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7

After you have established that the RHS and the LHS differ by a multiplicative constant, all you are left to do it plug in $z=1$. If you pair up the factors in the RHS as $$\Gamma \left( \frac{1+k}{n} \right) \leftrightarrow \Gamma \left( \frac{n-k-1}{n} \right) ,$$ and apply the reflection formula $\Gamma(z) \Gamma(1-z)=\frac{\pi}{\sin \pi z}$, things will ...


5

$\newcommand{\+}{^{\dagger}} \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{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


5

An entire function is a function that is holomorphic on the whole complex plane. Such a function can be expressed globally as its Taylor series, developed by ANY point of the complex plane. So you can take $a$. Hence you have that $f(z)=\sum_{n=0}^{+\infty}\frac{f^{(n)}(a)}{n!}(z-a)^n\;\;\forall z\in\mathbb C$. But $f^{(n)}(a)=0\;\;\forall n\geq0$ by ...


5

The definition of path integral implies that if $\gamma:[0,1]\rightarrow \mathbb{C}$ is a $C^1$ path, and $f$ is differentiable, then $\int_{\gamma} f'(z) dz = \int_0^1 f'(\gamma(t))\gamma'(t)dt= f(\gamma(1))-f(\gamma(0))$. Then it follows directly that if you have a piecewise $C^1$ closed curve, the integral of a derivative over it must be zero.


4

This is an application of Liouville's theorem. $|f(z)|\leq|(\sqrt{k}z)^2|\implies|\frac{f(z)}{(\sqrt{k}z)^2}|\leq1$ Notice that $f(z)$ and $(\sqrt{k}z)^2$ are both entire functions thus $\frac{f(z)}{(\sqrt{k}z)^2}$ is entire and it's also bounded therefore it's constant and the result follows.


4

Setting $\displaystyle e^{\dfrac{i\pi t}2}=u\implies u^2=e^{i\pi t};u^3=e^{\dfrac{3i\pi t}2};u^4=e^{2i\pi t}$ $$\frac{e^{\dfrac{it\pi}2}-e^{\dfrac{3it\pi}2}}{1 - e^{2\pi it}} =\frac{u-u^3}{1-u^4}=\frac{u(1-u^2)}{(1-u^2)(1+u^2)}=\frac u{1+u^2}$$ if $\displaystyle1-u^2\ne0\iff u^2\ne1\iff e^{i\pi t}\ne1=e^{2m\pi i}\iff t\ne2m$ where $m$ is any integer ...


4

You can see that if $z_0$ is a zero then so is $-z_0$, $\overline{z_0}$, $-\overline{z_0}$. It is straightforward to show that the restriction of $f$ to the real axis has a minimum $f({3 \over 2}) = {3 \over 4}$, so $f$ has no real axis zeros. If $f$ had an imaginary axis zero, then $f$ would have the form $f(z) = z^4 + c^2$ for some real $c$, hence $f$ ...


3

Actually this is an easy one, since $\sin(x) = \dfrac{-i}{2} (e^{ix} - e^{-ix})$, $$e^x \sin(x) = \dfrac{-i}{2} (e^{(1+i) x} - e^{(1-i) x}) = \dfrac{-i}{2} \sum_{n=0}^\infty \dfrac{(1+i)^n - (1-i)^n)}{n!} x^n$$ But in general, $$\left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty b_n x^n\right) = \sum_{n=0}^\infty \sum_{j=0}^n a_j b_{n-j} x^n $$ ...


3

I'm not sure if that is what you are after, but: If $\omega$ is a primitive $2k$-th root of unity, then the solutions of $z^k+1 = 0$ are $\omega^{2n+1},\, 0 \leqslant n < k$, so $$z^k + 1 = \prod_{n=0}^{k-1} (z-\omega^{2n+1}),$$ and hence $$\begin{align} \prod_{n=1}^{k-1} (z-\omega^{2n+1}) &= \frac{z^k+1}{z-\omega}\\ &= \frac{z^k - ...


3

Suppose there is an epsilon ball around $w$ which is disjoint from the image of $f$. Consider $g(z)=\frac{1}{f(z)-w}$ which is analytic except for at $z_n$ and $0$. Since $g$ is bounded above by $\frac{1}{\epsilon}$ these singularities are removable. Hence $g$ extends to an analytic function which is zero at $z_n$ (and at $0$). Hence $g$ is zero on a set ...


3

Using Cauchy Integral Formula, for $r\to 0$, we obtain that $f(0)=f'(0)=f''(0)=0$. Thus $f(z)=z^3g(z)$, where $g$ is also entire, and we get for $g$ that $$ \int_0^{2\pi} \lvert g(r\mathrm{e}^{i\vartheta})\rvert\,d\vartheta\le r^{1/5}, \quad r>0. $$ Now, for every $z\in\mathbb C$, let $r>2\lvert z\rvert$, and according to Cauchy Integral Formula $$ ...


3

Use the Cauchy-Riemann equations: $$ f(x + iy) = u(x, y) + iv(x, y) $$ Where $u$ and $v$ are real valued functions. We have the constraints: $$ \frac{\partial u}{\partial x} = \frac{\partial v }{\partial y} \\ \frac{\partial u}{\partial y} = -\frac{\partial v }{\partial x} $$ Giving: $$ f'(x + iy) = \frac{\partial u}{\partial x} + i\frac{\partial ...


3

You can just use the fact that $x \rightarrow \frac{1}{x}$ is continuous on $C^*$, $P$ is on $D$ and since $P(D) \subset C^*$ so is $z \in D \rightarrow \frac{1}{P(z)}$ by composition where $D=\{z$ $|$ $P(z) \neq 0\}$ is an open subset of $C$. If you really want to prove it by hand, you can say that $|\frac{1}{P(z_1)} - \frac{1}{P(z_2)}| = |\frac{P(z_2) - ...


3

I don't know the name of this formula but it is the two dimension counterpart for a well known integral identity in electrostatics. $$ -\frac{1}{4\pi}\int_{\mathbb{R}^3} \frac{1}{|\vec{x} - \vec{y}|} \vec{\nabla}^2 \phi(\vec{x})\; d\vec{x} = \phi(\vec{y})$$ The imporant point is $\;\displaystyle \log\frac{1}{|z - \xi|}\;$ is a solution to the two dimension ...


2

I was asked this exact question by my wife last night. She was looking for an everyday example of the use of complex numbers to explain to her 8th grade math class (whose knowledge of complex numbers consists of i = SQRT(-1) ). My response was this: Imagine an electronic piano. Each key produces a different tone. A volume control changes the amplitude ...


2

You're confusing three different things with each other: $$ \frac{df}{dz}, \qquad \frac{d}{dz}, \qquad\frac{d}{dt} $$ If you had written $$ \frac{d}{dt} \rho e^{it} = \rho ie^{it} $$ then it would be correct, but what you have written is at best a misunderstanding of notation. Applying the product rule, one gets $$ \begin{align} \frac{d}{dt} \rho e^{it} ...


2

Every torus are Kahler: the standard form $\omega = \frac{\sqrt{-1}}{2}\sum_{i=1}^n dz^i \wedge d\bar z ^i$ descends to a Kahler form on the torus. Let $\alpha$ be a 1 form. Write $$\alpha = \alpha^{1,0} + \alpha^{0,1}$$ so $$d \alpha = d\alpha^{1,0} + d\alpha^{0,1}$$ If $d\alpha$ has no $(0,2)$ component, then $d\alpha^{0,1}$ is a $(1,1)$ form and the ...


2

If you want $ij = 1$, then multiplying on the left by $-i$ you have $j = -i$, so that's not much use. On the other hand, $ij = -1$ leads to $j = i$, which is equally unexciting. In fact, no other choices will work either. Consider that whatever system you construct will form a vector space not only over the real numbers, but over the complex numbers as ...


2

If $f$ is holomorphic on $B_{\mathbb C}(a,r[$, and you know that $|f(z)|\leq M\;\;\forall z\in B_{\mathbb C}(a,r]$ then by Cauchy Integral Formula for dervivatives (the one written By Daniel) you have that $$\begin{align} |f^{(n)}(a)| &= \left|\frac{n!}{2\pi i}\int_{|\zeta-a|=r}\frac{f(\zeta)}{(\zeta -a)^{n+1}}d\zeta\right|\\ &\leq ...


2

The multiplication formula can be written in the form $$n^{nz-1/2} \prod_{k=0}^{n-1} \Gamma \left(z+\frac{k}{n} \right) = (\sqrt{2 \pi})^{n-1} \Gamma(nz)$$ Using the limit definition of the gamma function, $$ \Gamma \left(z +\frac{k}{n} \right) = \lim_{m \to \infty} \frac{m! \ m^{z+\frac{k}{n}-1}}{(z+\frac{k}{n})(z+\frac{k}{n}+1) \cdots (z+\frac{k}{n} + m ...


2

Hint: If we let $z=z_0+Re^{i\theta}$, then $$ \frac1{2\pi}\int_0^{2\pi}f(z_0+Re^{i\theta})\,\mathrm{d}\theta =\frac1{2\pi i}\oint f(z)\frac{\mathrm{d}z}{z-z_0} $$ where the path is the circle of radius $R$ counterclockwise around $z_0$. Next, look for the singularities of the integrand then use Cauchy's Integral Formula.


2

Consider a point $z_0 \in V$. Now consider any other point $w \in V$. Since $V$ is connected, it is possible to find a curve starting at $z_0$ and ending at $w$. Let's call this curve $C_w$. We then have $$f(w) - f(z_0) = \int_{C_w} f'(z) dz = \int_{C_w} 0 \cdot dz = 0$$ Hence, we have $f(w) = f(z_0)$. This is true for any $w \in V$. Hence, $f(w) = f(z_0)$ ...


2

With residue calculus, put $$ f(z) = \frac{P(z)}{Q(z)}\log z$$ where $log$ denotes the natural branch, i.e. with a branch cut along the positive real axis. Integrate over a keyhole contour: Assuming that $\deg Q \ge 2+ \deg P$ and that $Q$ has no zero on the positive real axis, it's not hard to show that the integral over the big and large circle vanish ...


2

This follows from Stokes' Theorem (these conditions are met by compactness and rectifiability which one needs to verify) once one shows that $dw = 0$, since closed implies exact in this case (star-convex). This is clear in the first case. Now, in the second case $$d(z\; dx + iz\; dy) = -i dx \wedge dy + i dx \wedge dy = 0,$$ which is just multivariable ...


2

$z$ is just the complex function on the complex plane and $\bar z$ is its conjugate. Writing in usual Euclidean coordinate, $z =x+ \sqrt{-1} y$ and $\bar z = x - \sqrt{-1} y$, so $$dz =dx+ \sqrt{-1} dy \text{ and } d\bar z = dx - \sqrt{-1} dy\ .$$ If you put this into the metric, $$\lambda dzd\bar z = \lambda (dx dx + dydy)\ ,$$ which means that ...


2

There is always some disagreement on such notation, but you should probably think of $dz$ as acting on tangent vectors (assuming we're in one dimension) by $dz(v) = v\in\Bbb C$ and $d\bar z$ as acting by $d\bar z(v) = \bar v$. (In general, in working with complex manifolds, one works with the complexified tangent bundle, which is spanned by the ...


2

The closed form is given in this paper by the same author. I'll transcribe it in this answer, I'm not sure I how to post the pdf directly in this answer. Define $$ r_{np} = \int_0^{\pi/2}(\log\cos x)^n (\log\sin x)^p\,dx $$ (I will use that paper's notation; you have $n$ and $p$ the other way around). Then $$ r_{np} = \frac{\pi n! p!}{2^{n+p+1}} ...


2

For $z \neq z_0$ and $z_0 \neq 0$: $$\begin{align} \sum_{k=0}^{n-1} z^k z_0^{n-1-k} & = z_0^{n-1} \sum_{k=0}^{n-1} \left( \frac{z}{z_0} \right)^k \\ & = z_0^{n-1} \frac{\left(\frac{z}{z_0}\right)^n - 1}{\frac{z}{z_0} - 1} \\ & = \frac{\frac{z^n}{z_0} - z_0^{n-1}}{\frac{z}{z_0} - 1} \\ & = \frac{z^n - z_0^n}{z - z_0} \end{align}$$ Therefore ...


2

Why do you think that $\Omega \setminus K$ is not connected? (It is.) On the other hand $K$ is not compact (or relatively compact) in $\Omega$, so it's not clear to me what you are trying to do. On the other hand, your $f$ doesn't extend to $\Omega$. I think the shortest way to see this is that $z_1 f = z_2$ on $\Omega \setminus K$, so if $f$ extends ...


2

You can do this as follows. Start as you did with $$\left\vert\int_{\gamma_2} e^{iz^2}dz\right\vert\leq R\,\int_0^{\frac\pi4} e^{-R^2\sin 2\theta} d\theta\, . $$ Then observe that on $[0\frac\pi4]$ you have $$\sin 2\theta \geq \frac4\pi \, \theta\, ,$$ thanks to the concavity of the sine function on $[0\frac\pi2]$. It follows that ...



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