Math Genius: Information/references/examples on fields $mathbb R^3 to mathbb C^3$ with divergence and curl free real and imaginary parts

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In the course of some physical considerations I came across a complex vector field
$$ mathbf u = mathbf v + i mathbf w, $$ with
begin{align}
mathbf v:& mathbb R^3to mathbb R^3\
mathbf w:& mathbb R^3to mathbb R^3
end{align}

and the special propert, that it has a divergence-free imaginary and
curl free real components, that means

begin{align}
vec{nabla}times mathbf v & = mathbf 0 tag{1}\
vec{nabla}cdotmathbf w & = 0 tag{2}
end{align}

In my attempts to better understand and interpret the quantitiy described by this field, I started to wonder if:

Question 1: This property has a special name/term in complex vector analysis.

Question 2: If there are any important prominent examples of such fields.

Question 3: If there are other properties that follow as a consequence of (1) and (2).

Question 4: If anyone can hint me at literature where such fields are investigated.

I would be very grateful for any hints at this. I would hope for answers like those to this question.

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Math Genius: Contour integration appears to be wrong. Why?

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I want to solve the following integral using contour integration:

$$I = int_0^{infty} frac{e^{i x}}{x^2 +1} dx. tag{1}$$

I built a contour consisting of three paths: $A$ (going from 0 to $infty$), $B$ (going from $-infty$ to 0 ) and $C$ the upper semicircle.

I define

$$f(z) = frac{e^{i |z|}}{z^2+1}, tag{2}$$

where $|cdot|$ stands for the modulus of a complex number.

I have that

$$oint f(z) dz = left( int_A + int_B + int_C right) f(z) dz. tag{3}$$

Now,

$$ int_{A} f(z) dz = int_0^{infty} f(x) dx = I, tag{4}$$
since $|x| = x$ if $x geq 0$.

Likewise,

$$ int_{B} f(z) dz = int_{-infty}^0 f(x) dx = – int_{infty}^0 f(-x) dx = int_0^{infty} frac{e^{i |- x|}}{(-x)^2+1} = int_0^{infty} frac{e^{i x}}{x^2+1} = I. tag{5}$$

The integral over the contour $C$ is 0 when the semicircle is “at” $infty$.

I evaluate the l.h.s of Eq. $(3)$ using Cauchy’s theorem, and I find

$$oint f(z) dz = 2 pi i frac{e^{i |i|}}{2 i} = pi e^i. tag{6}$$

Putting all together I find that

$$I = frac{pi}{2} e^i approx 0.848705 + 1.32178 i, tag{7}$$

whereas Mathematica says that

$$I approx 0.577864 + 0.646761 i. tag{8}$$

What is wrong with my reasoning?

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Math Genius: If $omega$ is a primitive cube root of unity, simplify ${omega}^{{2009}^{{2009}^{{2009}^{cdots 2009}}}}$

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Let $omega$ be a primitive cube root of unity. Let $x = {omega}^{{2009}^{{2009}^{{2009}^{cdots 2009}}}}$ (up to $2009$ times). Simplify the value of $x$.

My attempt: Let $m = {{2009}^{{2009}^{{2009}^{cdots 2009}}}}$ (up to $2007$ times). Then since $2009$ is odd so $2009^m$ is also odd. Let $k = 2009^m$. Now since $k$ is an odd integer so $2^k equiv 2 (text {mod} 3)$. Also $2009 equiv 2 (text {mod} 3)$. Therefore, $2009^k equiv 2 (text {mod} 3)$. Let $n = 2009^k$. Then $n = 3k’ + 2$ for some $k’ in Bbb N$. Therefore $$x = {omega}^n = {omega}^2$$

Am I right? Please verify it.

Thanks in advance for reading.

The answer is w^2 and ur process is right and check my process
2009^something odd can be written as (2010-1)something odd this 2010*m – 1 (m is some no.) as 2010 is divisible by 3 this is 1/w = w^2

$2009$ can be expressed as $6m-1$ where $m$ is any integer

Now observe that $(6m-1)^{6m-1}equiv-1pmod6,$ so again of the form $6m’-1$

If $w$ is a root of unity,

$$w^{6n-1}=(w^3)^{2n-1}w^2=?$$

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Math Genius: If $omega$ is a primitive cube root of unity, simplify ${omega}^{{2009}^{{2009}^{{2009}^{cdots 2009}}}}$

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Let $omega$ be a primitive cube root of unity. Let $x = {omega}^{{2009}^{{2009}^{{2009}^{cdots 2009}}}}$ (up to $2009$ times). Simplify the value of $x$.

My attempt: Let $m = {{2009}^{{2009}^{{2009}^{cdots 2009}}}}$ (up to $2007$ times). Then since $2009$ is odd so $2009^m$ is also odd. Let $k = 2009^m$. Now since $k$ is an odd integer so $2^k equiv 2 (text {mod} 3)$. Also $2009 equiv 2 (text {mod} 3)$. Therefore, $2009^k equiv 2 (text {mod} 3)$. Let $n = 2009^k$. Then $n = 3k’ + 2$ for some $k’ in Bbb N$. Therefore $$x = {omega}^n = {omega}^2$$

Am I right? Please verify it.

Thanks in advance for reading.

The answer is w^2 and ur process is right and check my process
2009^something odd can be written as (2010-1)something odd this 2010*m – 1 (m is some no.) as 2010 is divisible by 3 this is 1/w = w^2

$2009$ can be expressed as $6m-1$ where $m$ is any integer

Now observe that $(6m-1)^{6m-1}equiv-1pmod6,$ so again of the form $6m’-1$

If $w$ is a root of unity,

$$w^{6n-1}=(w^3)^{2n-1}w^2=?$$

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Math Genius: If $omega$ is a primitive cube root of unity, simplify ${omega}^{{2009}^{{2009}^{{2009}^{cdots 2009}}}}$

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Let $omega$ be a primitive cube root of unity. Let $x = {omega}^{{2009}^{{2009}^{{2009}^{cdots 2009}}}}$ (up to $2009$ times). Simplify the value of $x$.

My attempt: Let $m = {{2009}^{{2009}^{{2009}^{cdots 2009}}}}$ (up to $2007$ times). Then since $2009$ is odd so $2009^m$ is also odd. Let $k = 2009^m$. Now since $k$ is an odd integer so $2^k equiv 2 (text {mod} 3)$. Also $2009 equiv 2 (text {mod} 3)$. Therefore, $2009^k equiv 2 (text {mod} 3)$. Let $n = 2009^k$. Then $n = 3k’ + 2$ for some $k’ in Bbb N$. Therefore $$x = {omega}^n = {omega}^2$$

Am I right? Please verify it.

Thanks in advance for reading.

The answer is w^2 and ur process is right and check my process
2009^something odd can be written as (2010-1)something odd this 2010*m – 1 (m is some no.) as 2010 is divisible by 3 this is 1/w = w^2

$2009$ can be expressed as $6m-1$ where $m$ is any integer

Now observe that $(6m-1)^{6m-1}equiv-1pmod6,$ so again of the form $6m’-1$

If $w$ is a root of unity,

$$w^{6n-1}=(w^3)^{2n-1}w^2=?$$

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Math Genius: Change in argument of sum of complex numbers along a closed path

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Let $a(t)$ and $b(t)$ be complex variables that travel along closed paths that end up at their original points, i.e. $a(0)=a(T)=a_0$ and $b(0)=b(T)=b_0$, with $Delta Arg(a) = Delta Arg(b) = 0$ over the paths. Is it possible for their sum $c(t) = a(t) + b(t)$ to have $Delta Arg(a) = 2pi k$ where $k$ is a non-zero integer? Any example? Also, let’s assume $a(t) neq 0, b(t) neq 0$, and $c(t) neq 0$ for any $0 leq t leq T$.

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Math Genius: Consequence of maximum modulus principle [duplicate]

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Let $f$ be continuous on the closed unit disc $overline D={zin mathbb C~:~|z|leq 1}$ and analytic in the open unit disc $D={zin mathbb C~:~|z|<1}$. Suppose that $|f(z)|leq a$ on the set ${zin mathbb C~:~|z|=1, Im(z)geq0}$ and $|f(z)|leq b$ on the set ${zin mathbb C~:~|z|=1, Im(z)leq 0}$. Then prove that $|f(0)|leq sqrt{ab}$.

How maximum modulus principle plays here to get an estimate for $|f(0)|$, Is there any role for $f(-z)$?

Define $h(z) = f(z)f(-z)$. One can quickly note that $h(z)$ is holomorphic on the open unit disk and continuous on $overline{D}$ as it is a product of holomorphic/continuous functions. By the maximum modulus principle applied to $h$, we have that there exists $z_0 in partial D$ such that $|h(z)| leq |h(z_0)|$ for all $z in overline{D}$. Without loss of generality, one can assume that $z_0 in {zin mathbb C~:~|z|=1, space text{Im}(z)geq0}$ Then $-z_0 in {zin mathbb C~:~|z|=1, space text{Im}(z)leq 0}$. Can you finish?

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Math Genius: Determining a Möbius transformation image [closed]

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so I was attempting a question on Möbius transformations and I’ve encountered a problem in my workings. The question is “Determine the image of the strip $-1 < Re(z) < 1$ under $g(z) = (iz+1)/(z+1).$ How should I go about solving this?

A huge help when dealing with Möbius transformations:

The image of every circle-or-line is a circle-or-line.

So if you know the images of three points on one boundary line of the strip (say $1$ and $1pm i$), then you know the circle-or-line that is the image of that boundary line; similarly, if you know you know the images of three points on the other boundary line (say $-1$ and $-1pm i$), then you know the circle-or-line that is the image of this other boundary line; and if you know the image of a point inside the strip (say $0$), then you know which side of the image circles-or-lines is the interior of the image region.

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Math Genius: Asymptotic behavior of logarithm and some univalent functions

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For $z$ near infinity (sufficiently large $z$), I want to verify the following claims :

  1. If $f$ is univalent (an analytic injective function) such that $f(z)=z+O(1/z)$, then $f^{-1}(z)=z+O(1/z)$.

  2. $log|z+O(1)|=log|z|+o(1)$ (How does this identity involve both Big-Oh and little-oh?)

Well, I try to using power series expansion for $log(1+x)$ but it does not seem work.

For 1, let $|z|ge R>0$ with $R$ large tbd and $w=f(z)$

$|w-z| le C/|z|$ means $|w| le |z|+ C/|z| le 2|z|$ as well as $|w| ge |z|- C/|z| ge |z|/2$ if we choose $R^2 ge 2C$ hence we have $|w-z| le 2C/|w|, |z| >R$ as above and $O(1/w)=O(1/z)$

Since $z=f^{-1}(w)$ we get $|w-f^{-1}(w)|=O(1/w)$ or $f^{-1}(w)=w+O(1/w)$ Changing variables back to $z$ we are done!

For 2 we have $log|z+O(1)|=log |z|+log |(1+O(1)/z)|$

But $O(1)/z=o(1)$ for $|z|$ large and $log |1+x| le 2|x|, |x| le 1/2$, so $log |(1+O(1)/z)|=log |(1+o(1)|=o(1)$ and putting all together we get

$log|z+O(1)|=log |z|+o(1)$ so we are done!

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Math Genius: Is the sheaf of non-vanishing holomorphic functions soft?

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It is obvious that the sheaf of holomorphic functions is not soft, not every holomorphic function on a closed set (appropriately defined) can be extended to be an entire function. What about the sheaf of non-vanishing holomorphic functions $mathcal{O}_X^{ast}$? I assume that it is not soft either, but I’d like an example.

The case of non-vanishing holomorphic functions is really only easier. For instance, consider the function $f(z)=z$, defined on a non-discrete closed set that does not contain $0$. Since a holomorphic function on a connected set is determined by its values on any non-discrete set, the only possible extension to all of $mathbb{C}$ is $f(z)=z$, but this is not a non-vanishing entire function.

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