h(B(z))$ where $h$ is a normalized ($h(0)=0, h'(0)=1$) univalent starlike function on $D$ and $B$ ranges through the set of holomorphic functions on $D$ that satisfy $B(0)=0, B'(...


1

Let $g(z)=(1-i)f\bigl((1+i)z\bigr)$. If $x\in\mathbb R$, then $(1+i)x\in L$, and therefore $f\bigl((1+i)x\bigr)\in L$. But then, since $(1-i)(1+i)=2\in\mathbb R$, $g(x)\in\mathbb R$. So, $g$ is an entire function which maps $\mathbb R$ into $\mathbb R$. But then $(\forall z\in\mathbb C):\overline{g(z)}=g\left(\overline z\right)$. In particular, $g(z)=0\...


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