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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 $i 2 \pi$.

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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.

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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.

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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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