# Tag Info

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Yes. Consider that there is one unique solution to the differential equation $y^{(n)}(x) = 0$ where $y^{(j)}(0) = c_j\cdot (j-1)!$ for $0< j<n$ and $y(0) = c_0$. This solution is the polynomial $y = c_{n-1}x^{n-1} + c_{n-2}x^{n-2} + \cdots + c_1x+ c_0$.

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Hint: Can you do the case $n=1$? Induct on $n$.

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Without doing any complex integration, we can easily see that $$\lim_{\epsilon \to 0} \frac{1}{x + i\epsilon} = \lim_{\epsilon \to 0} \frac{x - i\epsilon}{x^2 + \epsilon^2} = \lim_{\epsilon \to 0} \frac{x }{x^2 + \epsilon^2} -i \lim_{\epsilon \to 0} \frac{\epsilon}{x^2 + \epsilon^2}$$ Applying the limit to the first term gives $\frac{1}{x}$, while the ...

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Fort the first part, the function $f(z)=e^z$ is a typical example. It is a non-constant entire function attaining every value with one exception - it is never zero. For the second part, typically $e^{1/z}$ is considered. One shows that arbitrarily close to the essential singularity $z=0$, all non-zero values are attained. You can "see" this in the plot here: ...

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Yes, your proof is correct. A bit more concisely: for every $z$ on the line segment between $1$ and $i$ $$|f'(z)|= 3|z|^2 \ge 3 (1/\sqrt{2})^2 = \frac32$$ whereas $$\left|\frac{f(i)-f(1)}{i-1}\right| = \left|\frac{-i-1}{i-1}\right| = 1$$

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First I think $\alpha$ must be positive. I think. P.S. I could not post comment so I post answer.

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Note that the integral can be written as $$\int_0^{\infty} \dfrac{x^4e^{-x}}{(1-e^{-x})^2} dx = \sum_{k=1}^{\infty} k\int_0^{\infty}x^4 e^{-kx} dx = \sum_{k=1}^{\infty}k \cdot \dfrac{24}{k^5} = 24 \zeta(4) = \dfrac4{15} \pi^4$$

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One option, in particularly if one knows (roughly) what to expect is to attack it from the other side. If we didn't already know what to expect, we could shape our expectation by considering the case where $k = -m$ is a negative integer. Then the plain semicircular contour works, and we obtain $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} e^{st}s^{-m}\,ds ... 3 The answer is -1/2. Indeed, let the considered integral be denoted by I. Clearly we have$$\eqalignno{ I&=\int_0^1\frac{\ln x}{(1+x)^3}dx+\int_1^\infty\frac{\ln x}{(1+x)^3}dx\cr &=\int_0^1\frac{\ln x}{(1+x)^3}dx+\int_0^1\frac{\ln (1/x)}{(1+1/x)^3}\frac{1}{x^2}dx\cr &=\int_0^1\frac{(1-x)\ln x}{(1+x)^3}dx\cr &=\left[\frac{x}{(1+x)^2}\ln ...

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Calculating the residues, $i$ and $-i$ which are inside the contour. Both are simple poles... This should give that $\Sigma Res = 1/(2i) + 1/(-2i) = 0$. We verify this now. Let $z = 2e^{it}$ and $dz = i z dt$, which gives $1/i\cdot \int_{[0,2\pi]} 1/(2z)\cdot \frac{i}{z+i} + \frac{-i}{z-i} = 1/2\cdot \int_{[0,2\pi]} \frac{1}{z^2+iz} - \frac{1}{z^2-iz} = i/2 ... 3 Let's try that decomposition again. In particular, we can find$A,B$such that $$\frac{1}{z^2+1} = \frac A{z-i} + \frac B{z+i} \implies\\ 1 = A(z+i) + B(z-i)$$ There are different ways to solve form this point. I like to solve by plugging in the roots of the denominator for$z. $$1 = A(-i + i) + B(-i -i) \implies 1 = -2Bi\\ 1 = A(i + i) + B(i -i) ... 1 Yes. We could do this another way, too.$$\begin{align} \frac{d}{dt}e^{it}&=\frac{d}{dt}\left(\cos{t}+i\sin{t}\right)\\ &=-\sin{t}+i\cos{t}\\ &=i\left(\cos{t}+i\sin{t}\right)\\ &=ie^{it} \end{align}$$Be careful, because you technically need a more rigorous definition of complex exponentials and derivatives for complex numbers. However, it ... 6 Because \gamma(t)=e^{it},\;t\in[0,2\pi]. Hence$$ \frac{1}{2\pi i}\int_0^{2\pi}\frac{\gamma'(t)}{\gamma(t)}\,dt =\frac{1}{2\pi i}\int_0^{2\pi}\frac{ie^{it}}{e^{it}}\,dt=1$$1 Hint: do it explicitly for the function f(z) = 1. 0 First of all, you can only assert that pluriharmonic functions locally can be written as the real part of a holomorphic function. Furhermore there seems to something missing in your decomposition: the right hand side vanishes at z=0. But, if you're happy with a local result, and assume that f(0) = 0 (or add a constant to the right hand side), then f = ... 1 I think what you're missing is that the issue isn't about choosing a branch to consistently define a function, but choosing branches to make an identity valid. It might help to visualize things dynamically. At the start, you have three points: a = 1, b = 1, and c = 1, and we choose \sqrt{a} = \sqrt{b} = \sqrt{c} = 1, and we have the identities c = ... 1 In the heart of the problem is the following problem: Given a region in \mathbb{C}, can we define a "logarithm"? Meaning, can we have a function g(z) such that e^{g(z)}=z? This is because, if we can do that, we can define a "square root" unambiguously: \displaystyle \sqrt{x} := e^{\frac{1}{2}\log(x)} and it is obvious why this is called THE ... 0 To see why they equate, just take the following contour integral around the unit circle:$$ \oint \frac{dz}{z^k}=\int_{0}^{2\pi}\frac{d(e^{i\theta})}{e^{ik\theta}}=\int_{0}^{2\pi}\frac{ie^{i\theta}d\theta}{e^{ik\theta}}=\int_{0}^{2\pi}ie^{i(1-k)\theta}d\theta. $$If k=1, then the integral is 2\pi i. If it's any other integer, then it's$$ ... 0 Iff$has an isolated singularity at$z_0$, then the coefficient of$\dfrac{1}{z-z_0}$in the Laurent Series is called the residue of$f$at$z_0$, denoted$a_{-1}=\operatorname{Res}_f(z_0)$1$1/\Gamma(z)$has order$1$but infinite type. See Wikipedia 0 Following Hans Lundmark's suggestion, we compose 3 maps:$\frac{z+1}{1-z}$,$z^2$,$\frac{-i+z}{i+z}$, to get the map $$h(z)=\frac{(1+z)^2-i(1-z)^2}{(1+z)^2+i(1-z)^2}$$ which maps the upper boundary of the semidisk to the upper boundary of the unit disk, and the lower line segment of the semidisk to the lower boundary of the unit disk. Now we solve the ... 1 I'll solve$(iii)$after which you should be able to handle$(ii)$. Let$z\in \mathbb C\setminus \{-2, 0, 1\}$. The following holds: $$\dfrac{1}{1-z}=-\dfrac 1 z\dfrac{1}{1-\frac 1 z}=-\dfrac 1 z\sum \limits _{n=0}^\infty\left(\left(\dfrac 1 z\right)^n\right)= \sum \limits _{n=0}^\infty\left(-\left(\dfrac 1 z\right)^{n+1}\right), \text{ if ... 0 The angle \theta that corresponds to x is$$\theta=-2\tan^{-1}\left(\frac{1-x}{1+x}\right)^2,$$so$$\frac{1-x}{1+x}=\sqrt{-\tan (\theta/2)}.$$Solving for x we get$$ x=f^{-1}(\theta)=\frac{1-\sqrt{-\tan (\theta/2)}}{1+\sqrt{-\tan (\theta/2)}}, -\pi<\theta<0.$$1 Part A):$$f(z) = \frac{\sin (z^2)}{z^2}$$is an even function, hence all coefficients of odd powers of z in its Laurent expansion are 0, in particular \operatorname{Res}(f;0) = 0 here. (Furthermore, this is an entire function, it has a removable singularity in 0.) Part B) (After the correction of the function):$$f(z) = z^3\sin \frac{1}{z}$$is ... 1 Let's agree explicitly that "\log" refers to the branch of logarithm defined on \mathbf{C}\setminus(-\infty, 0] whose imaginary part is between -\pi i and \pi i, and that "f(z) \leq a" means "both a and f(z) are real, and f(z) \leq a". If f is holomorphic in some region U, then \log(f) is holomorphic at z in U provided f(z) does ... 1 I'd suggest using the following theorem:$$ \psi_1(x) = \frac{1}{2\pi i} \int_{c - i\infty}^{c + i\infty} \frac{x^{s + 1}}{s(s+1)} \left( \frac{-\zeta'(s)}{\zeta(s)} \right) \mathrm{d}s$$where c > 1. A proof of this equality can, for example, be found in Complex Analysis by Stein and Shakarchi. It is on page 191 being proposition 2.3 of chapter 7. ... 1 You applied the chain rule incorrectly. We have$$\frac{d}{dz}\log f(z) = \frac{1}{f(z)}\cdot f'(z)$$0 Notice$$ \log(\frac{a}{b}) = \log a - \log b$$1 If |z^2-n^2\pi^2| \ge n^2\pi^2, then$$\left|\frac{1}{z^2-n^2\pi^2}\right| \le \left|\frac{1}{n^2\pi^2}\right|$$Therefore, if z is such that |z^2-n_i^2\pi^2|<n_i^2\pi^2 for some \{n_i\} \subset \mathbb{N}, then$$\sum_{n \not\in \{n_i\}} \left|\frac{1}{z^2-n^2\pi^2}\right| \le \sum_{n \not\in \{n_i\}} ... 0 The meaning of capitalized names such as$\operatorname{Log}$varies by source. I assume that$ \operatorname{Log}$has been defined so that it's continuous at$2i$and at$2$; this is the case for the common definitions I'm familiar with. Check your definition. Then$\operatorname{Log}((2/n) + 2i) \to \operatorname{Log}(2i)$as$n \to \infty\$ ...

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