Tag Info

New answers tagged

0

Complex analysis is used in 2 major areas in engineering - signal processing and control theory. In signal processing, complex analysis and fourier analysis go hand in hand in the analysis of signals, and this by itself has tonnes of applications, e.g., in communication systems (your broadband, wifi, satellite communication, image/video/audio compression, ...


0

I assume there is a typo in you question and the real question is how to show the maximum of $\Re\left[ \frac{zf'(z)}{f(z)}\right]$ over $\partial\bar{U}$ is greater than or equal to the number of zeros over $U$. Otherwise, your question is simply false. In any event, you don't really need anything complicated. You just need to rewrite the formula for ...


0

Starting with, $$\dfrac{H_n}{n} = \sum_{k=1}^{\infty} \dfrac{1}{k(k+n)}$$ Then,$$\sum_{n=1}^{\infty} \dfrac{H_n^2}{n^2} = \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty}\dfrac{1}{k(k+n)}\right)^2$$ $$= \sum_{n=1}^{\infty}\sum_{k,j} \dfrac{1}{jk(j+n)(k+n)} $$ $$= \sum_{n=1}^{\infty} \left(\sum_{k=1}^{\infty} \dfrac{1}{k^2(n+k)^2}+ 2\sum_{k<j} ...


3

Hint 1 Unless $f \equiv 0$, prove that $f$ cannot have any zero. If $z_0$ is a zero of order $m$ for $f$, what happens if you divide both sides by $(z-z_0)^{m-1}$? Hint 2 $\frac{f'}{f}$ is entire and bounded.


4

Hint: Use the general fact that if $f,g$ are entire functions with $|f| \leq |g|$, then $f=cg$ with $|c| \leq 1$ (which follows from Liouville's Theorem).


0

There cannot be more than one maximum (or minimum) value. There can be more than one place in which such values occur, but not more than one maximum (or minimum) value. The maximum occurs on the boundary (why? use a well known theorem). Try to maximize $z^2-1$ first on $D$ to see what is going on. This is asking, for $|z|=1$ how far can $z^2$ be from ...


2

Hint: write $$\text{Res}_{z=z_k}f(z)=\dfrac{1}{nz_k^{n-1}}=\dfrac{z_k}{nz_k^{n}}=\dfrac{z_k}{n}$$ and use the sum $$ \sum_{k=0}^n\mu^k =\frac{1-\mu^{n+1}}{1-\mu}, \quad \mu\neq 1 $$ with $\mu^k=z_k=...$. Then $$\sum_{k=0}^n\text{Res}_{z=z_k}f(z)= \dfrac{1}{n}\sum_{k=0}^n\mu^k=...$$


2

I already pointed to a related question in the comment above, but as I thought about this question a while ago, I'll give a slightly different proof here: For brevity, I'll call $\sum_{i\in I} \alpha_iz_i$ an additive combination of the sequence of numbers $(z_i \mid i \in I)$ if $\alpha_i\in\{1,-1\}$ for all $i\in I$. Claim1: If $z_1\in\mathbb{C}$ has ...


0

$$\sum_{r=0}^{n-1} (e^{2i\theta})^r=\frac{1-(e^{2i\theta})^n}{1-e^{2i\theta}}$$ $$=\frac{-e^{in\theta}(e^{in\theta}-e^{-in\theta})}{-e^{i\theta}(e^{i\theta}-e^{-i\theta})}$$ $$=e^{i(n-1)\theta}\frac{2i\sin n\theta}{2i\sin\theta}$$ $$=[\cos(n-1)\theta+i\sin(n-1)\theta]\frac{\sin n\theta}{\sin\theta}$$ As $\displaystyle\sin\phi=$ imaginary part of ...


1

No, the conjugate of a harmonic function that is continuous up to the boundary need not be continuous up to the boundary, or even bounded. This is related to the fact that the Hilbert transform does not preserve continuity (though it does preserve Hölder continuity of exponents $\alpha\in (0,1)$). Here is an example. Let $F$ be a conformal map of the ...


4

$$\begin{align} \left|\frac{1}{e^{i\omega t} - 1}\right| &= \left|\frac{1}{e^{i\omega t/2}(e^{i\omega t/2} - e^{-i\omega t/2})}\right| \\ &= \frac{1}{|e^{i\omega t/2}|} \left|\frac{1}{2i \sin(\omega t/2)}\right|\\ &= \frac{1}{2|\sin(\omega t/2)|} \end{align}$$


0

Let me present a way to get the close form. Noting $$ H_n=\int_0^1\frac{1-x^n}{1-x}dx, \frac{1}{n^3}=\frac{1}{2}\int_0^1x^{n-1}\ln^2xdx,$$ we have \begin{eqnarray} \sum_{n=1}^\infty\frac{H_n^2}{n^3}&=&\sum_{n=1}^\infty\iiint\frac{(1-x^n)(1-y^n)z^{n-1}}{(1-x)(1-y)}\ln^2zdxdydz\\ ...


3

No, it is not true. We just need that the Taylor coefficients of $f(z)$ in a neighbourhood of zero decay very fast: $$ f(z)=\sum_{n\geq 0}e^{-n^2}z^n $$ is an example of an order-$0$ entire function.


0

You are indeed almost there. Next write $$\sin\theta = {e^{i\theta}-e^{-i\theta}\over 2i}$$ and conclude $${1-e^{10i\theta}\over 1-e^{2i\theta}}={e^{10i\theta}-1\over e^{i\theta}{2i\over 2i}(e^{i\theta}-e^{-i\theta})}$$ $$={e^{-i\theta}(e^{10i\theta}-1)\over 2i\sin\theta} = {e^{9i\theta}-e^{-i\theta}\over 2i\sin\theta}$$ From there you just do ...


1

The key idea is geometric: the chord $\operatorname{Re}z=1/2$ subtends the angle of $2\pi/3$, which is one third of the full angle. So, if we take it together with rotations by $\pm 2\pi /3$, it will completely surround the origin, and the maximum principle can be applied to the resulting triangle. To formalize the above, let $g(z) = f(z) f(e^{2\pi ...


1

Since $\| e^z\|=e^{\Re(z)}$, your integral equals: $$\int_{0}^{\pi/2}e^{-r \sin t}dt\leq \int_{0}^{\pi/2}e^{-2rt/\pi}dt\leq\int_{0}^{+\infty}e^{-2rt/\pi}dt=\frac{\pi}{2r},$$ as wanted. Notice that a more careful estimation is given by: $$\int_{0}^{\pi/2}e^{-r \sin t}dt\leq\int_{0}^{\pi/2}\frac{dt}{1+r\sin t}=\int_{0}^{1}\frac{2\, ...


0

Based on the context, I think you are being asked to verify that the Cauchy-Riemann equations hold at every point. You should expect this because $f$ is a composition of two analytic functions.


1

Hint: The quantities $$|z|, \quad |w|,\quad \bar z\,w,\quad |z-w|$$ all remain unchanged when $z$ and $w$ are replaced by $e^{i\alpha} z$ and $e^{i\alpha} w$.


1

The open unit disk $\{z \in \mathbb{C} : |z|<1\}$ has rotational symmetry. Rotation is an isometry. The same is true for the unit circle $\{z \in \mathbb{C} : |z|=1\}$. Let $\mathrm{R}_{\theta} : \mathbb{C} \to \mathbb{C}$ be an anti-clockwise rotation about the origin through an angle of $\theta$ radians. $$\begin{eqnarray*} |z| < 1 &\iff& ...


1

The notation $w=\sqrt z$ is used to denote one of the two solutions to the equation: $$ w^2 = z. $$ If $w=\sqrt z$ is one such solution, the other one is $-w=-\sqrt z$. Now if $w=\sqrt{a+ib}$ you have $w^2 = a+ib$ and $(\bar w)^2 = \overline{w^2} = \overline{a+ib} = a - ib$. So it is true that the conjugate of $w$ is one of the two square roots of $a-ib$: ...


0

You first have to define what is meant by "the" square root of $\sqrt{a+ib}$, because there are two candidates. We can see this by writing $a+ib$ in polar form $re^{i\theta}$. Then $r^{1/2} e^{i(\theta / 2)}$ and $-r^{1/2}e^{i(\theta / 2)} = r^{1/2}e^{i(\theta/2 + \pi)}$ both square to the correct value. One can resolve this ambiguity by choosing, for ...


1

I'm sure this example will solve your problem: $$f(z) = |z|^{2}\text{.}$$ Considering this one we can clearly see that this is only differentiable at $0$ while the criteria for analyticity is that you have some $k>0$ such that this still stays differentiable. Hence differentiability does not imply analyticity.


2

$\lim_{n\to\infty} f_n(x)/n$ exists for $x$ in the open unit disc, and is equal to the partition generating function $P(x)$. Convergence is uniform on closed unit balls. To prove this, we use the following lemma: Let $a_{n,m}$ be a double sequence of numbers such that $a_m= \lim_{n\to \infty}a_{n,m}$ exists and $|a_{n,m}| \leq |a_m|$. Let $R$ be the ...


1

It is a little subtle. Even if a function satsifies Cauchy-Riemann's equations, it's not necessarily $\mathbb{C}$-differentiable. Take for example $$ f(z) = \begin{cases} \exp(-1/z^4), & z \neq 0 \\ 0 & z = 0 \end{cases}. $$ As an exercise, you can check that $f$ satsifies C-R everywhere (it's pretty much obvious outside the origin), but $f$ is ...


3

An other way: Firstly $$\int_0^{\pi/2}\ln(\cos t)dt\underset{t=\frac{\pi}{2}-u}{=}\int_{0}^{\pi/2}\ln\left(\cos\left(\frac{\pi}{2}-u\right)\right)du=\int_0^{\pi/2}\ln(\sin u)du \tag 1 $$ Then, $$\int_0^{\pi/2}\ln(\sin t)dt=\frac{1}{2}\left(\int_{0}^{\pi/2}\ln(\sin t)dt+\int_0^{\pi/2}\ln(\cos ...


3

Hint: with $\cos x=u$ $$\int_0^{\pi/2}\log\cos x\mathrm{d}x=-\int_0^1\frac{\log u}{\sqrt{1-u^2}}\mathrm{d}u$$


1

I assume that you intend the definition to be $$ M(r) = \max_{|z| = r} |f(z)| $$ as otherwise $f(z)\in\mathbb{C}$ and I am not sure how you are ordering $\mathbb{C}$. In this case observe that since $f$ is a polynomial of degree $n$, there exists $n+1$ complex constants such that $$ \frac{f(z)}{z^n} = \sum_{j = 0}^{n} c_j z^{-j} $$ Let $w = z^{-1}$ we have ...


1

If the system is linear in $b_k\cos(a_k t), a_k,b_k\in\mathbb{R}$, then you can use the complex replacement. $$\sum_{k} b_k \cos(a_k t) \implies \sum_{k} b_k Re(\exp(i a_k t)) \implies Re(\sum_{k} b_k \exp(i a_k t) )$$ If it is not linear in $b_k\cos(a_k t)$, then you usually can not use the complex replacement. For example $$(2\cos t)^3=(e^{i t}+e^{-i ...


1

$(Re(e^{it}))^3$ is NOT equal to $Re((e^{it})^3)$


2

The usual way to do this is to combine Mittag-Leffler's theorem with the Weierstrass factorization theorem. Weierstrass gives you an entire function $g(z)$ which has a simple zero at each $z_k$. Mittag-Leffler gives you a meromorphic function $h(z)$ with a simple pole at each $z_k$ where $w_k \ne 0$, and residue $w_k/g'(z_k)$ there. After removing the ...


1

As long as the circle and the triangle enclose exactly the same singularities of the integrand, the two integrals will be equal. Changing the direction of the path changes the sign of the integral.


1

Multiplying by a power of $z$ can never get rid of an essential singularity. Nor can composition with an integer power (which is what you're doing in going from $\exp(1/z)$ to $\exp(1/z^3)$). So you certainly do have an essential singularity. As others said, the Laurent series is the way to go to find the residue: fortunately it is easy to find using ...


1

As I commented a few days ago, I think that many answers to this question were already given by the person who asked it. However, it occurred to me that there is one very simple fact which may be relevant to this discussion, and wasn't mentioned yet. As we all know, if $F,f:\mathbb{R}\to\mathbb{R}$ are two real functions such that $F'=f$, then ...


0

$V$ is the sector consisting of points with $3\pi/4\le\arg(z)\le 5\pi/4$ and $|z|\le 1/2$, as such it is symmetric with respect to the negative $x$-axis, so it suffices to show the claim for the upper half plane only. The upper ray of the sector can be parametrized as: $$z_1(r)=r\exp\left(\frac{3\pi}{4}i\right)\text{, with } 0\le r\le \frac{1}{2}$$ The ...


1

There isn't any general way to solve all problems of this sort. However, if the given function is easy to write as a power series, then this is probably the best approach. In this case, as I think you noticed, it is pretty easy to write $f$ as a power series, and what you suggested is indeed true - the coefficient of $z^{-1}$ is the residue. The type of the ...


1

Recall that, for any $h$ holomorphic on an open set $U$ which contains $D=\{z:|z| \leq 1\}$, $\int_{C} h(z)dz=0$. Now note that $|z|<1$ implies $|\frac{1}{\bar{z}}|=\frac{1}{|z|}>1$. Hence $$\frac{f(s)}{s-\frac{1}{z}}$$ is holomorphic on an open set containing $D$, and you can apply the first observation to conclude that the second integral is $0$.


2

By Cauchy-Goursat, $$\int_k\frac{f(z)}{(z-z_0)^{n+1}}dz=\frac{2\pi i}{n!}f^{(n)}(z_0)$$ and so $$\int_k \frac{\cos(z)}{z^4}dz=\frac{2\pi i}{3!}\sin(0)=0$$


1

Let $$ I(a)=\int_{0}^{\infty}\ln(1-e^{-ax})\cos(bx)dx. $$ Then \begin{eqnarray} I'(a)=&=&\int_{0}^{\infty}\frac{xe^{-ax}}{1-e^{-ax}}\cos(bx)dx\\ &=&\int_{0}^{\infty}\sum_{n=0}^\infty xe^{-a(n+1)x}\cos(bx)dx\\ &=&\sum_{n=0}^\infty \frac{(a(n+1)-b)(a(n+1)+b)}{(a^2(n+1)^2+b^2)^2}\\ &=&\sum_{n=1}^\infty ...


2

$a_n\to a$ follows from $$|a_n-a|=||z_n|-|z||\leq|z_n-z|<\epsilon$$ and $\theta_n\to\theta$ can be proved from $$\lim_{n\to\infty}\frac{z_n}{z}=1$$


1

This is not true. Take for example $f(z) = 1+z^2$. Then $M(0) = 1$ and $M(r) \approx r^2$ for $r$ large. Hence $M(r)/r^2$ blows up at $r=0$, and tends to $1$ as $r\to\infty$. In fact, for this example, the function $M(r)/r^2$ is actually decreasing:


1

Since $g$ is smooth, we see that ${\partial g(s,z) \over \partial z}$ is bounded on $[0,1] \times U$, where $U$ is a bounded neighbourhood of $\Omega$. The mean value theorem gives $|g(s,z+h)-g(s,z+h)| \le \left( \sup_{(s,z) \in [0,1] \times U}|{\partial g(s,z) \over \partial z}| \right) | h| $. Hence ${1 \over h} |g(s,z+h)-g(s,z+h)| \le \left( ...


1

The motivation for the question seems to be: Why does the author integrate $f(z)=(1-e^{iz})/z^2$ over the complex plane, rather than the original integrand $g(z)=(1-\cos z)/z^2$ ? You're right that $g$ is an entire function, and it could be integrated over a semicircle without having to dodge the origin. However, the integral of $g$ over the arc ...


1

I don't see a direct application of the standard estimates, so let's use a deviation. The circle $\lvert z-\pi\rvert = \pi\sqrt{2}$ intersects the imaginary axis in $\pm \pi i$. Let $\gamma$ be a (piecewise smooth) path in the left half-plane connecting $-\pi i$ and $\pi i$; we shall see later what kind of path might be convenient. Then $C-\gamma$ is a ...


1

You can think of the derivative as expressing the local/instantaneous stretching and rotation. For example, suppose a real function $f$ has $f'(a)=2$, for some $a$. We can think of this as saying that near $a$ the function is (approximately) doubling distance, up to adding a constant. This is what the tangent line $2(x-a)+f(a)$ expresses. Now for a complex ...


1

We have the convergence for all $z\in\Bbb C\setminus\{2\}$ such that $$\operatorname{Re}\left(\frac{z}{z-2}\right)\le0$$ Can you see why? The domain of convergence by Wolfram


3

Hint: What is the radius of convergence of $\sum \frac{1}{n^2} x^n$? Plug in $x = \exp(\frac{z}{z-2})$.


3

This is a straightforward logarithmic integration, in your region $|f(z)-1|<1$, which means that the values of $f$ lie in a ball of center $1$ and radius $1$. There is a well defined branch of the log defined on this region, (in fact on any simply connected region not containing the origin). Therefore $$\int \frac{f^{\prime}(z)}{f(z)}dz=\ln f(z)$$ a ...


1

Use principle of argument: $$\int\limits_{\gamma}\frac{f'(z)}{f(z)}dz=2\pi i (N_{zeros}-N_{poles}) .$$ As was said, $$|f(z)-1|<1 \Leftrightarrow -1<f(z)-1<1\Leftrightarrow 0 <f(z)<2 $$ Since $f(z)>0$, it has no zeros in any $\operatorname{Int}\gamma$. As $f(z)<\infty$, it has no poles. Therefore, $$N_{zeros}=N_{poles}=0$$ ...


2

Often, it is easier/more efficient to use the Taylor expansions of the involved functions, when one looks at compositions. Here, we have $$z\cos z - z = z\left(1 - \tfrac{z^2}{2} + \tfrac{z^4}{24} - \dotsc \right) - z = - \tfrac{z^3}{2} + O(z^5),$$ and inserting that into the Taylor expansion of $e^w$, we find $$e^{z\cos z - z} = 1 - \frac{z^3}{2} + ...



Top 50 recent answers are included