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

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.

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

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?

Tagged : / /

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

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.

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,

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

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

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.

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,

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

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

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.

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,

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

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

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 $$0 leq t leq T$$.

Tagged : /

## Math Genius: Consequence of maximum modulus principle [duplicate]

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?

Tagged :

## Math Genius: Determining a Möbius transformation image [closed]

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.

## Math Genius: Asymptotic behavior of logarithm and some univalent functions

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!

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

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.