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0

There is a simple pole at the origin ($z^2/\sin^2{z}$ is defined to be $1$ at the origin). No other pole of the integrand is within $C$ so by the residue theorem, the integral is $i 2 \pi$.


1

As has been discussed in the comments, your solution is correct except you need to specify that $\left|\dfrac{z}{4}\right| < 1$, or equivalently $|z| < 4$, so that the geometric series converges.


0

Squaring both sides you'll get $$\cos^2(z) = 2 \\ \implies 1-\sin^2(z) = 2 \\ \implies \sin^2(z) = -1 \\ \implies \sin(z) = \pm i \\ \implies z = \arcsin(\pm i)$$ You can see this works by the identity $$\cos(\arcsin(x)) = \sqrt{1-x^2}$$ (assuming the identity holds for complex numbers) And you could convert $\arcsin(\pm i)$ into $$\arcsin(\pm i) = ...


1

First separate to cases $y=0$ and $y \neq 0$. You will find that the first has no solution. The latter implies $x=k\pi$ that implies $y = (-1)^k \textrm{arcosh}(\sqrt{2}) = (-1)^k \ln(\sqrt{2} + 1)$.


0

Boundness inequality $|g(z)|<M$ needs to take place only on a neighborhood of $z_0$ (not the whole domain of definition, $D$, say common to both $f$ and $g$), say for all $z\in (z_0-\delta_1, z_0-\delta_1) \cap D$, with given $\delta_1>0$. On the other hand, given $\epsilon>0$, there is $\delta_2>0$ such that $|f(z)|<\epsilon/M$ for all ...


0

Suppose $T$ is a map which sends $(-1, 1, 0)$ to $(\infty, 0, 1)$. Now let $\lambda \in \mathbb{C}\setminus\{0\}$. Note that $\lambda\infty = \infty$, and $\lambda 0 = 0$ so $\lambda T$ sends $(-1, 1, 0)$ to $(\infty, 0, \lambda)$. Now choose $\lambda$ appropriately.


2

Here’s how I learned to do it from Ahlfors: First write the first step in standard form $z\mapsto (-iz+1)/(z-i)$, just for uniformity’s sake. Now see where the five crucial points of the (closed) unit disk go: the center, and the four intersections of the circumference with the axes. You see that $0\mapsto i$, $1$ and$-1$ are fixed, $i\mapsto\infty$, and ...


0

(We replace $\theta$ throughout by $x$.) Let's look at the full Fourier series of $f(x) = x(\pi-x)$ on $[-\pi,\pi]$ first, where $f$ is defined outside of this interval by periodicity, i.e., $f$ is of period $2\pi$ outside $[-\pi,\pi]$. In this case $\displaystyle a_0 = \frac{1}{\pi}\,\int_{-\pi}^{\pi}\,x(\pi-x)\, dx = -\frac{2\pi^2}{3},$ where the first ...


-2

Here is an approach. Recalling the Mellin transform $$ F(s) = \int_{0}^{\infty} x^{s-1}f(x)dx.$$ We have the Mellin of our function $f(x)=\frac{1}{x^2-1}$ is given by $$ F(s) = -\frac{\pi}{2} \cot(\pi s/2). $$ So our integral can be evaluated as $$ I =\lim_{s\to 3/2}F(s) = \frac{\pi}{2} $$


0

The classical upper bound $\big|\zeta(\sigma+it)\big|\le O\big(t^{1-\delta}\big)$ for $\sigma\ge\delta$ provides $\big|\zeta(\sigma+it)\big|\le O\big(\sqrt{t}\big)$ which can be sufficient for you. See Titchmarsh's book, formula (9) on page 3. Instead of the trivial bound $\big|\Gamma(\sigma+it)\big|\le\Gamma(\sigma)$ we need something better, for example, ...


0

You can't say $f(z) \le f(w)$, because these are complex numbers; what you want is $|f(z)| \le |f(w)|$. Now use the fact that $|P(z)| \to \infty$ as $|z| \to \infty$.


1

Algebraically, you can take the modulus of $e^{i\theta}-1$ and solve it that way. That is, $$|e^{i\theta}-1| = \sqrt{(e^{i\theta}-1)(e^{-i\theta}-1)} =\sqrt{2-e^{i\theta}-e^{-i\theta}}.$$ If you make the "$\textrm{cis}(\pm\theta)$" substitution now, you should be able to find $\theta$ without too much trouble. Since you're supposed to use "geometric ...


1

look at the unit circle $|z| = 1$ and consider the points $O = 0, A = 1$ and the point $P = e^{i\theta}.$ the number $e^{i\theta} - 1$ represents the complex number $AP$ in the isosceles triangle $OP = OA = 1, AP = \sqrt 2.$ that means $\angle AOP=90^\circ$. if you want it in the negative half plane $\theta = -90^\circ$


1

Hint: $(\cos \theta - 1)^2 + \sin^2 \theta = 2$


2

If $z = x + yi$, then $|\Re(z)| = |x| = \sqrt{x^2} \le \sqrt{x^2 + y^2} = |z|$.


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


1

Clearly false. Take $f(t) = -t$. Then your would-be identity boils down to: $$\sin(t) e^{-it} = 0 \implies \sin(t) = 0,$$ which is valid for only some values of $t$. (Here I used that the exponential never vanishes)


0

Not true. Try $t = 0$, $f(t) = \pi/2$.


2

The last limit is $0$, it is not indeterminate. An indeterminate form arise when you have two functions that tend to $0$ and you take their quotient; I think it is important to understand that what this "indeterminate" form tells you is that you don't have enough information to understand if the limit is $0$, $\infty$, or any other constant. Indeterminate ...


2

$$\frac{z+\bar{z}}{2}\frac{z-\bar{z}}{2i} + i((\frac{z-\bar{z}}{2i})^2/2-(\frac{z+\bar{z}}{2})^2/2 +C)$$ $$\frac{z^2-\bar{z}^2}{4i} + i(\frac{z^2-2z\bar{z}+\bar{z}^2}{-4}/2-\frac{z^2+2z\bar{z}+\bar{z}^2}{4}/2+C)$$ $$\frac{z^2-\bar{z}^2}{4i} + i\frac{z^2-2z\bar{z}+\bar{z}^2}{-4}/2-i\frac{z^2+2z\bar{z}+\bar{z}^2}{4}/2+iC$$ $$\frac{z^2-\bar{z}^2}{4i} + ...


5

What you have is $-\frac{i}{2} z^2 + iC$, and this seems fine. The issue ought to be in the step you did not detail.


2

The problem is that complex powers are multi-valued functions. By definition, $a^b = e^{b \log(a)}$, but there are different branches of $\log(a)$, each of which may give a different value to $a^b$. So it's not true in general that $(a^b)^c = a^{bc}$. What you can say is that $(a^b)^c = \exp(c \log(a^b)) = \exp(c \log(e^{b \log a}))$, and $\log(e^{b \log ...


2

First let me correct a mistake in your question: is it WRONG to think of this as an astroid. It is just 4 arcs of circles, NOT an astroid, though it looks somewhat similar. Second. The conformal map is certainly NOT fractional-linear (which you call Mobius). This conformal map is written explicitly in the paper I already referred to on MO: arXiv:1110.2696. ...


5

The integral as stated does not converge. On the other hand, its Cauchy principal value exists and may be computed using the residue theorem. Consider the integral $$\oint_C dz \frac{\sqrt{z}}{z^2-1} $$ where $C$ is a keyhole contour of outer radius $R$ and inner radius $\epsilon$, with semicircular detours of radius $\epsilon$ into the upper half plane ...


4

Let $\delta_1>0,\delta_2<1$ real numbers close to zero and one, respectively, and I=$(\delta_1,\delta_2)\cup(\delta_2^{-1},\delta_1^{-1})$. We have: $$ \int_{I}\frac{\sqrt{x}}{x^2-1}\,dx = \int_{\delta_1}^{\delta_2}\frac{\sqrt{x}}{x^2-1}\,dx +\int_{\delta_1}^{\delta_2}\frac{\sqrt{1/x}}{1-x^2}\,dx = ...


1

There is an example of such a function on that very page. Namely the one given in #4.


4

Mobius transformations are bijective Such a transformation maps the extended complex plane to itself in such a way that every point gets covered, and each point gets covered only once. "Extended" means we also consider the point "at infinity" (To formalize this, we need to give the one point compactification a complex manifold structure, aka the Riemann ...


1

Mobius transformations correspond to rigid motions of the sphere in a very natural way. Mobius Transformations Revealed (also on YouTube) is the definitive illustration of this. Here's simpler illustration: In this animation, the graticule of lines that we see on the plane and on the sphere form the image of a square in the complex plane. If we project ...


2

For $p\leq 0 $ we have that $f(x)=|x|^{-p}$ is continuous and thus in the compact $[a,b]$ it has a maximum value let's say $M>0$. This means that $I\leq M(b-a)$. Now if you mean $p>1$ , in order for $I<\infty$ we must have that $0 \not \in [a,b]$. Because if $0\in [a,b]$ then $I\geq \int_{0}^{b}|x|^{-p} dx=\int_{0}^{b}x^{-p} dx=\infty$. Think if ...


3

Here is a fairly simple and geometrically intuitive argument based on the following fixed point theorem. Lemma: Let $R \subset \mathbb C$ be a solid rectangle and suppose that $f:R\rightarrow\mathbb C$ is continuous. If $f(R)\supset R$, then $f$ has a fixed point in $R$. Note that the lemma holds for any compact, simply connected set. It's stated for ...


0

The function $\psi:\>X\to{\mathbb R}$ defined by $$\psi(f):=\int_0^1\left|{\rm Im}\bigl(f(x)\bigr)\right|\>dx$$ is continuous on $X$: $$|\psi(f)-\psi(g)|\leq\int_0^1|f(x)-g(x)|\>dx=\|f-g\|\ .$$ Therefore $S:=\psi^{-1}(0)$ is closed in $X$.


4

As mollyerin says, you need to be more careful. I would use the following argument. Let $u,v:D\to\mathbb{R}$ be the real and imaginary parts of $f$, and let $x,y,$ be real coordinates on $D$. We have $$f(z)dz=(u(z)+iv(z))(dx+idy)=u(z)dx-v(z)dy+i(u(z)dy+v(z)dx).$$Since $f$ is holomorphic, the Cauchy-Riemann equations imply that both real and imaginary part ...


1

Just an outline of an answer. I assume that all the parameters $\gamma$, $\alpha_1$, $\alpha_2$, $\beta$ and $s$ are positive. All the terms with exponential factors have integrals that converge. The only term left to consider is $$[1- 1/(1 + sv^{-1})](1/\alpha_2)v^{2/\alpha_2 - 1} \sim (s/\alpha_2)v^{2/\alpha_2 - 2}.$$ The integral therefore converges if ...


0

I think it diverges. If $\beta, \alpha_1, \alpha_2 > 0$, then the exponential terms all have integrals that converge, so the only term that matters is $v^{\frac{2}{\alpha_2}-1} $. But $\frac{2}{\alpha_2}-1 > -1$, so its integral diverges by comparison with $\frac1{v}$ (i.e., $v^{\frac{2}{\alpha_2}-1} =v^{\frac{2}{\alpha_2}}v^{-1} > \frac1{v} $).


0

Let $P(z)=\prod_{i=1}^n(z-\alpha _i)$ with $\Re \alpha _i<0$ $(i=1,2,...,n)$ and $$g(z)=\frac{1}{z-\alpha_1}+\frac{1}{z-\alpha _2}+...+\frac{1}{z-\alpha _n}.$$ Then $P^\prime (z)=P(z)g(z)$, and hence every root $z_0$ of $P^\prime(z)=0$ satisfies $P(z_0)=0$ or $g(z_0)=0$. Suppose that $g(z_0)=0$ and $\Re z_0\ge 0$. It is easy to see that $\Re ...


7

Following is an elementary and mundane approach which count/bound the roots of the equation $$\sin z = z\tag{*1}$$ using winding number. For any $n \in \mathbb{N}$ and $r > 0$, let $R_n = (2n+\frac32)\pi$, $C_n$ be the square contour centered at origin with side $2R_n$. $S_1, S_2, S_3, S_4$ be the following $4$ line segments whose union is $C_n$. ...


1

Hint: Show the complement is open. Show that the nearest element of $X$ to $f$ is $\mathrm{Re}\,f$.


3

I would like to suggest "different" approach. First off as noted in one of the comments $\lambda_i$ must be real since $\mathbb{C}$ is not an ordered field. On the another hand $\mathbb{C}$ is a vector space over the field of real numbers. So I am going to translate your original statement into a geometric interpretation. Imagine that you pick $n$ points in ...


3

By the triangle inequality we have $$|\lambda_1 a_1 + ... +\lambda_n a_n|\leq |\lambda_1||a_1|+...+|\lambda_n||a_n|$$ Since each $|a_i|<1$ and each $\lambda_i$ is non-negative, we have $|\lambda_i|= \lambda_i$. Then $$|\lambda_1||a_1|+...+|\lambda_n||a_n| < \sum_i \lambda_i =1$$ Since you tagged this as complex analysis, I assume you meant for the ...


1

Let $|z| = 1$. If $z$ is real, $z = \pm 1$, in which case $|z^{10} - 9| = 8 < 10 \le 10|z|e^{\text{Re}(z) + 1} = |10ze^{z+1}|$. If $z$ is not real, $\text{Re}(z) > -1$, and thus $$|z^{10} - 9| \le |z|^{10} + 9 = 10 < 10|z|e^{\text{Re}(z) + 1} = |10ze^{z+1}|.$$ Therefore, $|z^{10} - 9| < |10ze^{z+1}|$ for all $|z| = 1$. By Rouche's theorem, ...


12

The great Picard theorem says that $f(z) = \sin(z)-z = c \in \Bbb C$ infinitely often for all but possibly one value of $c$. Note that $f(z+2\pi) = f(z) - 2\pi$. Suppose some $c_0$ is not hit infinitely many times; then $c_0+2\pi$ is, say $f(z_k) = c_0+2\pi$ for some infinite sequence $(z_k)$. By the above functional equation, then, $f(z_k+2\pi) = c_0$, ...


0

Hints: $$|e^{ikl/n} - e^{ikl}e^{-il/n}| = |e^{ikl/n}||1 - e^{-il/n}| = |\frac{1}{e^{il/n}}||e^{il/n} - 1| = |e^{il/n} - 1|$$ We have that $\displaystyle{\sum _{k=1}^{n}} \ 1 = n$. Hence $$\lim_{n\to\infty}\sum _{k=1}^{n}|e^{ikl/n} - e^{ikl}e^{-il/n}| = \lim_{n\to\infty}\sum _{k=1}^{n}|e^{il/n} - 1| = \lim_{n\to\infty}n|e^{il/n} - 1|$$ Remember that ...


0

$e^{i(k-1)l/n}$ is a complex number with modulus one, hence it can be factored away from the LHS of the first equation, giving identity. For the same reason, the second sum equals: $$ n\cdot|1-e^{il/n}| = 2n\cdot\sin\frac{l}{2n}$$ and since $\lim_{x\to 0}\frac{\sin x}{x}=1$, the limit equals $l$.


0

The function $f:\mathbb C\rightarrow\mathbb R$ prescribed by $z\mapsto|z+i|$ is continuous so the preimage under $f$ of open set $(-\infty,2)$ is open. The function $g:\mathbb C\rightarrow\mathbb R$ prescribed by $z\mapsto Im(z)$ is continuous so the preimage under $g$ of open set $(-\infty,0)$ is open. $S$ is the intersection of these open sets, hence ...


1

To minimize the right hand side of (1), first consider all $\lambda$ with a fixed modulus. The expression will be minimized among those choices of $\lambda$ which make the last term as large as possible. Let $\lambda=z_1$ and $\sum a_ib_i=z_2$ to simplify the notation. Multipling $z_2$ by $\overline{z_1}$ will rescale $z_2$ to have modulus $|z_1||z_2|$ and ...


0

It's implicit that the period is $P$ as you can gather from the partial sums of the FS (read line 2 under "Definition"). The important thing here is that the Fourier coefficients are always defined by integrating over a symmetric interval of length $P$, the period of $f$. Here you are looking for a so-called "half range expansion", so you need to extend ...


0

Suppose $\;|z|<r\;$ , then $$|a_nz^n|\le |a_n|r^n\;\;\;(**)$$ but $$\lim_{n\to\infty}\sup \sqrt[n]{|a_n|}=\frac1r\implies \;\text{for almost all index}\;\;n\;,$$ $$\sqrt[n]{|a_n|}\le\frac1r=r^{-1}\implies |a_n|\le r^{-n}$$ and thus in (**) we get $$|a_nz^n|\le1\;,\;\;\;\text{for almost all}\;\;n$$ and we get absolute convergence. There are proofs ...


2

There is no extra $\varepsilon$ in the last limit. De l'Hopital rule just gives $-\pi$ as the value of the limit. This limit is twice the integral since $$\int_{0}^{+\infty}\frac{\cos x-1}{x^2}\,dx = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{\cos x-1}{x^2}\,dx.$$ Also notice that this problem can also be solved through integration by parts: ...


1

Keep in mind that, when $f(z)$ has an $(n+1)$th-order pole at $z=z_0$, the residue may be computed using $$\operatorname*{Res}_{z=z_0} f(z) = \frac1{n!} \left [ \frac{d^n}{dz^n} \left [(z-z_0)^{n+1}f(z)\right ] \right ]_{z=z_0} $$ Thus in your case, as you are interested in the residue of the function $f(z) = (1+z^2)^{-(n+1)}$ at $z=i$, you must compute, ...


0

Shouldn't there be moduli symbols around the denominators in the third line? If you mean $\displaystyle \frac{1}{| \sqrt{z^2+3}|+2}< \frac{1}{\frac{3}{2} + 2}$ then this is clear because $0 < A < B$ implies that $\displaystyle \frac{1}{A}>\frac{1}{B}$ along with the basic theory of inequalities. The other is proved similarly provided one is ...



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