Worked Examples of Exponential Polar Decomposition of Split-Complex Numbers

Introduction

This article computes the exponential polar decomposition of eight split-complex numbers, one after another. It is the computational companion of Exponential Polar Decomposition of Split-Complex Numbers. The decomposition, its existence and uniqueness, the component sign and the exponential and logarithm are established in that article and in Split-Complex Elementary Functions; no definition or proof is repeated.

Each case is read from the two coordinates $q$ and $q'$ of $Q=q+jq'$. The modulus, the angle and the component sign are computed from them by the formulas of §The Method, and the word is checked against the reconstruction

$$ Q=r\,\epsilon\,e^{\phi j}. $$

The eight cases cover the four open components, the pure discrete unit, the two scalar directions and the boundary.

Group Case What it shows
Future cone $5+3j$ the generic case, $\epsilon=1$
Right wedge $3+5j$ the same modulus and angle as $5+3j$, $\epsilon=j$
Past cone $-5-3j$ the same modulus and angle, $\epsilon=-1$
Left wedge $1-2j$ a negative angle, $\epsilon=-j$
Discrete unit $j$ a pure component sign, $\phi=0$
Positive scalar $2$ a pure scale, $\epsilon=1$
Negative scalar $-2$ the same exponent as $2$, $\epsilon=-1$
Null cone $1+j$ $(q+q')(q-q')=0$, but no polar word

The plan. The method is stated once (§The Method). The eight cases follow (§The Eight Cases), each with its coordinates, its polar data, its word and its check. The summary collects the results in two tables and reads the patterns across the cases (§Summary). The eigenbasis, the idempotents and the spinors of the same eight cases are collected in Worked Examples of Exponential Polar Decomposition with Eigenbasis of Split-Complex Numbers.

The Method

The Coordinates

For $Q=q+jq'$ the two real coordinates are $q$ and $q'$. The modulus and the angle are

$$ r=\sqrt{\lvert q^2-q'^2\rvert} =\sqrt{\lvert(q+q')(q-q')\rvert}, \qquad \phi=\tfrac12\ln\left\lvert\frac{q+q'}{q-q'}\right\rvert. $$

The component sign is read from the quadrant of the pair $(q,q')$:

Condition $\epsilon$ Component
$q>\lvert q'\rvert$ $1$ future cone $F$
$q<-\lvert q'\rvert$ $-1$ past cone
$q'>\lvert q\rvert$ $j$ right wedge $R$
$q'<-\lvert q\rvert$ $-j$ left wedge $L$

The four conditions are exclusive and cover every unit. In every row the product $\epsilon Q$ lies in $F$.

The two combinations are recovered from the polar data by

$$ \boxed{\; r\,e^{\phi}=\lvert q+q'\rvert, \qquad r\,e^{-\phi}=\lvert q-q'\rvert \;} $$

These two identities are the check used in every case.

The Reconstruction

The polar word is

$$ \boxed{\; Q=\epsilon\,r\,e^{\phi j}=r\,\epsilon\bigl(\cosh\phi+j\sinh\phi\bigr), \qquad Q_{\log}=\ln r+j\phi \;} $$

The exponent of $\epsilon Q$ and the unit factor $\epsilon e^{\phi j}$ are read separately, and the reconstruction is the scale $r$ times the unit factor. The check is the multiplication.

The Matrix Picture

The matrix of multiplication by $Q$ in the basis $(1,j)$ is

$$ M_Q=\begin{pmatrix}q&q'\\q'&q\end{pmatrix}. $$

Its trace is $2q$ and its determinant is $q^2-q'^2=(q+q')(q-q')$. It is invertible exactly when $Q$ is a unit. The product $\epsilon Q$ acts on $F$ by a positive scale and a hyperbolic rotation.

The Null Cone

For $q^2-q'^2=0$ the element is a nonzero zero divisor, of the form $Q=q(1\pm j)$ with $q\neq0$, and no polar word exists: the modulus vanishes, the angle diverges and the unit factor is undefined. The matrix has rank one, its determinant is zero, and $Q^2=2q\,Q$. Case 8 treats the boundary explicitly.

The Eight Cases

Case 1: A Generic Element of the Future Cone

Coordinates. For $Q=5+3j$,

$$ q=5, \qquad q'=3, \qquad q^2-q'^2=25-9=16, \qquad q+q'=8, \qquad q-q'=2. $$

Since $q>\lvert q'\rvert$, the component sign is $\epsilon=1$.

Polar data. The modulus and the angle are

$$ r=\sqrt{16}=4, \qquad \phi=\tfrac12\ln\frac{8}{2}=\ln2, \qquad \epsilon=1. $$

The two combinations check: $r e^{\phi}=4\cdot2=8=q+q'$ and $r e^{-\phi}=4\cdot\tfrac12=2=q-q'$.

The word. The exponent and the unit factor are

$$ Q_{\log}=\ln4+j\ln2, \qquad \epsilon e^{\phi j}=\cosh(\ln2)+j\sinh(\ln2)=\tfrac54+\tfrac34j. $$

Therefore

$$ Q=4\left(\tfrac54+\tfrac34j\right)=5+3j. $$

Check. The exponential of the exponent is

$$ \exp(Q_{\log})=4\bigl(\cosh(\ln2)+j\sinh(\ln2)\bigr)=4\left(\tfrac54+\tfrac34j\right)=5+3j, $$

and the error is $0$. The matrix

$$ M_Q=\begin{pmatrix}5&3\\3&5\end{pmatrix} $$

has trace $10$ and determinant $16$.

Case 2: A Generic Element of the Right Wedge

Coordinates. For $Q=3+5j$,

$$ q=3, \qquad q'=5, \qquad q^2-q'^2=9-25=-16, \qquad q+q'=8, \qquad q-q'=-2. $$

Since $q'>\lvert q\rvert$, the component sign is $\epsilon=j$.

Polar data. The modulus and the angle are

$$ r=\sqrt{\lvert-16\rvert}=4, \qquad \phi=\tfrac12\ln\left\lvert\frac{8}{-2}\right\rvert=\ln2, \qquad \epsilon=j. $$

The two combinations check: $r e^{\phi}=8=\lvert q+q'\rvert$ and $r e^{-\phi}=2=\lvert q-q'\rvert$.

The word. The exponent is the same string as in case 1,

$$ Q_{\log}=\ln4+j\ln2, $$

but the unit factor is different:

$$ \epsilon e^{\phi j}=j\left(\tfrac54+\tfrac34j\right)=\tfrac34+\tfrac54j. $$

Therefore

$$ Q=4\left(\tfrac34+\tfrac54j\right)=3+5j. $$

Check. The product $\epsilon Q=jQ=5+3j$ lies in $F$, and

$$ \exp(Q_{\log})=5+3j=\epsilon Q. $$

The exponent is a logarithm of $\epsilon Q$, not of $Q$. The error in $Q$ is $0$. The matrix

$$ M_Q=\begin{pmatrix}3&5\\5&3\end{pmatrix} $$

has trace $6$ and determinant $-16$.

Case 3: A Generic Element of the Past Cone

Coordinates. For $Q=-5-3j$,

$$ q=-5, \qquad q'=-3, \qquad q^2-q'^2=25-9=16, \qquad q+q'=-8, \qquad q-q'=-2. $$

Since $q<-\lvert q'\rvert$, the component sign is $\epsilon=-1$.

Polar data. The modulus and the angle are again

$$ r=4, \qquad \phi=\ln2, \qquad \epsilon=-1. $$

The two combinations check: $r e^{\phi}=8=\lvert q+q'\rvert$ and $r e^{-\phi}=2=\lvert q-q'\rvert$.

The word. The exponent is again $\ln4+j\ln2$, and the unit factor is

$$ \epsilon e^{\phi j}=-\left(\tfrac54+\tfrac34j\right). $$

Therefore

$$ Q=4\left(-\tfrac54-\tfrac34j\right)=-5-3j. $$

Check. The product $\epsilon Q=-Q=5+3j$ lies in $F$, and $\exp(Q_{\log})=5+3j=\epsilon Q$ with error $0$. The matrix

$$ M_Q=\begin{pmatrix}-5&-3\\-3&-5\end{pmatrix} $$

has trace $-10$ and determinant $16$.

Case 4: A Generic Element of the Left Wedge

Coordinates. For $Q=1-2j$,

$$ q=1, \qquad q'=-2, \qquad q^2-q'^2=1-4=-3, \qquad q+q'=-1, \qquad q-q'=3. $$

Since $q'<-\lvert q\rvert$, the component sign is $\epsilon=-j$.

Polar data. The modulus is $\sqrt3$ and the angle is negative:

$$ r=\sqrt{\lvert-3\rvert}=\sqrt3, \qquad \phi=\tfrac12\ln\left\lvert\frac{-1}{3}\right\rvert=-\tfrac12\ln3, \qquad \epsilon=-j. $$

The two combinations check: $r e^{\phi}=\sqrt3\cdot\tfrac{1}{\sqrt3}=1=\lvert q+q'\rvert$ and $r e^{-\phi}=\sqrt3\cdot\sqrt3=3=\lvert q-q'\rvert$.

The word. The exponent is

$$ Q_{\log}=\ln\sqrt3-\tfrac12\ln3\,j=\tfrac12\ln3-\tfrac12\ln3\,j, $$

and the unit factor is

$$ \epsilon e^{\phi j}=-j\left(\frac{2}{\sqrt3}-\frac{1}{\sqrt3}j\right) =\frac{1}{\sqrt3}-\frac{2}{\sqrt3}j. $$

Therefore

$$ Q=\sqrt3\left(\frac{1}{\sqrt3}-\frac{2}{\sqrt3}j\right)=1-2j. $$

Check. The product $\epsilon Q=-j(1-2j)=2-j$ lies in $F$, and

$$ \exp(Q_{\log})=\sqrt3\left(\frac{2}{\sqrt3}-\frac{1}{\sqrt3}j\right)=2-j=\epsilon Q. $$

The error is $2.2\times10^{-16}$. The matrix

$$ M_Q=\begin{pmatrix}1&-2\\-2&1\end{pmatrix} $$

has trace $2$ and determinant $-3$.

Case 5: The Discrete Unit

Coordinates. For $Q=j$,

$$ q=0, \qquad q'=1, \qquad q^2-q'^2=-1, \qquad q+q'=1, \qquad q-q'=-1. $$

Since $q'>\lvert q\rvert$, the component sign is $\epsilon=j$.

Polar data. The modulus is one and the angle vanishes:

$$ r=1, \qquad \phi=0, \qquad \epsilon=j. $$

The two combinations check: $r e^{\phi}=1=\lvert q+q'\rvert$ and $r e^{-\phi}=1=\lvert q-q'\rvert$.

The word. The exponent is $Q_{\log}=0$, and the unit factor is $\epsilon e^{\phi j}=j$. Therefore

$$ Q=1\cdot j=j. $$

Check. $\exp(0)=1$, so $\exp(Q_{\log})=1=\epsilon Q$, and the error is $0$. The matrix

$$ M_Q=\begin{pmatrix}0&1\\1&0\end{pmatrix} $$

has trace $0$ and determinant $-1$. The element is the pure component sign: no scale and no angle.

Case 6: The Positive Scalar

Coordinates. For $Q=2$,

$$ q=2, \qquad q'=0, \qquad q^2-q'^2=4, \qquad q+q'=2, \qquad q-q'=2. $$

Since $q>\lvert q'\rvert$, the component sign is $\epsilon=1$.

Polar data. The modulus is two and the angle vanishes:

$$ r=2, \qquad \phi=0, \qquad \epsilon=1. $$

The two combinations check: $r e^{\phi}=2=q+q'$ and $r e^{-\phi}=2=q-q'$.

The word. The exponent is

$$ Q_{\log}=\ln2, $$

and the unit factor is $1$. Therefore

$$ Q=2\cdot1=2. $$

Check. $\exp(\ln2)=2$, with error $0$. The matrix $M_Q=2I$ has trace $4$ and determinant $4$.

Case 7: The Negative Scalar

Coordinates. For $Q=-2$,

$$ q=-2, \qquad q'=0, \qquad q^2-q'^2=4, \qquad q+q'=-2, \qquad q-q'=-2. $$

Since $q<-\lvert q'\rvert$, the component sign is $\epsilon=-1$.

Polar data. The modulus is two and the angle vanishes:

$$ r=2, \qquad \phi=0, \qquad \epsilon=-1. $$

The two combinations check: $r e^{\phi}=2=\lvert q+q'\rvert$ and $r e^{-\phi}=2=\lvert q-q'\rvert$.

The word. The exponent is the same as in case 6,

$$ Q_{\log}=\ln2, $$

and the unit factor is $-1$. Therefore

$$ Q=2\cdot(-1)=-2. $$

Check. $\exp(\ln2)=2=\epsilon Q$, with error $0$. The matrix $M_Q=-2I$ has trace $-4$ and determinant $4$. Cases 6 and 7 show that the exponent does not determine the element: the component sign does.

Case 8: The Null Cone

Coordinates. For $Q=1+j$,

$$ q=1, \qquad q'=1, \qquad q^2-q'^2=1-1=0, \qquad q+q'=2, \qquad q-q'=0. $$

The product $Q\bar Q$ vanishes, so there is no modulus, no angle and no component sign: the element is a zero divisor.

No polar word. The exponential has image $F$ and is injective, so $Q\notin F$ has no logarithm. The matrix

$$ M_Q=\begin{pmatrix}1&1\\1&1\end{pmatrix} $$

has rank one and determinant $0$. Its square is

$$ Q^2=2Q. $$

The boundary. The element is the limit of the future-cone family

$$ Q_\delta=(1+\delta)+j, \qquad \delta>0. $$

Its coordinates give

$$ r=\sqrt{\delta(2+\delta)}, \qquad \phi=\tfrac12\ln\frac{2+\delta}{\delta}, \qquad \epsilon=1, $$

and the two combinations are $r e^{\phi}=2+\delta$ and $r e^{-\phi}=\delta$. The table gives the approach.

$\delta$ $r$ $\phi$ $r e^{\phi}$ $r e^{-\phi}$
$0.1$ $0.4582575695$ $1.522261219$ $2.1$ $0.1$
$0.01$ $0.1417744688$ $2.651652454$ $2.01$ $0.01$
$0.001$ $0.0447325385$ $3.800701167$ $2.001$ $0.001$

As $\delta\to0$, the modulus tends to $0$, the angle diverges, and the larger combination tends to $2$; the word tends to the boundary value $1+j$. The polar decomposition exists on the open components and stops exactly at $q^2-q'^2=0$.

Summary

Main Results

Case $Q$ $q$ $q'$ $q^2-q'^2$ $\epsilon$ $r$ $\phi$ $Q_{\log}$
$1$ $5+3j$ $5$ $3$ $16$ $1$ $4$ $\ln2$ $\ln4+j\ln2$
$2$ $3+5j$ $3$ $5$ $-16$ $j$ $4$ $\ln2$ $\ln4+j\ln2$
$3$ $-5-3j$ $-5$ $-3$ $16$ $-1$ $4$ $\ln2$ $\ln4+j\ln2$
$4$ $1-2j$ $1$ $-2$ $-3$ $-j$ $\sqrt3$ $-\tfrac12\ln3$ $\tfrac12\ln3-\tfrac12\ln3\,j$
$5$ $j$ $0$ $1$ $-1$ $j$ $1$ $0$ $0$
$6$ $2$ $2$ $0$ $4$ $1$ $2$ $0$ $\ln2$
$7$ $-2$ $-2$ $0$ $4$ $-1$ $2$ $0$ $\ln2$
$8$ $1+j$ $1$ $1$ $0$ — — — none

The Reconstruction Table

Case $Q$ Unit factor $\epsilon e^{\phi j}$ $r$ times the unit factor Error
$1$ $5+3j$ $\tfrac54+\tfrac34j$ $5+3j$ $0$
$2$ $3+5j$ $\tfrac34+\tfrac54j$ $3+5j$ $0$
$3$ $-5-3j$ $-\tfrac54-\tfrac34j$ $-5-3j$ $0$
$4$ $1-2j$ $\tfrac{1}{\sqrt3}-\tfrac{2}{\sqrt3}j$ $1-2j$ $2.2\times10^{-16}$
$5$ $j$ $j$ $j$ $0$
$6$ $2$ $1$ $2$ $0$
$7$ $-2$ $-1$ $-2$ $0$
$8$ $1+j$ — — no word

What the Cases Show

  • Cases 1 to 3 share the modulus $r=4$ and the angle $\phi=\ln2$. The two combinations $(q+q',q-q')$ are $(8,2)$, $(8,-2)$ and $(-8,-2)$, and the exponent is the same string in all three; only $\epsilon$ changes. Case 4 completes the four quadrants, with a different modulus and a negative angle.
  • Cases 2 and 3 have $\exp(Q_{\log})=\epsilon Q$, not $Q$. The exponent is the logarithm of the element pushed into the future cone.
  • Case 4 has $q>0$ but $q+q'<0$, so the component is the left wedge, not the future cone. The sign of $q$ does not determine the component.
  • Cases 6 and 7 have the same exponent $\ln2$ and differ only by $\epsilon$. The discrete factor is not a period: the kernel of the exponential is $\{0\}$.
  • Case 5 is the pure discrete factor, with $r=1$ and $\phi=0$.
  • Case 8 is the boundary. The matrix has rank one, while the polar word does not exist. The family shows the rate $r\to0$, $\phi\to\infty$ with $r e^{\phi}\to2$.

Summary of Notation

Symbol Meaning
$\mathbb{D}$ the split-complex algebra, $j^2=+1$
$Q=q+jq'$ a split-complex number
$q,q'$ the two real coordinates
$q^2-q'^2=(q+q')(q-q')$ the product $Q\bar Q$
$\epsilon\in\{1,-1,j,-j\}$ the component sign
$r=\sqrt{\lvert q^2-q'^2\rvert}$ the modulus
$\phi$ the hyperbolic angle
$Q_{\log}=\ln r+j\phi$ the exponent, a logarithm of $\epsilon Q$
$F,R,L$ the future cone and the two spacelike wedges
$M_Q$ the real symmetric matrix of multiplication by $Q$
$\delta$ the parameter of the boundary family of case 8

Further Reading