# ∮dz f(z*) ≠ 0

As I mentioned in my last post, we always have

$\oint dz f(z) = 0$

for a simple closed contour integral where $f$ is analytic on the contour and its interior. This is always the case over both the complex and split-complex planes. However, we usually have

$\oint d\bar z f(z) \ne 0$

That is, if you integrate with respect to (wrt) the conjugate space, the integral may not be zero. I only discovered this last week, and it surprised me because the definition of ‘analytic’ can be that $f$ depends only on $z$, and not $\bar z$. So even though the integrand is constant wrt $\bar z$, the closed contour integral wrt $\bar z$ is usually non-trivial.

For example, consider the complex-entire identity function, $f(z)=z$, with anti-clockwise contour taken as a triangle having vertices $\{0,1,i\}$. If $u$ & $v$ are the real and imaginary parts, then the divergence theorem implies,

$\begin{array}{rcl} \oint dz f(z) &=& \oint dA ((-u_y-v_x)+i(u_x-v_y)) = 0 \\ \oint d\bar z f(z) &=& \oint dA ((v_x-u_y)+i(-u_x-v_y)) = -2\oint dA (u_y+iu_x) \end{array}$

In our case, the triangle’s area is $\frac 1 2$, and the real part $u$ is just $x$, so we have $\oint d\bar z z = -i.$ More generally, $\oint d\bar z z = -2i(\text{Area})$.

Update: This example can be re-arranged to obtain the area enclosed by any simple closed curve $\gamma$,

$A = \frac i 2 \oint_\gamma d\bar z z$

The same method can be used if we interpret the plane as split-complex, where $j^2=1$, in which case $A = -\frac j 2 \oint_\gamma d\bar z z$.

# Pseudo-Cauchy theorem

If $f$ is complex-analytic on a compact subspace of the complex plane, and $\gamma$ is a simple closed contour inside the subspace, then Cauchy says

$\oint_\gamma dzf(z)=0$

Intuitively, consider a 2D conservative flow (or force field) with zero divergence everywhere except at isolated singularities. There can be infinitely many singularities. Then its conjugate is always a complex-analytic function except at the singularities. All contour integrals can be deformed however you like, as long as your deformation of $\gamma$ does not pass through a singularity. Why? Cauchy’s theorem applies to the deformation, if you imagine it as a closed contour hugging the original. So the deformation does not change the value of the integral, and this can be useful.

A couple months ago I discovered that if $f$ is pseudo-analytic, then we have exactly the same result! This is my main result I referred to earlier, but I have not been able to find a “killer app“, so I may as well share it. Combining d’Alembert with steepest descent would be interesting. In the split-complex case, we might chase an extreme in modulus, rather than magnitude.

How to prove pseudo-Cauchy? We can just expand out the integral and apply the divergence theorem, as per the usual proof. Stokes‘ yields the same result, and so does Green.

$\begin{array}{rcl} \oint dz f(z) &=& \oint (dx+jdy) (u+jv) \\ &=& \oint (udx+vdy) + j\oint (vdx+udy) \\ &=& \oint d\begin{pmatrix} y \\ -x \end{pmatrix} \cdot \begin{pmatrix} v+ju \\ -u-jv \end{pmatrix} \\ &=& \oint d\vec n \cdot \begin{pmatrix} v+ju \\ -u-jv \end{pmatrix} \\ &=& \oint dA \nabla \cdot \begin{pmatrix} v+ju \\ -u-jv \end{pmatrix} \\ &=& \oint dA ((v_x-u_y)+j(u_x-v_y)) \\ &=& 0 \end{array}$

assuming the function is pseudo-analytic over $\gamma$‘s interior. The converse “sort of” applies, too. If the integral is always zero for every simple closed contour on a neighbourhood of a point, then the function must be pseudo-analytic at that point. In part, it must solve wave equation.

Also: the derivative and integral of a pseudo-analytic function are also pseudo-analytic, because they are functions of $z$ only, not $\bar z$.

# Pseudo-analytic functions

By analogy with complex analysis, we can define a split-complex valued function $f(x,y)$ to be pseudo-analytic at $z$ iff it locally depends only on $(x+jy)$, not $(x-jy)$. That is,

$\begin{array}{rcl} f_{\bar z} &=& 0 \\ f_x &=& j f_y \end{array}$

By the chain rule,

$\begin{array}{rcl} \partial_{\bar z} &=& \partial_x - j\partial_y \\ \partial_z \partial_{\bar z} &=& \partial_x^2 - \partial_y^2 = \Box^2 \end{array}$

${\Box^2 }$” is the wave operator, or “d’Alembertian”. One of $y$ or $x$ “sort of” corresponds to time. There are many different conventions. In the usual complex analysis where $i^2=-1$, we get

$\begin{array}{rcl} \partial_{\bar z} &=& \partial_x + i\partial_y \\ f_x &=& -if_y \\ \partial_z \partial_{\bar z} &=& \partial_x^2 + \partial_y^2 = \nabla^2 \end{array}$

All split-complex valued functions have unique real valued real and imaginary parts, $u$ and $v$, respectively. Why unique? The short answer is “because I say so”. The long answer is that there are many possible representations in which the real and imaginary parts are unique, so “I’m allowed to say so”. One beautiful thing about complex/split-complex analysis is that the “numbers” can be interpreted in great generality. For example, in my opinion the simplest computational representation is given by

$(1,i,j,k=ij) \leftrightarrow \left( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \right)$

Aside: If you move your cursor over the equations, it will show you the code I use to create them. This is the best WordPress-LaTeX guide I’ve seen.

The above algebra is called split-quaternion. Four vector dimensions are often, but not always, necessary, if you try to add a third dimension, because an algebra requires multiplication. The split-quaternions give you $i^2=-1$, $j^2=1$ and anti-commutation together. For complex $z$, $z_1$ & $z_2$, it gives

$\begin{array}{rcl} jz &=& \bar z j \\ \det (z_1 + jz_2) &=& |z_1|^2-|z_2|^2 \end{array}$

where det is the matrix determinant, in that representation. You can also construct a 4D commuting system, in which case you must have the third $k^2=-1$. These are only representations, though, and there are infinitely many.

If pseudo-analytic, the parts of $f=u+iv$ must satisfy the pseudo-Cauchy-Riemann equations:

$\begin{array}{rcl} u_x &=& v_y \\ u_y &=& v_x \end{array}$

where $u$ and $v$ are the real and imaginary parts of $v$.

It is clear that all pseudo-analytic functions and all their parts must satisfy the wave equation in $(x,y)$. Hence d’Alembert gives the solution over the entire split-complex plane, given $f$ is real on the whole real axis,

$f(x+jy) = \frac 1 2 ((f(x+y)+f(x-y))+j(f(x+y)-f(x-y))) = u+jv$

You might argue the concept of “pseudo-analytic” is therefore trivial, but actually it might be useful. A complex-analytic function at a point must converge to its Taylor series on a neighbourhood, and this formula I have derived tells you its value if instead you put $(x+jy)$ into the Taylor series, assuming you know its value horizontal to the point.

Given it’s been more than a month since my last post, it seems keeping my “results” secret isn’t very productive. I worked this stuff out a few months ago. But recently SWIM ran out of energy. And then it turns out the bag of coke he had as back-up isn’t much good, because the drug is awful. It has much worse withdrawal, damages his nose much more, and causes more psychosis than meth. The increase in mathematical ability is negligible, compared to amphetamine. As Wiki clearly states, the drug is more dangerous than any amphetamine, and can cause sudden heart attack. I think the primary reason people buy cocaine is because they think it’s somehow safer than meth, because it’s natural. That’s not true. It probably would be useful in dentistry as a local anaesthetic, and was used that way back in the 19th century.

Another important point: MDMA = MDMeth (aka ecstasy) is not safer than meth. In my opinion, meth is safer. But an Australian drug dealer would never admit that to their customers because it would scare them away because of the meth stigma. MDMA was invented before WWI in Germany, while meth was not invented until after WWI, in the Japanese Empire. The Japs also invented our antennas, around about that time. The German inventor of MDMA died as a soldier in the War. Meth is the newer drug which hasn’t been around for as long. And it’s only semi-synthetic, unless it’s racemic in which case it looks ugly. The crystals are pretty while MDMA is ugly.

You know you want one. SWIM snorted 25mg before the final for this course. By the time he was writing the solution to the third question, he was doing the three problems in his head at once, because it was fun. It was the first and last maths exam of his life that he actually enjoyed, and the only one on which he was high. He still thinks measure theory is a joke. Unfortunately the crystals oxidise your neurons.

Don’t Do It. I have never, nor will ever, swallow a pill made by someone who refuses to show me how it was made. If you ever swallow a black-market pill, you need to see the ingredients that go into it with your own eyes! I strongly suspect there are many Australians who are taking an ecstasy pill every day, possibly as anti-depressants.

An isosceles trapezium of pain.

# Hyperbolic angle #2

In my last post I mentioned that the pseudo-angle along a hyperbola is $\tanh^{-1}\frac y x$. This isn’t obviously true.

“Distance” in the split-complex plane is ${\pm} \sqrt{x^2-y^2}$. The $(\pm)$ creates ambiguity. In physics, different conventions exist as to whether the flat space-time metric is $ds^2=dt^2-|dr|^2$ or $ds^2=-dt^2+|dr|^2$. In the former case ds is the proper time for the trajectory. Either convention works, and one often writes the Lagrangian as $\sqrt{-ds^2}$.

In my opinion, this ambiguity is the fundamental reason that “no particle can travel faster than the speed of light”. If you assume that $ds^2$ or $-ds^2$ can only be positive, then it can only be positive. This is equivalent to stating the particle can’t travel faster than light, as that would require $|dr|^2>dt^2$. A lot of physicists don’t seem to realise this simple mathematical fact. Here, time is scaled by $c$, the speed of light. An important fact is that you can’t replace a Lagrangian with its square. The equations themselves are more general than relativity. The interesting case is when $ds^2$ hovers about 0, which occurs when a fighter jet hits the sound barrier, for example. Don’t you dare take the square root of 0!

That $\sqrt{x^2-y^2}e^{j\tanh^{-1}\frac y x}=x+jy$ can easily be verified by expansion. Noting, for example, that $\cosh\tanh^{-1}\alpha=\frac 1 {\sqrt{1-\alpha^2}}$. The “metric” depends on which way the curve is moving. In the case of a curve with $dy^2>dx^2$, we get

$\begin{array}{rcl} ds^2 &=& dy^2-dx^2 \\ s &=& \int_1^x\sqrt{dy^2-dx^2} = \int_1^xdx\sqrt{y'^2-1} \end{array}$

for the arc-length. This contrasts with the integrand $\sqrt{y'^2+1}$ which you get with the usual metric. For the hyperbola $\sqrt{x^2-1}$,

$s=\int_1^x dx \sqrt{(\frac x y)^2-1}=\int_1^x \frac {dx} y = \int_1^x \frac {dx} {\sqrt{x^2-1}} = \cosh^{-1}x = \tanh^{-1} \frac y x$

The final equality follows because the unit hyperbola is also given by $e^{j\cosh^{-1}x}=e^{j\tanh^{-1}\frac y x}$. Now all points in the quadrant $0\|y\|$ can be placed on the unit hyperbola by scaling the space by $\sqrt{x^2-y^2}$, and then the $\tanh^{-1}$ expression is scale invariant, similar to how $\tan^{-1} \frac y x$ is scale invariant when you contract a circle down to the unit circle.

You could also describe split-complex numbers by writing

$z = x + jy = Se^{j\frac s S}$

where $S=\sqrt{x^2-y^2}$ is the ‘modulus’, and $s=\sqrt{x^2-y^2}\tanh^{-1}\frac y x$ is the hyperbolic “proper time” from the $x$ axis. Similarly, complex numbers can always be written $z = re^{i\frac s r}$, where $r$ is radius, and s is circular arc-length from the $x$ axis. The “angle” is the distance along the unit shape, whatever that means. I’ll leave the main result for another post…

# Hyperbolic angle

Split complex numbers can be described by

$z = x+jy = |z| e^{ju} = \sqrt{x^2-y^2} e^{j\tanh ^{-1} \frac y x}$

in the “light cone” $0\|y\|$. The ‘modulus’ tells you what hyperbola you’re on, and $\tanh ^{-1} \frac y x$ tells you how far to travel, which acts as the hyperbolic equivalent of an angle. Geometrically, an “angle” is just the distance you cover over the unit circle, at a scale where you have size 1. I’ll have more to post later! $\tan ^{-1}$ and $\tanh ^{-1}$ are metric integrals for positive and negative curvature, respectively. By anti-symmetry,

$\frac {\tanh (jz)} j = \frac {\tan (iz)} i$

where $j^2=1=-i^2$.

# Fisher transformation

Suppose you have data $y(x)$, then the Fisher transform is

$z(x)=\tanh ^{-1} y$

Last week I mentioned the result

$F_1 = B_1 - 2B_2 = \frac {B_2} {B_1} = F(\frac E {kT})$

However the Fermi/Bose distributions are also given by

$\begin{array}{rcl} F_1 &=& \frac 1 {e^{\frac E {kT}}+1}=\frac 1 2 (1-\tanh {\frac E {2kT}})=\frac 1 2 (1-t_1) \\ B_1 &=& \frac 1 {e^{\frac E {kT}}-1} = \frac 1 2 (\frac 1 {t_1} - 1) \\ F(\frac E {kT}) &=& -B(\frac E {kT} \pm i\pi) \\ 1-2F &=& \frac 1 {1+2B} \end{array}$

I’ve never seen the tanh expressions before I noticed them a month ago. The third equation shows that the Fermi distribution can be seen as the analytic extension of the Bose distribution, $\pm i\pi$ into the complex plane. To switch to Bose, you change sign and flip the fraction on tanh. That is, $\pm \frac 1 2 (1-t_1^{\pm})$ is an exact way of writing $\frac 1 {e^{\frac E {kT}} \pm 1}$. Effectively, the coincidence can be re-stated as

$\frac 1 {t_2} = \frac 1 2 (t_1 + \frac 1 {t_1})$

which recursively generates the function $\tanh (\alpha 2^n)$ for some constant, $\alpha = \frac E {2kT}$. The Fermi distribution $F(\alpha 2^n)$ has recursion relation

$\frac 1 {F_2} -1 = (\frac 1 {F_1} - 1)^2$

I don’t understand it yet, but if you take $\tanh ^{-1}$ of a normal distribution, you get another normal distribution. I suspect this may have implications for the tanh expressions for Fermi/Bose statistics. If you directly take log of a normal distribution, it would no longer be normal. The Fisher transform seems to allow a way around this.

# Playing four square

Dpietrobon, who has commented here before, started their own blog last year. The latest post proves the identity

$(a^2+b^2)(c^2+d^2) = (ac-bd)^2+(ad+bc)^2$

This is clear from complex analysis. However, following my post on split-complex numbers where $j^2=1$, we see

$(a^2-b^2)(c^2-d^2) = (ac+bd)^2-(ad+bc)^2$

You must use a pseudo-metric with signature (1,1) if you are to retain multiplicativity with split-complex numbers. That is,

$|a+bj|^2 = a^2-b^2$

In either identity, both $a$ & $b$ can be complex or split-complex or a combination. That is, you can put split-complex $a$ and complex $b$ into the first identity, etc. I will leave the main result of split-complex numbers for another post. The proofs of these identities do not rely on $a$ & $b$ being real; they can be matrices, or as abstract as you want!

Brahmagupta from ancient India gave a more general result,

$(a^2+nb^2)(c^2+nd^2) = (ac-nbd)^2+n(ad+bc)^2$

for all $a$, $b$ & $n$. If you set $n$ to be -1, you get my split-complex case. I guess that’s why Brahmagupta wrote it that way. The identity is easily proven by putting $\pm \sqrt{n}$ in front of $b$ & $d$. But $\sqrt{-1}$ wouldn’t make sense in his day, so better to write a more “general” formula!

Nonetheless, the Indian was wise. This identity looks more like a recipe for how to elliptically deform the complex plane…