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1

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 $y = c_{n-1}x^{n-1} + c_{n-2}x^{n-2} + \cdots + c_1x+ c_0$.


2

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


2

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 ...


2

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: ...


0

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 $$ \left|\frac{f(i)-f(1)}{i-1}\right| = \left|\frac{-i-1}{i-1}\right| = 1 $$


2

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


3

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


1

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 ...


0

Mr. Fischer has given a very effective hint in his comment. The following approach will give the same result. $z = x + iy$. So the circle $|z - 1| = 1$ can be written as $(x-1)^2 + y^2 = 1$ or by $x = 1 + \cos(\theta), y = \sin(\theta)$. $z^2 = x^2 - y^2 + 2ixy$. Write $u = x^2 - y^2$ and $v = 2ixy$. Represent $u$ and $v$ by function of $\theta$. $u = 1 + ...


1

As far as I can see, you are trying to find and upper bound for $$\frac{f(x)}{g(x)}=\left(\frac{1+|x|^2}{1+|x|}\right)^{\frac s2}$$ and a lower positive bound for its inverse: $$\frac{g(x)}{f(x)}=\left(\frac{1+|x|}{1+|x|^2}\right)^{\frac s2}$$ If $s>0$, the first tends to infinity and the second to $0$ when $|x|\rightarrow \infty$, so there is no such ...


0

For every holomorphic $f\colon U \to \mathbb{C}$, the function $z\mapsto \overline{f(\overline{z})}$ is holomorphic on the domain you obtain from $U$ by conjugation. Thus $$z\mapsto T(z)\cdot \overline{T(\overline{z})}$$ is an entire meromorphic function (a rational function, since $T$ is a Möbius transformation). Apply the identity theorem to reach the ...


2

According the Maximum Modulus Principle we have $\vert f(0)\vert\leq 1$. But since $f$ does not vanish, the function $z\mapsto1/f(z)$ satisfies the same conditions as $f$ and consequently we have also $\left\vert \frac{1}{f(0)}\right\vert\leq 1$, or equivalently $\vert f(0)\vert\geq 1$. Combining these two inequalities we get $\vert f(0)\vert = 1$, and ...


1

Note: I'm a taking $\theta \in [0, 2\pi)$, i.e. $0 \le \theta < 2\pi$. This doesn't restrict $z(t)$ in any way, and by eliminating a possible ambiguity in $\theta$, makes things a little more clear. To my mind, in any event. Write $z_0 - w_0 = re^{i\phi}$ for some $\phi \in [0, 2\pi)$. Note that $\vert z_0 - w_0 \vert = r$, the ordinary Euclidean ...


2

The first paragraph is not meant to show $\int_\sigma f \,dz=0$; its purpose is to show that there exists an analytic function $F$ in $\Delta$ such that $F'=f$. Once this is obtained, the statement about $\int_\gamma f\,dz$ follows from the fundamental theorem of calculus for line integrals: the integral of $f$ is equal to the difference of the values of $F$ ...


1

Because the function $z\mapsto\dfrac{1}{z-a}$ is not analytic at $z=a$. In fact, if $a$ is in the exterior of the disk then Cauchy's theorem is applicable and the integral is $0$. It is only if $a$ is in the interior of the disk, i.e. the curve over which one integrates winds around $a$ one in a counterclockwise direction, that the integral is $2\pi i$


1

Being $a$ near of far from the origin is irrerevant. The important fact is $\gamma$ enclose $a$? (the singular point)


0

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1

Try applying the formula another time: $$ \frac{1}{|z-a|^{2}}=\frac{1}{(z-a)(\overline{z}-\overline{a})}= \frac{1}{(z-a)(\rho^{2}/z-\overline{a})}=\frac{z}{(z-a)(\rho^{2}-\overline{a}z)}. $$ Then consider the cases where $\rho < |a|$ and $\rho > |a|$. For example, the integral can now be written as $$ ...


0

From the triangle inequality we have $|z|\leq|z-a|+|a|$, then \begin{align} |z|-|a|\leq|z-a|\leq|b||z-\overline{a}| & =|bz-\overline{a}b|& \text{by properties of $|\cdot|$} \\ & \leq |bz|+|-\overline{a}b| & \text{by triangle inequality} \\ & = |b||z|+|\overline{a}||b| & \text{by properties of $|\cdot|$}\\ & =|b||z|+|a||b| ...


0

From $\frac {|(z-a)|}{|(z-\bar a)|} \le |b|$, we have $ {|(z-a)|}{} \le |b||(z-\bar a)|$ Apply triangle inequality $|p - q| \ge |p| - |q|$ to the LHS, we have $ |z|- |a| \le |b||(z-\bar a)|$ On the RHS, replace the $-\bar a$ by $ + {(-\bar a)}$, we have $ |z|- |a| \le |b||z+ (-\bar a)|$ Let Apply triangle inequality $|p + q| \le |p| + |q|$ to the RHS, ...


0

(Note that if $a$ is real, then the left-hand side of the first inequality equals $1$, so the inequality can never be satisfied.) As long as $a$ is not real, the function $f(z)=(z-a)/(z-\bar a)$ is a linear fractional transformation; the first inequality says that the image of $z$ lies inside the circle of radius $|b|$ centered at $0$. The inverse of $f(z)$ ...


1

"Integrand" only means "something that you have to integrate". It can be the function $f(z)$, or the differential form $f(z)\,dz$, depending on which point of view you prefer. An "exact differential" is a particularly nice differential form, so I guess that the point of view is the latter. The object $p\,dx + q\,dy$ is exactly a differential form. Click on ...


2

Prove the series converges (absolutely) for $|z|<1$, then the Cauchy product formula says $$\left(\sum_{n=0}^\infty\frac1{4^n}\binom{2n}nz^n\right)^2=\sum_{n=0}^\infty\sum_{k=0}^n\frac1{4^n}\binom{2(n-k)}{n-k}\binom{2k}kz^n$$ Now prove $$\sum_{k=0}^n\binom{2(n-k)}{n-k}\binom{2k}k=4^n$$ So the square of the sum is $$\sum_{n=0}^\infty z^n=\frac1{1-z}$$ ...


1

From a purely algebraic point of view, if $z=re^{i \theta}$ and $(1-z)(1- \overline{z})=|1-z|^2<1$, then you get $r<2 \cos(\theta)$. Therefore, $\cos(\theta)>0$ that is $\theta \in (- \pi/2,\pi/2)$.


3

The set of complex numbers which satisfy $|1- z| < 1$ is the open disk of radius 1 centered at 1. This is contained in the open right half of the complex plane. For complex numbers in the open right half plane, you have $$\text{Arg}(z) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right).$$


1

Write $z(t) = (x(t), y(t)) = w_0 + t\cdot e^{i\theta} = (x_0 + t\cdot cos\theta, y_0 + t\cdot sin\theta)$ with $w_0 = (x_0, y_0)$ and $z_0 = (a_0, b_0)$. Consider $f(t) = (x_0 - a_0 + t\cdot cos\theta)^2 + (y_0 - b_0 + t\cdot sin\theta)^2 = t^2 + (x_0 - a_0)^2 + 2(x_0 - a_0)\cdot t\cdot cos\theta + (y_0 - b_0)^2 + 2(y_0 - b_0)\cdot t\cdot sin\theta$. Taking ...


2

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3

You don't really need to do any partial fraction decomposition. The integrand $\frac{1}{z^2+1}$ has two poles at $\pm i$, both of them are inside the contour $|z| = 2$. Since the integrand is analytic on the rest of complex plane, one can evaluate it by deforming the contour to a large circle of radius $R$. We have following estimate: $$\left|\oint_{|z| = ...


1

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

If $f$ 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 $$\frac{d}{dz}\log f(z) = \frac{1}{f(z)}\cdot f'(z)$$


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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