On the geometry of zeros of a class of functions of a quaternion variable

ON THE GEOMETRY OF

ZEROS OF A CLASS OF FUNCTIONS

OF A QUATERNION VARIABLE

PROEFSCHRIFT

TER VERKRijGING VAN DE GRAAD VAN DOCTOR IN DE TECHNISCHE WETENSCHAP AAN DE TECHNISCHE HOGESCHOOL TE DELFT, OP GEZAG VAN DE RECTOR MAGNIFICUS DR R. KRONlG, HOOGLERAAR IN DE AFDELING DER TECHNISCHE NATUURKUNDE

VOOR EEN COMMISSIE UIT DE SENAAT TE VERDEDIGEN OP WOENSDAG 22 JUNI 1960, DES NAMIDDAGS TE 2 UUR

DOOR

PIETER ANTONIE JOHAN SCHEELBEEK GEBOREN TE AMSTERDAM · ...

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TECHNISCHE

HOGESCHOOL

DELfT

DIT PROEFSCHRIFT IS GOEDGEKEURD DOOR DE

PROMOTOR: PROF DR L. KUIPERS

De samensteller van dit proefschrift werd in 1925 gebo-ren te Amsterdam, bezocht aldaar de 2e H. B. S. 5 j. c. B (Spinoza Lyceum) en de Gemeentelijke Universiteit, enis sedert 1950 werkzaam als leraar in de wiskunde bij het voorbereidend hoger en middelbaar onderwijs.

Aan mijn Vader, Moeder Aan MaDolores, Ramón

T ABLE OF CONTENTS

Chapter I

1. 1. Definitions and properties of quaternions . 1. 2. The characteristic equation.

1. 3. Some remarks ab out the argument. Chapter 11

2.1. The function f(q) has no derivative with respect to q.

p

2.2. Special functions of the type 1: mi (q-qi)'

1 13 13 15 16 19 19 20 2.3. Evaluation of zeros. 24

2.4. Geometrie interpretation of the zeros. 28

2.5. The location ofclass-surface

'I

in relation to the points 30 q. (i=I, 2, •• , p).

2.6. The geometrie meaning of the zeros of

Chapter III

1: ak!

!;,.

(~.i _)(q-lqi+Omi-al. 1 1

3.1. The Lucas Theorem. 3.2. The Jensen Theorem. 3.3. Some generalizations. 3.4. Systems of functions. Chapter IV

4.1. Non singular linear transformations. 4.2. Physical interpretation.

4.3. TheLaguerreTheorem.

Chapter V m

5.1. The function -q

t

1/(q-Qi)+!3· 5.2. Linear combinations of functions.

p

5.3. The function I-À

t

mi/(q-qd. Chapter VI

6. 1. The two hyperspheresproblem. 6.2. The general case.

6.3. The special case p'"'2, n=O. Samenvatting Bibliography 32 35 35 36 39 44 50 50 53 56 60 60 62 67 69 69 72 74 77 79

lNTRODUCTlON

Let

m p.

(1) f(z}=(Z-ZI}I(z-Z2}m2 .••• (Z_Zp}mp, n=l; mi;mi>O, integral,

I

be a polynomial in the complex variabIe Z of degree n.

Let the complex numbers ZI, Z2, .. , zp be the zeros of fez) with the multiplicities mi ~ 1 (i: 1,2, •. ,p) respectively.

The zeros of the derivative fl (z) of the polynomial (1) fall into two classes. First there are the points ZIJ for which mi

>

1. Second there are the p-l zeros of the logarithmic

deriva-tive .

Pm F(z}=d [log fez} ] /dz.l;_i

I z-zi (2)

The problem ofthe location of the zeros of fl (z) in the com-plex plane being given the zeros of the polynomial fez} is in-vestigated extensively (see [6] eh. II). In view of the above mentioned it is easily understood that this investigation led to thestudyofthelocationofthe zeros of functions of type (1) and more general to those of functions of the type

n mi(z-ail)(z-aiÛ .••.•• (Z-aip) I:----,.---~---"--I _{(z-b il )(Z-b i2 ) .••..• (z-bir )}

In this thesis the analogy bet ween the above mentioned problem in the complex plane and that of similar functions of a quater-nionvariabIe is discussed.

For, apolynomialf(q}, where q is a quaternionvariabIe, does not possess a derivative with respect to q. Therefore the cen-tral question asked in Chapter II is: which is the location of the zeros of the function

(3 )

p

( ) - ~ mi F q = " ' - - ,

I q-q~

where mi(i=I, 2, ... , p) is real. qi(i=I, 2, .•• , p) and q are quaternions, in relation to the points qi?

P

The function (3), with

f

m +0 has at least one zero (accord-ing to a not yet published theorem from N:H.Kuiper).

Moreover in Chapter II some calculations of zeros of special functions of the type (3) are c arried out.

In Chapter l a summary of the most important properties of the quaternions is given supplied by some remarks ab out the argument of a quaternion.

In Chapter IIl special attention is paid to generalizations of the Lucas and Jensen theorems. (see [6] eh. II)

In Chapter IV af ter some introductory remarks on linear ra-tional transformations the function

(4)

p

F ( )=~ - ~ mi,

1 q q-qo 1 q-qi

where

!:

mi=n,· _{m i}

U-l,

2, .. , p) is real, is considered. This

1

function is the analogy of the function n P m·

F1(z)= _ _ -:l: __ 1 , z-zo 1 Z-Zi

the logarithmic derivative of f(z)(z-zofn • The Laguerre

theo-rem for the zeros of functions of the type (4) is deduced. Finally in the Chapters V and VI linear combinations of func-tions of the type (2) are studied. In general this investigation furnishes results similar to those concerning analogous func-tions of a complex variable.

CHAPTER I

QUATER NIONS.

A treatment ofthe quaternionalgebra can be found in chapter 10 of Louis Brand Vector and Tensoranalysis. John Wiley &

Sons, lnc. New York.

In this introduction we restrict ourselves to a summary of those definitions and properties, which we shall use in the fol-lowing chapters.

1,1. A quaternion is by definition a hypercomplex number of the form

q

= w

+ xi + yj + zk; w, x, y and z are re al numbers.

The quaternionunits i, j and k are defined by the multiplica-tiontable:

x i j k

i -1 k -j

j -k -1 i

k j -i -1

Fundamental definitions and operations are the following. Eguality: q = ql, if and only if,

w

=

w I, X

=

X I, Y

= Y land z

= z

I

Addition: q + ql = (w + w l ) + (x + x')i + (y + yl)j + (z + z')k Multip-lication by a scalar À:

Àq = Àw + Àxi + Àyj + Àzk The QPP~ of a quaternion

-q = (-l)q Substraction: q - q' = q + (-q' )

From the definitions it follows that addition is associative and commutative, and the multiplication by a scalar associa-tive, commutative and distributive with respect to the addi-tion.

/

Multip-lication. The product of quaternions

qq'

=

(w + xi + yj + zk) (w' + xli + y'j + z'k), is obtained bymultiplying the factors as two algebraic quadri-nomials with observance of the order.

By this definition we find:

q q' = w w'-xx'-yy'-z z' + w (x'i +y'j+ z'k) + w' (xi +y j + zk)+

i + _X Xl j k y, z Y z'

In order to determine q' q ~ we only need commuting the cor-responding coefficients wand w' etc in the above mentioned product. Hence q q' =1= q 'q.

Clearly q q' = q' q, if the determinant is equal to zero, or x : x' = y : y' .= z : z' .

In this case q and ql are called complanar (denoted by q

111 q').

Property (1,1,1) if qi

1/1

qz, then q1 + qz 11/ q1 and 111 qz. Property (1,1,2) if q1 1/1 qz, then q1 qz /1/ q1 and 11/ qz.

The multiplication of quaternions is associative and dis-tributi ve with respect to the addition, but in general not com-mutative.

If x = Y = z = 0, then q = w is a real number.

If y = z = o;-then q = w + xi is a complex number.

q

= w-xi-y j-zk is called the conjugate quaternion of q = w+xi +yj+zk.

Property (1, 1, 3) q q' = q'

q.

The norm of q is the positive real number N (q)

=

q

q

=

WZ + xZ + yZ + zz.

Property (1,1,4) N (q q')

=

N (q). N (q')

=

N (q'). N (q). Property: (1, 1, 5)

Is q = 0, then the inverse of q is -1 _ 1 1 - _ 1 q -

q

q.

q - N(q)·

q.

The equations aq = 1 and qb = 1, where a, band care quaternions, are resolvable.

In this algebra division is also possible.

Is in an algebra a

f

0, b

f

0 and ab = 0, then a and bare called divisors of zero.

Theorem (1,1,1). In an algebra w.ith a unit element, every element has aright or aleft inverse, if andonly ifthisalgebra does not have divisors of zero. (see [2] p 40, 41, 47);

Property (1,1,6): The quaternionalgebra has no divisors of zero.

Proof. We infer this from the possibility of division and Theorem (1,1,1)

Second Proof. We suppose

(a +-bi + cj + dk) (e + fi + gj + hk) = 0, or { ae - bf - cg - dh = 0 be + af - dg + ch = 0 ce + df + ag - bh = 0 de - cf + bg + ah = 0

This system of four linear homogeneous equations in the unknowns e, f, g and h has a non-trivial solution, when

a - b - c - d b a - d c

₌

_{(a 2 + b 2 +} _9. + d2 )2

=

0 c d a - b c- , d - e b a or if a = b = c = d '" o. Property (1,1,7) (q ql

r

l = (ql t I . q-l. Property (1,1,8) q ql + qlq = N (q) + N (q'). N (q - q'). 1,2. We derive from [10] p. 13 the following:

Each quaternion satisfies an algebraic equation of the second degree with real coefficients, the characteristic equation of the quaternion.

q = w+xi + yj + zk satisfies

q2 + aq + b = 0, (a and bare real), if

w2_{(j)- x 2 - y2 - z2 + aw + b = 0} 2 wx + ax = 0

2 wy + ay = 0 2 wz + az = o.

Now we suppose that one of the unknowns x,

r'

z is unequal to zero. This leads to -2w = a and w2 +

r

+ y + z2 = N (q)

= b.

Conversely it is easily seen that an equation of the second degree has infinitely many quaternionroots, if the discriminant is negative.

For, if we have q2 + aq + b = 0, then w + xi + yj + zk is a root, if we choose w =

-8/2

and the numbers x, y and z such that x2+ ~ + z2 = (4b 2-a2

Y4.

Is the discriminant of the charac-teristic equation non-negative, then there are two real roots

only. In general we have: An algebraic equation with real coef-ficients, which has at least one imaginary root, has infinitely

many quaternionroots. An algebraic equation can be written

as follows:

F (z) = A ?T.(z - zt )hi (zz _{+ P j z + qj )kj ; E h i + E} 2 kj = m, zi' Pj and qj arJ'Jreal and PjZ - 4 qj

<

o.

1, 3 A quaternion mayalso be written in the form:

q = N(q)! (cos 9'+ esin 9'), where 9' is called the argument of q; o!: 9' '=. 7f.

Here e =(xi -:- yj + zk) / (x z + y2 + zZ) hence eZ = -1, and cos 9' = w/ N(q)!, sin tp::(x z + yZ + zZ )!/ N(q)!.

If _{a = a o + al i + a} 2 j _{+ a 3 k and b = b o + bI i + bzj + b3 k}

two quaternions, then the real part of ab is: and hence ao bo - al bI -

az

bz -

a 3

ba '

cos arg (a b) = (aobo - al bI -

az

bz - aa b 3)/ N(ab)! =(aobo - al bI - az~ - a 3b3 )/ N(a)!. N(b)!.

are

Furthermore cos (a + f3 ), ( a and

fJ

are the arguments of a and b) = cosa cosfJ -sinasinf3 = a o/ N(a)! . bo / N(b)! - (ai + a~ + a~)

(bi +

b~

+

b~

)/ N(a)! • N(b)!.

n n n 2

Now I: ar E bf ~ (E ~ bi ) , or

T I I

cos (a + (3) ~ cos arg (ab), or (1,3,1) a + f3 ~ arg (ab).

If the angle tIJ (0' tIJ !: 7f ) between a and b is defined by

cos 1/1 = (~l:b _{+albl +aZb2 +a3ba)/N(a)!} N(b~ andif a is the argument of a then the an,gle betweenb and ab is equal toa. Pro of: cos (b, ab) = [bo (aob o - albl - a2 b2 - a3 b3) + bI (aob! + albo + a 2b3 - a 3bz ) + b 2 (aobz + azbo

+

a 3b l - a l b 3 )

+ b 3 (a ob 3 + a3bo + albz - a2 bl)] / N(b)! [N(a)! N (b)]! ao/N (a)! = cos a.

Finally the argument of qP (p is a positive illteger) has the following property:

qP= (N(q)!)P (cos9'+esin9')P = (N(q)!IP(cosP9'+esinp9').

since e 2 = -1. we apply the lIDe Moivre' S" theorem. As to P9' we distinguish two cases.

Case 1. Case 2.

o '= P9' ~ 7r (mod 2,.. ). 7r '=.P9' '=21/(mod 27r ). In the latter case - P9' satisfies

-,..~ -P9' à -27r(mod. 27r), or of:-p" '= 7r (mod 2.,,).

Since cos P9' + e sin P9' = cos (-P9') - e sin(- P9'>. the e-vec-tors corresponding to qP and q respectively in the second case are equal and opposite.

For instanee, we take q = -2 + i + 3 j + 2 k.

Then is arg q = 118°7' 30" ; eq = 1. 14-i. i + 3. 141j + 2. 14-i. k q2 = (-2 + i + 3j + 2kf = -10 - 4i - 12j - 8k.

argo q2 = 123"45' = 3600-2 x 118°7' 30" • e q2 = -4. 224-ii -12. 224-ij - 8. 224-i k. Hence e q2 = -e q•

We can deduce the argument of qP directly from pand the argument 9' of q.

Theorem (1,3,1) If 9' is the argument of the quaternion q, if P is a positi ve integer. if further k is a positi ve integer satis-fying P9'

I

27r

-t

'= k '= P9'

I

27r, then arg qP is equal to p" -2k 7r. If however k satisfies

p,,1

27r ~ k'=.p9' / 27r +

t

thenarg q is equal to 2 k7r - P9'.

Proof. 1. Let P9'

I

27r -

t kkbp9'

I

27r (kis apositive integer). P9' - 7r ~ 2 7r k '=. P9', or

o ~p" - 2 7r k'=7r.

Since p" - 2 7r k

=

P9' (mod 27r) and also satisfies o l: p9'-2 7rk~ 7r, p9' - 27rk is the argument of qP.

2. Let P9'

12."

!: k !:

p,,1

27r +

t (kis apositive integer).

P9' i::2 7r k '=. P9' + 7r, or

0'=

2 7r k - P9'6: 7r •

Since 2,.. k - P9' :-P9' (mod 2,.. ) and also satisfies o '=.2 7r k -P9' b. 7r, 2."k - P9' is the argument of qP.

Examples 1. 9' =

Tri

6, p = 7 The positive integer k= 1 satisfies

7.7r16'=k"=.7.7r16 +1. 2

Hence arg q'1 = 21T - 7. 1T/ 6 = 51T/6

2. rp

=

2.,,/ 3, p

=

10.The positive integer k

=

3 satisfies 10.2.,,/3 _

t

'k~lO· 21r/ 3

2". 21T

CHAPTER II

2, 1. Extensively investigated is the location of the zeros of the first derivative f' (z) of a given polynomial f(z) in relation to the distinct zeros of f(z), where z is a complex variabIe (see

[ 6] chapters 1, IIL

Now it is possible to define formally a derivative of a function f(q), where qis a quaternionvariabIe. Since multipli-cation of quaternions is not commutative,this formal derivative is impracticable.

Let

q-qo

be the first derivative of f(q) at the point qo'

The set of quaternions can be put in one-to-one correspond-ence with the points of a four dimensional space R oriented by a rectangular co-ordinatesystem. Everyquaternion q=w+xi+ +yj+zk corresponds to the point with co-ordinates (w,x.y.z). As to the functions \Jo (w, x, y, z). UI \w, x, y, z), u2 (w, X. y,z) and u 3 (w. x, y, z) the meaning, of the requirement that f(q)

==

uo+u1i+u2j+u3k be differentiabIe at the point qo= Wo +xoi+YoJ + +zok can be expressed as follows. The difference quotient

{M.

(ó.q)

-1}

q=~

:: [(uo+u₁i+u_2j+u3k) -(uo+u1i+~j+U3k)

1

w.x,y.z Wo.Xo'YO'Zo

)( [(w+xi+yj+zk)-(wo+Xoi+Yoj+zok» -1

must always tend to a single definite number as limit, how-soever q-qo' In particular. the limit must exist if q is al-lowed to approach qo once along aline parallel to the w-axis. another time along a line parallel to the x-axis etc; that is, if for fixed x=Xo. y=Yo' z=zo. w is made to approach wo. for fixed w=w • y=Yo' z=zo. x is made to approach Xo etc.

Is u(w) îhe partial derivative with respect to w. then in the four cases belonging to the different ways we have

f I (qo )=u oCw) +u I (w) i+uiw) j +u 3(w.' k. f'(<:Jo ) = (llo (x) +ll.J(X) i+u

2(x} j

+uP)

kW

I • f' (qo )=(uo(Y) +u 1 (Y) i+~(Y) j +u₃ (Y) k

)r

l

Consider the following example.

f(q)= q2=W2 _x2_y'2 _z2+ 2 wxi+2wyj+2wzk.

The value of the "formalIl derivative with respect to q, at the point q=qo' is

ft (qo)=2qo=2wo +2xoi+2Yoj+2zok.

If qapproachesqo along the paths parallel to the axes, then we have

fl (qo)=2w,o +2xoi+2Yoj+2zok. fl(qo)=(-2x o+2w oi)i-1

fl(qo}=(-2yo +2w oj)fl

fl(qo)=(-2zo +2w ok)k-1 , respectively. The zeros ofthe derivative fl (z) of the polynomial in a

com-p

plex variable z. f(Z):::(Z-ZI) mI (Z-Z2) ~~ (Z-Zp )mp,

r:

II\ = n, I

are first the points Zj , for which mj

>

I, and second the p-l zeros of the logarithmic derivative F(z)=d [log f(z)] / dz, or

(2 1 1) , , F(z>=

J!

IJ mi 1 Z-Zi Analogous to (2, 1, 1) we consider p mi (2,1,2) F(q)=I:--. 1 q-qi P

where the mi U=I, 2, ... , p) are real numhers, ~ mifo and the qi (i=I, 2, .•. , p) are quaternions and q is quaternionva-riable. (2,1,2) has at least one zero. We shall show that there are special functions of the type (2,1,2) with exactly p-l zeros. Further we shall investigate the location of a zero of (2,1,2) in relation to the points qi (i=l, 2, .. , pl.

2,2. Theorem (2,2,1). If the points qi (i=I, 2, .. , p) lie in a plane V of R4 then the function tp(q)=t, Illi/(q-qi), where

t

Il\

f

0, has p-1 zeros in

v..

I . 1

To prove Theorem (2,1,2) We use

Lemma (2, 2, la) If q.' (i=l, 2, .• , p) are the points into which the points qi (i=l, 2/ .. , p) of (2, 1,2) are transformed by the non-singular linear integral transformation

then a zero of (2,1, 2) transform into one of the function

P m ·

(2,2,2) F' (q)

=

1 : _1

1 q_q~

Proof. On proving the lemma we substitute Q' = aQ+{3 in (2, 2, 2) P 1: 1 = P 1: 1 m. 1 aQ+{3 - aCli -{3 ( p mi) ~-- a-I = o.a- I = o. Q-qi P = 1: I a(Q-qi}

Lemma (2,2, 1b). There is a transformation (2, 2, I) whieh two points A and B of R4 transforms into two given points A' and B'.

Lemma (2,2, Ic). A linear set of R4 is transformed by (2,2,1) into a similar one.

The proofs of the Lemma's (2,2, 1b) and (2,2, Ic) are left to the reader.

Proof of Theorem (2,2,1). There is a transformation (2,2,1) whieh transforms the plane Vinto a plane V' through the real-axis (according to the Lemma' s (2,2, 1b) and (2,2, Ic

L

By this operation the points q i (i=l, 2, .. ,p) are transformed into points which correspond to complanar quaternions q.' (i=1,2,

... ,p). If we allow q to be eommutative with the quitternions q: the transform of .,(q) ean be written as a polynomial of the degree p-1 with eomplanar eoefficients. Sueh a polynomial has p-l zeros. From Lemma (2,2, la) we infer that .,(q) has p-1 zeros in V.

Remark. Not only polynomials with complanar coeffieients have zeros. There are also non-eommutative polynomials whieh have at least one zero. See (31 , (4] and (7).

Now we pay attention to the function (2,2,3) 2::: _ _ _ P Il\ --=-____ _

0, 1 w+xi+yj+zk-wi -xii

In stead of (2,2,3) we write P mi [(W-Wi )-(x-xt )i-y j-zk] (2, 2,4) 1: I = 0, where Ni = N (q-qi) or af ter reduction: (a) _E P I (b) E I> I (2.2,5) mi (W-W_{i )} Ni mi (x-Xi) Ni p (c) E ffit y = 0, 1 _Ni =0 =0 or y (d) _{E _ _} P m·z _{l_ = 0, or z} I _Ni p _m. E_l_=O 1 Ni P mi I ; - - = O. 1 Ni

If furthermore we suppose mi> 0 (or < 0) (i = 1,2, .•• ,p), th en is y

=

z

=

o. On substituting Y

=

z

=

0 into (2,2,5) (a)

and (2,2,5) (b). We have reduced the problem to one in the WOX-plane. Consequently the function has p-l zeros in V.

If not all the coefficients are of the same sign, then also y

=

z

=

0 satisfy (2,2,5) (c) and (2,2,5) (d) etc. Hence there

are p-l zeros in V. In this case there may also be zeros, lying outside V. The following examples show that there are functions which have just p-l zeros in V, while on the other hand there are functions which have p-l zeros in V and also zeros outside V.

Example (2,2,1) We take four points:

(0,0,0,0);(1,0,0,0);(0,1,0,0, );(1, 1,0,0,) lying in the WOX plane; mi

=

_{m 2} = _{m 3}

=

_{1 m 4} = -1. Af ter reduction we obtain a system of four equations.

(a) wN 2N3N4 +(w-l)NI N 3N4 +wNI N2N4 -(w-l )N₁ N_2N3=0 (2,2,6) (b) xN 2N3 N4 _{+xNI N 3N4 +(x-l )N1 N 2N}ç (x-l)NI N2N 3"'0

(c) yN2N3 N4 +yNIN3N4+yNIN2N4 -yNI N2N 3 =0 (d) zN 2N3N4 +zNI N 3N4+zN IN 2N4 -zN1 N 2N3 =0

If y

f

0, Z

f

0

(2,2,7) _{Nz N 3N4 +NI N3N4+NI NzNçNINz N 3=0.}

On substituting from eq. (2,2,7) into (2,2,6) (a) and

(2,2,6) (b) we find

(2,2,8) (a) -NIN3 N4 + NI NZN3 = 0 (2,2,8) (b) _{-N I Nz N4 +NI N zN3 =0} From (2,2,8) (a) and (2,2,7) it follows (2,2,9) _{Nz N3N4 + NI Nz N 4 = o.} No q =1= qi (i = 1,2,3,4) satisfies (2,2,9)

Example (2,2,2) We consider once more the term (2,2,10)

mi mi (q-<ii) mi

(q-ëii)

1

(2,2,10) _{- - =}

Hence this term may be represented by a vector of the direction from qi to

q

and having the magnitude of Int times the reciprocal of the distance from qi to

q.

If IIlj.

<

0 then the vector has the direction from

q

to qi.

D +-~

_ _

-._c

A B

We choose five points lying in the WOX-plane A(-l, 1, 0, 0); BU, 1,0,0); CU, -1, 0, 0); D-1, -1,0,0); E(o, 0, 0, 0).

Further we take m. = 1 (i = 1, .•. 4), _{m 5} = -1.

The above describJd vectors, are denoted by vi (i=l, 2, ... ,5).

The point S lying on the perpendicular in E on the WOX-plane

is a zero of the function, ti

(2,2,

11)

{4

VI cos €X =v5 VI : v5 = ES:

AE.

Is ES=P and EB= d, then we find after reduction

(2,2,12)

so that 4p2= d2 + p2 and q = 1r / 3.

Hence the points (0, O,~. 6 i, 0), (0,0,

-1-

6i , 0), (0,0, O,~. 6i ), (0,0, 0, -~. 6i ) lying outside the WOX-plane are zeros of the function. Moreover this function has zeros in the WOX-plane.

.

.

2,3. Calculation of zeros of functions of the form (2,1, 2)

Example (2,3, I). p= 2. (2,3,1) or, mI _{m 2} - - + - - = 0 q-ql q-q,z

--=---q-ql q-q 2 Since the numerators are real, is

q-ql q-q2 =

-or,

(2,3,2) q =

mI + m2

Evidently the root (2,3,2) of (2,3,1) lies on the line, joining the points ql and q2'

Example (2, 3,2) p= 4.

We select the following four points:

(1,0,0,0);(0,1,0,0 );(0,0,1,0);(0,0,0,2), or

I I I I

F(q): + + +

On solving the equation F(q) = 0 we are led to the investigation of the following system of equations

w-l w w w (a) - - + - + - + - = 0 NI _{N 2 N3} N₄ x x-I x x (b) + - - + - - + - = 0 NI _{N 2 N3} _ N4 (2,3,3) y Y y-l Y (c) - - + - - + - - + - - = 0 NI _{N 2 N 3} N₄ z z z z-2 (d) - + - + - + - = 0 NI _{N 2} N₃ N₄

The system (2,3,3) in the four unknowns 1/ NI, 1/ N2, 1/ _{N3 ,} 1/ N₄ has a non-trivial solution, if and only if

w-l w w w x x-I x x (2, 3,4) _y-l = 0 Y Y Y z z z z-2 or af ter reduction (2,3,5) 2w + 2x + 2y + z = 2,

Hence the zeros of the considered function are points of the linear R3 through the points qi (i= 1,2,3,4)

Remark. Since the factors 1/ Ni, (i = 1,2,3,4) of (2,3,3) are definitely positive, the system (2,3,3) can only be satisfied by the values w,x,yandzlying intheintervalsO<w< 1, O<x < 1,

o

< y < 1 and 0 < z < 2.

In a following chapter we go deeper into the meaning of these intervals.

From (2, 3, 3) (a) en (b) we deduce (2,3,5) wN_I = xN₂

or (2,3,6)

So that (2, 3, 3) can be replaced by the system (a) (w-x)(W Z+xz +y2 +zz -2w-2x+1) = 0 (b) (W_y)(wz+xz+yZ+zz -2w-2y+1) = 0 (2,3,7) (c) (x-y)(w2.rx z+y2+zZ-2x-2y+1) = 0 (d) 2w N₁ = z N 4 (e) 2w + 2x + 2y + Z = 2

Now we distinguish the following cases: a. w=x=y

From (2,3,7) (d) and (e) it follows (2,3,8) 156w z - 65w + 5 = 0

Denoting the roots of(2, 3,8) by v₁ and _vz' we find the following zeros:

v

1 + 1111+ 1I1j+ (2-6111 )k and Vz+ 1121+ 112j+ (2-6 v2 )k.

b, w=x=/=y.

From w:fyand (2,3, 7)(b) it follows: w 2 +xz+y. +Z2 -2w-2y+1= O. We replace (2, 3, 7) by (2,3,9) (a) w = x '(b) w 2 +~+y2+z2-2w-2y+1 = 0 (c) (d) 2w+2x+2y+z = 2.

2y (w z+~ +yz+z 2_ 2y+1)= z (wZ+r+y2+z2-4z+4). On substitutingfrom WZ +X2+y2+ZZ= 2w+2y-1, w=x and z=2-4w:

2y into (2,3, 9)(d), and from (2,3, 9)(a) and (c) into (b), we obtain

(a) {10y2+40YW+36W2-15Y-28W+5 = O. (2,3,10) (b) 5y 2+16yw+1Bw2 -10y-18w+5 = O. On eliminating y,

576w4-368w3_{+145w 2-25w= O.}

w= 0, x= 0, y= I, z= 0, is no zero (see above). Let us consider f(w)=: 576w3-368w2+145w-25.

f(O)

<

0 and f(l)

>

0, thus f(w) has a real zero _{11 3 ,} between 0 and

1. fl (w) = 1728 w2-736w+145 has no real zeros, hence f(w) has only one real zero.

v₃ furnishes three zeros more, namely

-8v₃

+

5 -8 v₃

+

5

113+ v₃i + j + (2-4v -2 ----=----)k,

&/ 3 + - - - - i + &/ 3 j + (2 -4 &/ 3- 2 8v 3 + 5 -8"3+ 5 - - = - - ) k, 8 v₃ + 5 -8&/3+ 5 -8"3+ 5 ----"'----+1 &/3i + V₃j + (2-4v₃-2 ) k . ..J.. =/= 8&/3+ 5 BV3+5 c. w T X y.

From (2,3, 7)(a), (b) and (c) it follows w = x s y.

Hence a contradiction, so that the function has no zeros with

w =1= x =1= y.

Example (2,3,3) p=5; mi=1 (i=I,2 .• , 5). We select the following points:

(0,0,0,0);(1,0,0,0);(0,1,0,0);(0,0,1,0) and (0,0,0,1), or

I I I I I

F(q) - - - - + + + +

-w+xi+y j+zk (w-I)+xi+y j+zk w+(x-I)i+y j+zk w+xi+(y-I)j+zk w+xi+y j+(z-I)k

Solving F(q) = 0 we are here led to the investigation of the following system of equations

w w-l w w w (a) + - - + - - + - - + - = 0 NI N_{2 N 3} N₄ N

_s

x x x-I x x (b) + - + - - + - - + - - = 0 (2,3,11) NI N_{2 N3} N₄ N_s y y y y-l y (c) - + - - + - - + - - + - = 0 NI N_{2 N3} N4 Ns z z z

z

z-1 (d) - + - + - - + - - + - - = 0 NI N₂ N₃ N₄ N_s From (2, 3, 11) we de duce w N_{2=xN3 =yN4= zNs .} Now we distinguish the following cases: a. w= x=y= z.

On substituting into '(2,3, l1)(a), we find 20w2-6w + 1 = O. This equation has no real roots.

b. w=x=y=/=z.

From wN₂ zN_s it follows

or

(2,3,12) 3w 2+z 2 = 2w+2z-1

Substituting (2,3, 12) in (2, 3, 11) (a) we find

(2,3,13) 6w 2+10wz+2z 2-5w-3z+1 = O.

From (2,3,12) and (2,3,13) we get

150w4-70w3 +22w2-w = O.

w= 0, x= 0, y= 0, z= 1 is n~ zero of the given function.

Consider now f(w)

=

150w -70w 2 +22w-1.

f(O)

<

0 and f(l) > O. Hence f(w) has a re al zero p between

o

and 1. Moreover f' (w) has no real zeros, in other words,

p is the only re al zero of f(w). Thi s gives four zeros:

p-tPi+pj+~

k,

P+P-i+~ j+pk,p+~

i+pj+pk, P+1 +pi-tp

10p+1 10p+1 10p+1 10p+1

c. w=x, w=/=y, w1=z. Then is

(2,3,14) (a) {W2+X2+y2+Z2-2W-2Y+1 = O.

(b) W 2-tx 2+y2+z2 -2w-2z+1 O.

or y = z

Substitution from 2w2+2y2 = 2w+2y-1 into (2,3,11) (a) leads

::,t:,e1:~~:~m{4W2+10YW+4l_4W_4Y+1

= Q

2 2

(b) 2w +2y - 2w-2y+1 = 0

Eliminating z, we find

(2,3,16) 100w4-100w3+50w2 -10w+1 = 0

(2,3,16) has no real roots between 0 and 1.

d. w=f:x, w::ly, w::lz, xjy, xfz, yfz.

From w='l"2=X~3=y~4=zN5 it follows: w=x=y=z.

Hence a contradiction.

2,4. Investigating the location of the zeros of (2, 1, 2), we shall apply Morris Marden' s method (see [61 p 9, 10).

We write

p

k - - - = 0 in the form

p mi 1 (2,4,1)

t

= 0, t =

-t -1 - (wi+xi i+Yd+zi k) w+xi+yj+zk

Multiplying from the right both sides of (2,4,1) by -t-1 and applying Property (I, 1,7) we find

p mi

(2,4,2) ~ = 0

1 t(wi+xii+Yd+21 k)-l

Let us compare (2,4,2) with the equation

P mi (2,4, 3) ~ - - = 0

1 Ei

(2,4,3) represents a class-surface

r

(of class p-1) Eq.(2,4,2) tells us that eq (2,4, 3) is satisfied by the surface with the co-ordinates 1 1 À= / . l = - - - i , w+xi+yj+zk w+xi+yj+zk (2,4,5) 1 1 11

=

j, C1

=

k. w+xi+y j+zk w+xi+y j+zk Consequently the linear space

(2,4,6) À W +jJ X + IIY +C1 Z - 1 = 0

where À, jJ, 11 , C1 are defined by (2,4,5) touches (2,4,3).

The zero of (2, 1,2) with co-ordinates w,x,y and z satisfies (2,4,6); hence the space (2,4,6) passes through the zero

(w, x, y, z) of (2, 1,2).

Now in R4 the equation of the isotropic hypersphere with center (w,x,y,z) is:

(W-W)2+(X-x)2+(Y-y)2 +(Z-z) 2 =

o.

We write the first member of the equation as a product. I(W -w)+(X-x)i+(Y -y)j~(Z-z)k)

x ((W-w)-(X-x)i-(Y-y)j-(Z-z)k) .

By definition a focus of a class- surface is the center of an isotropic hypersphere which touches that surface.

Theorem (2,4,1). A zero of 'the function (2, I, 2) is a focus of the class-surface (2,4,3).

For, multiplication fr om the left of (2. 4. 6) by w+xi +y j+z k gives:

(2.4.7): W-w+ (X-x)i+(Y-y)j+(Z-z)k= O.

The left side of (2.4.7) is one of the linear factors of the left side of the equation of the isotropic hypersphere with center (w,x.y. z). So (w,x.y. z) is a focus of the surface (2,4.3). 2.5. We pay now special attention to the case of real space-co- ordinate s À. IJ, tJ, a. When cleared of fractions equation

(2,4,3) has the form:

(2,5.1)

I

(À,,...tJ,a) _{mlEz •... E p+ ... +mpEI·· .. E p_1} =0.

(2, 5. 1) is the equation of a three dimensional surface of the class p-l.

More general:

(2,5,2)1(À_{1 • • • •} ,Àn)=mlEz ..• Ep+ .... +mpEI .... Ep-l =0.

is the equation of a (n-l) dimensional surface of the class p-l, where

Jl. . j

Ej = f=1 Ài Jxi - 1,(j=1,2, ... ,p)and x

1(j=1,2, ... ,p) is a

point of Rn with co-ordinates

ei

xl' •..•.• jx n).

Let us consider now a linear Hn-l which goes through n of the given points, with co-ordinates °À₁(i=1, 2, ... , n), then the space-co-ordinates of Rn-l satisfy n equations,

El = O. E 2 = 0, . . . , En = 0 or

n o .

(2,5,3) l:: ÀiJXi-1=0. j=I.2, ••.. ,n.

1

Since Ej = 0 (j= 1, 2, ...• n) satisfies (2,5,1), Rn-l touches the class-surface.

~ow the point of contact ofthe tang~nt-space

Ho-I

(ij, i= 1.2,

.... , n) has the space-equation.

n

°

al

(2,5, 4) ~ (Àj- Àj) (aT)o= O.

where the subscript 0 indicates values at

(À

1 ' o~ , ••• , \ \ ).

As Ej = 0 (j= I, 2, ..• , n) the first side of (2,5,4) is identically zero. Consecjuently the point of contact is in':lefinite.

For further investigation about the order of contact we have to consider in the Taylor expansion the first terms that don' t vanish identically. These are the (n-l) th derivative terms in the expansion.

(2,5,5)

B

p ,(n-l)! p

,G\-~I>P!

.•. (À-Ànfn P1~o •• ·Pn~O l' • • • • • n' Epi=n-1 where,

a

n-l

I

_

2 3 P1+1 n-Pn+1 Pn À P P - _{mI { Xl Xl'" Xl • . •. Xn···· Xn} + ....•. + }l

a

11 ...

a

Ànn + m { } 2 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 2

The terms ofthe expression { }. are obtained by permuting the upper-indices. A term does not1 change when we permute the upperindices of that term in such a manner that each of them remains accompanied by the same lower index. Clearly the expression {

h

consists of terms, of which PI! .... Pn! are equal. Hereby the factor l(Pl! •...• Pn! ) vanishes.

Evidently we can write (2,5,5) in the following way:

n

(2,5,6) L m. _{El' .. E. 1 E. l ' ... En} = 0,

1 1 1- 1+

n . n o '

for from Ej = LÀ/JG. -1 and L ~ JXi -1

=

0 it follows

1 1

(2,5,7)

n Q '

EJ'

=

L (À. - A· ) lX. , and thus 1 1 I I

n n o

Qn-1 == l: mk rr l: (~C _{Ài ) jX i =} O. k=1 j=1,*k i

Consequently the linear Rn-l 's going thro 19h n of the gi ven points are (n-l )-fold spaces of contact of the class- surface (2,5,1). (2,5,7) is the equation of a surface of the class n-l which is the locus of the points for which the Rn-1 is a n-fold tangent-space of (2,5,1). The linear (n-1) dimensional spaces going through n-1 of the considered points, touch the surface Qn=1 = O. Hence the intersection of these linear R n-1 's, a Rn-2 , touches Qn-1 = O.

Calculating the points, in which this R n-2 touches Q n-l =0, we find n-1 n-1 (2,5, 8) Qn-2=k~1 mk 7r E. j=1,,* k J 0, etc. So n-i+1 n-i+l (2,5,9) ~ . = l: mk 7r E· - I - k

=

1 j=1, =1k 1 0,

the locus of those points of ~-i , going through n-i+l of the given points, for which Rn-i is a (n-i+l)-fold linear

tangent-space of the surface Qn-i+l= O.

Finally we consider the case i = n-2. Then, 3 3

Q2 == ~ mk j=~=Fk~ = 0,

the equation of a conic lying in a R2 going through the points ~ (i= 1, 2,3). This conic touches the line segments (~, q2), (q2, q3) and (q3, ql). The point of contact of a tangentline has the equation:

Ql = m1E2 + m2 E l = O.

The line segment qi qk is di vided by the point of contact in the ratio mi:m k (i =f k; i, k= 1,2, 3).

Substitution n= 4 in Qn-i = 0, i= 1, 2,3 furnishes the desired result for quaternionvariables.

2,6. The kth derivative of the polynomial

p

fez)

=

a Ir (Z-Zi )mi is: 1

(2,6,1)

In order to study the location of the zeros of (2,5,1) we write (2,6,1) into the form

(2,6,2) P

m-a },; ak! ~

<r::l)

(-zt / z+l) i i

Now we substitute z by the quaternionvariable q and zi(i= 1, 2, .•. , p) by ~ (i= I, 2, .•• ,p) and we investigate the location of the zeros (if any) of

(2,6,3)

where the qi (i= 1, 2, ..• ,p) are points in R4. These points de fine a class- surface:

(2,6,4) F (Àl> •••• ,À5 )= P 7f (I: 4. ~ Ài +À5 ) m j

=

0 j=l 1

(~, i= 1, 2, .. ,5 are homogeneous co-ordinates, ~ is real, Àl' .. , À4 are quaternions).

Further we have

(2,6,5)

a

kF(Àt.····,À5)

aÀ k

5

Lemma (2,6,1). Iff(ql •.•.•• qn.À) andg(ql, ..•.•• <lo, À)are functions of the quaternionvariables qi (i= 1. 2, .•.• n) and the real variable À, then

a

ag

éH

- ( f g ) = f - + - g.

aÀ aÀ aÀ

a

.

f(ql····Cln,À+.!U)g(ql> .•• cm,À+.D.À)-fg Proof: - (fg) = Llm

-äÀ .!U-O .!U

= Lim [f(ql, ••• qn' À + .!U)g (ql' .• , qn' À + .!U)-f(ql, •• qn, À+.D.À)g + f(ql"'. qn' À+.!U)g - fg

J/

.!U.

= etc. Hence

(2,6,6)

E

k!

al'

o.

,a_p D lli=k

rp

k = 0 is the equation of a class surface.

In section (2. 4)P • .t9 . the focus of a surface has been defined. As to the foci (if any) of '" k = O. we shall prove a theorem. Theorem (2.6, 1). If the surface '" k = 0 possesses a focus then this focus is a zero of (2. 6, 3).

Remark. This theorem is ageneralization of a theorem of M. Fujiwari [5} p 108.

Proof. On calculating the foei of'" k = 0 we are led to the fol-lowing system of equations. We choose in homogeneous co-or-dinates:

{

"'k~r, À~. À~.

À4 >=O{"'k(ÀI. Àz.

~, ~

)=0 ,,1

=

À' i À'

=

-i À' A2' 1 2 1 (2,6,7 À' = À' 0 À' = _ 0" , 3 lJ 3 J Al À' ₄

=

À ₁ i k À' ₄

= -

kÀ ' _{1 •}

We observed (seep.2j) that Xl + x2 i + x3j + X4 k

=

0 and Xl -x 2i--x3j--x4k = 0 are the equations of two isotropic spaces.

Substituting from À

L

À~ and Àl into

tt

k = 0, we have:

c

a ){

~k (À~, À~

i,

À~j,

À

~

k) = 0 (2,6,7)

(b) flik (À~, -iÀ~, -j:~, -kÀ~) =

o.

Is Àh a root of (2,6, 7)(a) and À ~ a root of (2,6, 7)(b) then a focus is a solution of the system

(a){À~XI +À~x2i

+

~x3j

_{+ \ x 4k + 1 = 0}

(2,6,8) \I-

*

*

>t

(b) À kXI - i À~x2 - jÀ~Xa - kÀ~x4 + 1 = 0

Since Xi (i= I, .•• ,4) are real numbers, we find the JG. on solving the equation (2,6,8) (a), or

xl +x 2i+x3j +x4k =

-1/

À~.

The ~ (i= 1, ••• , 4) have also to satisfy the second equation (2,6, S")(b), or

*

x - x i - x j - x k = -I/À'

1 2 3 4 k

11-Hence

>{

and À~ must be conjugate. Af ter having chosen a Àh WhlCh satisfies (2,6, 7)(a) it is always possible to find a root ~k of (2,6, 7)(b) th at is the conju~ate of Àh. .

Now we show that ~ satisfies ~k (À, -:1,\, -jÀ, -kÀ) = 0, whenisgiventtk(~' ~i, \'j,Àh'k) = O.

tt k (~, -~ , -j~ , -~' )= tt k (\' , ~ i,~j, ~ k) = tt k (~

,Àh

i,~ j, À\N IfÀ

_h

= -1/~, then

r

_{= Xi + x 2i + Xaj + x4 k satisfies}

(2,6,9) _{tt k [-(rr}l ,-(rf\,-(r>lj,-(r}-lk) = O.

Evidently the f~ci of (2,6,6) are the same as those of (2,6,3) with qi= l~ _{+ lx 2i + lX3j + lx4 k and q = Xi + x2i + x3j + x4k.} We applied an argument, due to T. Takagi

l

9 ] p. 27.

CHAPTER III

3. 1. Generalizations of the theorems of Lucas and Marden. Let fez) be a polynomial in a complex variabie. Lucas [6]

pp 14, 15 has proved:

Any convex polygon K which contains all the zeros of a poly-nomial fez) also contains all the zeros of the derivative fl (z). In the followingwe consider an analogous problem in R4. First we define a convex polyspace.

Definition. If a space is bounded by linear -three dimensional spaces and if every point of the linesegment joining two points lying within the polyspace, also lies within the space, then that polyspace is call-ed convex.

Theorem (3,1,1). Any convex polyspace K which contains the points qi (i = 1, 2, ... , p) also contains all the zeros of the fune-t i o n p

F(q)=~

mil

(q-qd,

1

where mi (i=l, 2, •.• , p) are positive.

Proof. In stead of F(q) let us introduce its conjugate F(q)=

f

~

= ;, mi(q-qi)::

t

mi (q-qd .

-.==1===-1

q-qj

1 N(q-qi)

\j

N(q-qi) YN(q-qi)

p mi

=

t

yN(q-q

i~q-qi.

Hence, itmaybe represented by a vector having the direction fr om qi to q and having the magnitude of mi times the recipro-cal of the distance fr om qi to q. In other words, the ith term may be regarded as the force with which a fixed mass (or electric charge) mi at qi repels (attracts if rni< 0) a movable unit mass (or charge) at q, the law of repulsion being the inverse distance law. This means that we may interpret the conjugate of F(q) as the resultant force at q, in a field due to a system of p masses (charges). The zeros are the positions of equilibrium in the given force field. Hence, from a physical point of view, the zeros of F(q) must lie in any convex poly-space enclosing the points Qi(i=l, 2, .... p).

Theorem (3,1,2). Anyhypersphere H which encloses the points

Qi(i=l, 2, ••. , p) also encloses all the zeros of

p

F(q)

=t

mï!(q-qi),

I

where ml(i=l, 2, ... , p) are positive.

Proof: For, if K is the smallest convex polyspace enclosing

the points q.(i=l, 2, .•• p), then K lies in Hand therefore by

Theorem (3: 1,1) all the zeros of F(q), being in K, also lie

in H.

Conversely Theorem (3.1,1) follows from Theorem (3,1,2).

For, ü Theorem (3,1,2) were valid, through each linear

side-space of the polyside-space K of Theorem (3, 1, 1) we could draw a

hypersphere Hk, which contains K and hence the points qi (i =

1, ~, .•• , p) and the zeros of F(q). The region Dl common to

.all these hyperspheres Hk would also contain the points q i

(i=l, 2, .•. 1 p) and the zeros of F(q). Since this holds for all

choices of hyperspheres Hk passing through side-spaces of K

and containing K, all the zeros of F(q) must lie in the region

common to all the possible Di' that is, in the polyspace K. 3.2. Jensen [6] pp 16,17 announced in 1913 a theorem, that everynon-realzero of the derivative of a real polynomial fez)

lies in or on at least one of the Jensen circles of f(z).

To the point qi with co-ordinates wi ,xi,yi,zi' in R corresponds

the quarternion wi +xii+Yd + Zik. Besides this point qi we shallconsider seven other points with co-ordinates(wi' -xi'Yi'

zd • (wi xi' -Yi' zd, (wi' Xi Yi' -zd, (wt • -Xi' -Yi, ziJ, (Wi,

-Xi' Yi. -zd, (Wi' Xi, -Yi, -Zi) and (VIt, -Xi, -Yi, -zt}.

Now we are interested in the problem, which is the location of the zeros of:

p s 1: 1 _ _ _ _ --=1 _ _ _ _ + 1: __ 1 _ where q-(wi~xii:!:y d:!:Zik) P+ I q- ri ' r i(i=p+1, p+2, ... , s) is real,

in relation to the location of the points (wi,:!:"ï':!:Yi,~zi) (i=l,

2 ... p) and ri (i=p+1,'p+2, ... , s).

Definition. The Jensen surfaces ofthepoints (Wi,~Xi'~Yi,~Zi)

are:

iJx : (w-wd2 + x2_x1 .2 = 0

(3,2,1)

_J

_{y : (w-w i )2 + Y -Yi2 = 0}

Remark. In R3 these Jensen surfaces are cylinders. By defi-nition a point lies within (outside) such a Jensen surface i JJ."

if (w-wi)2+J.'2-J.'r <0, {(w-wi)2+tl2_tl2i>0}.

Theorem (3,2,1). If Q is a zero of the function

(3,2,2)

where the 8p quaternions wi

!

xii 2:.Yij 2:. Zik (i=1, 2, ... , p) are not re al, and the s numbers ri are real, then holds for each. set iJ tl (i=1, 2" ..• , p), that Q is not lying outside all the p Jensen surfaces (3,2,1).

Proof. We shall prove theorem (3,2,1) by a reasoning ana-logous to the proof of Jensen's Theorem.

(3,2,3)

+ (w-w i)- (x+xdi - (y+Ydj- (z+z i)k (w-w i)2+(xtxi)2+(Y+Yd2+(z+zi)2

(w-w. )- (x+x. )i- (y-Yi )j- (z-zi)k (w-w i)- (x-xi)i- (Y+Yi)j- (z+zi)k

+ 1 1 + _ _ -=-:::~_.::....--::-_-=---=-_-=--_

(w-w )2+(x+x i)2+(y_Ji )2+(z-zi)2 (w-Wi)2 +(x-x i) 2+ (y+Yi )2+(Z+Zi )2

+ (w-w

t

(X-Xi )i- (Y:-Yi )j-(Z+Zi )k + (w-w i)- (x+xdi- (Y+Ydj- (z-zdk ] (w-wi )2+(x-xt ) 2+(y_y i) 2+(Z+Zi ) 2 (w-w i ) 2 + (x+xt )2+(y+y i)2 +(Z- zJ2 The denominators of the fractions in ~he expressiOI~ [ ] of (3,2,4) shall be assigned furthermore by1N_{1 , . , lN.s} respective-ly. Now we consider the coefficients of i, j and k.

- (x - xi) [iNZ iNa __ -. iN a + iN 1 _iNZ ~ a iN s- - iN a +

-

.t

i

N

1 __ iN4iN6iN7iNa+iNl ••• i:N6 i_{N a]} (3. 2. 5 (3.2.6)=-x [A i +B i

1

+xi[Ai-Bi ].

whère Ai and B~ are the expressions. by which -(x-xi) and -(x+Jq) -ärè mulhplied respectively_ The coefficient Ai+Bi of -x is

.

>

0_ Now we investigate first Ac Bi:

i i iN iN ._{iN ) iN iN iN} iNa(iNz_iN)

:: N_z N_{4 " - -} a ( a - I + 1 a s" - .'4

(3. 2. 7)

iN iN iN iN (iN iN) iN iN ,i..~ (iN iN )

+ 1- - - 4 6 7 a- s + 1 - - - 5 -.N a 6- 7

(3.2.8) = 4xxi [i Nz iN4 __ iN8+ iNI iNa iNs __ lN8+iN1" _iN4 iN lN7

i i · ]

+ NI -"" N_s iN_a

On adding -xl A-+B i

ï

to x i[Ai-Bt) and assigning the ex-pression [ ) of

(3,

_{2. 8) by C i •}

we

may write (3,2.6) now in

the foUowing manner:

(3.2.9) -x

[~iNliNziNa

_{iN4 iNs iN /N7 - 4xt}

z

Ci] ( 3 . 2 . 1 0 ) : : .-x [ i~ iN

4" " _ iNa (iNI +iN a-4xiz)

+

iN

(w-w_i) 2t-x ~Xi 2 = 0 is t.l}e equation of a quadratical surface V ; in R4. On investigating the coefficients of j and k we meet in their coefficients the expressions (w-wd2+y2_y 2i and (w-w_{i )} 2

+z2_Zj2 respectively. .

So the term 1/ (w+xi+yj+zk-wl) corresponding to the real point

w_{1 '} has the imaginary parts:

x y z

-(W-Wl)2+x2+y 2+z 2 ,- (w-wl )2+x2+y2+z2 ,- (w-wl )2+x 2+y2+z2.

1 1

Thus sg I [ ( . . k) + ( . + . . . ] q- wi +Xil+YiJ+zi q- WcXil+Yd+zik )

= -

sg x, etr. "for every point q outside all the Jensen surface s and sg I ( - - ) = - sg x etc., for everypoint q. In otherw'ords,

q-wl

outside all the Jensen surfaces:

sg I [ r/I (q) ] = -sg x, sg J [rjJ (q)]= -sgy, sg K [ rjJ (q) ] =-sg z. In particular for each of the three sets of Jensen surfaceswe see that, if qis a non-real point outsJde all the Jensen surfaces,

rjJ (q)

+

0, a re sult which proves the theorem.

Remark. Incidentally we see that a non-real common point of a Jx,a Jyand a Jz cannot be a zero of r/I (q)if itlies outside all the other Jensen surfaces.

Example of a function rjJ(q) which has at least one zero.

We choose four points: (0,2,2,0), (0,2, -2, 0), (0, -2, 2, 0) (0, -2, -2,

Ol.

The equations of their Jensen surfaces are:

w 2 + x 2 - 4 = 0,

v?

+ y2_ 4 = 0, w 2 + z 2 - 4 = O.

The point (0,1,1,0) is lying in these Jensensurfaces. Now we determine two points with co-ordinates (a, 0, 0, 0) and (-a, 0,

o,o), so that

1 1 1 1 1 t

i+j-2i-2j + i+j-2i+2j + i+j+2i-2j + i+j+2i+2j + i+j-a + i+j+a = O.

After reduction we find a_l =

\[13,

a2= - \(13. Conversely q=i+j is a zero of the function

1 1 1 1 1 1

VI (q) =-q---2-i--2t q-2i+2t q+2i-2t q+2i+2t q- '{f3+ q+ \fî3·

two sections it is clear that the Theorems (3,1,1) and (3,2,1) are essential results regarding the zeros of the function

n

F(q) = 1: md (q-CU), 1

and that these results are valideven when the positive numbers m i are not integers. This expression however is only a special

case of the linear combination (3,3,1)

n

F(q) = 1: mifi (q) 1

where [

]-1

fi(q) = (q-ail) ...• (q-a_{ip )} (q-bn)···(q-bir ) and where the mi are quaternions such that

(3, 3, 2) 0_{- arg mi -}

<

<

'V _{, < "2'} lr _1-'-1 2 _{, , • • • , n.}

We ask now whether or not Theorem (3,1,1)maybe generalized to functions F(q) of type (3,3,1).

If K is a convex region in R 4 and ~ a point outside K then we may consider ~ as the top of a hypercone enveloping K. The descriptive lines of this hypercone two by two enclose angles. By definition the largest of these angles is the angle subteI).ded by K at the point ~ .

We shall first prove

Lemma (3,3,1). If each quaternion qi,(i=1, 2, .•. , n) has the properties: qi

+-

0 and

(3,3,3)

then their suro

o

~ arg qi

<

11/2, i=1, 2, ... , n, n

1: qi cannot vanish. 1

Proof. Theqi are vectors drawn from the origin to points in the half-space R3' 9l (q»O. From (3,3, 3,)it follows that the resultant vector cannot vanish. In other words 1: qi f O. Theorem (3,3,1) If K is a convex region which enclöses the zerosa_ik and poles bik of eachfi(q) of eq (3,3,1), then F(q)f 0 at any point ~ at which K subtends an angle less than rp

=

(f- "(

)/(p+r).

Proof. Since ~ is necessarily exterior to K, we may find in K two points Cl and

IJ

such that

arg (~ - f3 ) (~ - a ) -1 <

,p.

Moreover we suppose that a and f3 are chosen such that for given ~, arg (~ - f3) (~ - a ) -1 is maxima!.

Further is for all i and k

(3,3,4) arg (Tik <

,p ,

arg Tik <

,p ,

Let us now set

(3,3,5)

In section 1,3 has been proved that arg q1q2 ~ arg q1 +arg q 2 Hence

(3,3,6) arg

~i ~

arg mi+arg [(q-a_il) . . . (q-a ip )_1_ ..• _1_ .

q-bir q-~

.(~-f3)~

1

'J

(~-a)p .

Since by definition cos arg(q) = w / N(q)t, the argument of the product of some quaternions is invariant to commuting of the factors. Hence

(3,3,7) arg

~i ~

arg mi + arg [q-au . . .

~J

t -

a q-b_ir

On use of eqs (3,3, 2) and (3,3,4) it follows from (3,3, 7) that

(3,3,8)

0':::

arg ~i

<

'Y+ (p+r) , = t1/'.

FromLemma (3,3,1), eqs (3, 3, 1) and (3,3,5) wemayconclude

n

F(O[

(~-f3)r(~-arpJ =1: ~i=l=o,

1

as required in Theorem (3,3,1).

Remark. If we define 21/' to be the angle subtended by K at a point interior to K, we may say that the zeros of F(q) !ie in the region S(K, t/J ) comprised of all points at which K subtends an angle of at least t/J. For example, if K is a hypersphere of radius r, then S(K, t/J) is a concentric hypersphere of radius r csc(

,/2).

Definition. The region K* is star- shaped with respect to K, when the entire line-segment PQ lies in

K*.

if Pis any point af K and Q is any point of K . The region SCK, 1/» is star-shaped with respect to K. The fact is obvious when Q is also a point of K. Let now P be a point of K and let Q be a point of

S(K,~ )-K. Is '" thetop-angle corresponding to Q, then I/>'.~

'"

< 7f. Q' on PQ lies in the projecting cone of K with top Q.

Hence fr om Q 1 an angle '" I> '" subtends. Consequently I/>

<

ril';

in other words Q 1 lies in the region S(K, I/> ).

Theorem (3,3,2). If all the zeros and poles of each rational function ftCz) entering in eq (3,3,1) lie in a closed convex region K and if the mi(i=1, 2, .• , n) are constants satisfying ineq (3,3,2), then all the zeros of the linear combination F(q) = 1; mifi(q) lie in S(K,·t), a region which is star-shaped with respect to K and which consists of all rOints from which K subtends an angle of at least I/> =

(t

71 - y) (p+r).

We remark that in Theorem (3,3,2) the region SCK, 1/» cannot be replaced by a smaller region.

Proof. If _{s is any point in SCK, ,), two points t l and t 2 may be}

found in K such that L tlst2 = , = (71 /2-y )/p+r). There is a transformation which transforms s into the origin and t l into a point of the real-axis. Then

(3,3,9) _{arg t 2 =} I/> = (7I/2-y)/ (p+r). Further we define

(3,3,10)(a) and

(3,3,10)(b) k₂= (s-t2)/d2 ] p+r (cos(Y+"./2)+esin(Y+1I/2) ] where d_l=

I

s-tlland d₂=

I

s-t₂

1;

cos (Y+7I/2) + esin (Y+1I/2)

is a unit _{vector lying in the plane through st l and st2}• (t₂ ·has a e-vector which is equal to e or -el. Now is

(3,3,11) argk 2 = (p+r) ('1r/2- Y)/ (p+r)+y+ '1r/2 = 7(.

Since I k l I

=

I k₂1 =1, arg k ₁ = 0, the vectors k l and k₂ are equal and opposite and thus

(3,3,12) k_l + k 2 = 0, or

(3,3,13) (s-t_l)/ dl] p+r + (s-t₂/d

21

pH [cos( Y + 11' /2) +

as k 1 is real we may deduce (3,3,13) in the fdllowing way

Now

cos ('Y+7r/2)+esin ('Y+"'/2)=[cOS ('Y/2+"./4)+esin ('Y/2+1(/4) J~ Hence (3,3,14) may be written

(3,3,15) d 2 r

(s-t 1)P cos [('Y/2+7r/4)-e sin ('Y/2+7r/4)] + d₁P(s-t₂)r

This means that the function d r

G(q)

=2

_{(q-t 1)P(q-t 2}

r

r [cos ("1/2+

"./4)~e

sin ('Y/2+".14)]+

dl

d r

+_1 (q-t 2)P(q-t 1)-r [cos (1'/2+"'/4) + e sin ('Y/2+7r/4)]

d P

2 .

has a zero at the point s. In other words, every point s of S

(K,i/) ) is a zero of at least one function F(q) of type (3, 3, 1). Finally we make an application of theorem (3,3,2). If r = 0, then

= f(q)-(m 1 +m2+'" +m n

r

1 [m1h 1(q)+m2 h 2(q)+ ... +mnhn(q)] The a ip are clearly the points at which f(q) = ~ (q) and the zeros of F(q) are the points where

n n

f(q) = (1: mir1 1: mihi (q) 1 1

Hence we have

Theorem (3,3,3). Let f(q) and hi(q), (i=l, 2, .. , n) be functions such that f(q)-hi(q)

=

(q-an)(q-ai2) .. (q-aip)' i=l, 2, .. , n. and mdi=l, 2 •.. , n) quaternions satisfying ineq (3,3,2). If all the points aij(i=l, 2, .. n; j=l, 2, .. p) lie in a convex region K, all

the points at which

lie in the star-shaped region S

[K,e

1r/2-'Y )/p ] .

3,4. In (6) pp 25,26 attentionis paid to the locationofthe zeros

of the polynomial solutions of the generalized Lamé differential equation (3,4,1) d 2 P d

q,

(z)

--.!!..

+ (l: ~)~+----w=o. dz 2 1 z-a. dz P 1 1r(z-a.) I 1 n

If S(z) = 1r (z-zd is a Stieltjes polynomial then every zero zk

of S(z) isleither a point ai or a solution of the system P

(3,4, 2) l:

i=l

k=l,2, ... ,n.

Likewise a zero h of the Van Vleck polynomial V(z) corre-sponding to S(z),is either a zero of S'(z)=n(z-zl') .... (z-z 'n-I)' or a solution of P a · n-1 1 _ ~ __ 1_+ ~ 0 ~ t a~

t - , - ·

i=1 k - i i=l k zi (3,4,3)

We ask now the question as to the location of the zeros (if any) of

(3,4,4)

where k=l, 2, .. , n; al' a 2

nions, r is real and a p' ql' q 2' ~ .. ,

Cln

are

quater-(3, 4, 5 )

p 1 P r'

1: a .- - - + 1:

=

0

where q 1 I, q21' • • • • q

t

are solutions of

p n

1: _1_ + 1:

q~qk

= 0, i=l q- ai \ k=l

(3,4,6)

in relation to the location of ai(i=l, 2, .. , pl·

Th~ z~ros _{qk and t k have a physical interpretation. The term}

a i(aCqk)-l in the conjugate of (3,4,4) may be regarded as the force upon a unit mass at the variable point qk ~u~ to the mass·

a ) r situated at the fixed point ai' The term (qi -qkt1 _{may be}

regarded as the force upon the unit mass at q k. due to the unit mass at the variable point qi' The system (3,4,4) defines the qk as the points of equilibrium of n movable masses in a field due to p fixed particles ai of mass a dr.

So eq (3,4,5) defines the tk as the points of equilibrium of a movable unit mass in a field due top fixed particles ai ofmass ai

Ir'

and to N fixed particles qi' of unit mass.

Theorem (3,4,1). If

(3,4,7) arg ai~ 'Y<'lr12, i= l,2, ... ,p

and if all the points ai !ie in a hypersphere H of radius r, then the roots of (3, 4, 4) and (3, 4, 5) !ie in the concentric hype r-sphere Hl of radius r l = r sec 'Y.

Proof. We may consider a point of Hl as top of a projecting hypercone of H. The top-angle of such a cone is 'Ir - 2 'Y. Let

us suppose that (3,4,4) has some roots outside Hl and tha-t among these the one farthest from the center of H is q •

Through ql we draw the hypersphere

H*

concentric with H arld we consider the three-dimensional linear tangent space to H* at q1' By the assumption concerning qi all the poin!.s Cl· !ie in or on the hypersphere H*and hence the quantities (qi-q11)-1 are represented byvectors drawnfrom ql to points on the side

of T con~irgng the hypersphere H. Furthermore, since the

quantity (ai-ql) -1 maybe represented by a vector drawn from q 1 and lying in a projecling hypercone of H with top q l' due to

(3,4, 7), the quantity ai (ai-q 1J-1maybe represented by a vector drawnfrom q, and lying in a hypercone with top angle

<

'Ir -2'Y

+'Y+'Y= 'Ir. In short both types terms (qi-qd-1 and.ai(ai-ql)-l

are representable by vectors drawn fr om q1 to points on the

same side of T. This means according to Lemma (3,3,1) that

the left side of eq (3, 4, 4) cannot vanish. Since this result con-tradicts eq (3,4,4), our conclusion is that the point q1 and consequently all qi must !ie in H!

zeros of (3,4,5). Since we know that all the zeros qk(k=l, 2, ... , n) of (3, 4,4) He in hypersphere Hl, we .may infer from Theorem (3,1,2) that all the zeros qil _{of (3,4,6) also He in} Hl. The prooI' is similar to that of the first part.

We consider now the case, that a i(i=l, 2, .. , n) and r a r e positive. As in the case of Theorem (3,2,1) we shaU consider eight quaternions

i=l, 2, .• ,p. We add to eachofthem3pellipticsurfacesdefined by (3,4,8) (w-a )2 ;cVrn(a) : Q ma_I2 yVrn(a) : (w-ao )2 nla 2 2 zVrJa) : (w-ao )2 ma32 = 1 = 1, m=I, 2, .. , p.

Apart from m=1 the surfaces (3,4,8) are clearly elliptic Jensen surfaces. The curves of intersection of these surfaces and a R 2 parallel to the wx-, wy-and wz-planerespectively are eliipses. We consider now the circles whose diameters parallel to the w-axis are chords of these eliipses. Let these sets of circles successively be

(3,4, 9) f(w, x, p)=o; g(w, y, q)=o; h(w, z, r)=o,

then we define next to (3,4,9) the set of cylinders, also given by (3,4,9).

Lemma (3,4, 2a). The elliptic surfaces (w-ao)2 Ima l2+x

21

a l2 = 1

is the envelope of the set of cylinders f{w, x, p) = 0, corre-sponding to the elliptic surface (w- a o )2

1

(m -1 lal 2 +x2 / al 2:::; 1. Lem.ma (3,4, 2b). If the non real ~ero.~J of system (~, 4, 4) He.s outslde the Jensensurfaces 1-1 VI ( a), 1-1,2, .. ,p, 1-1 -x,y, z, lt lies inside at least one Jensen surface IJ. VI (qi)' 1-1 =x, y, z ;

2:S

i

Sm.

Proof. For k:::;l (3,4,4) becomes

p ai r

~ + =

m r n r

+ 1: + t = 0,

2

q1-(wi~xii!Yij~zik)

m+1 q1 -

~

qi(i=m+1 ••.•.•• n) is rea!.

Except. for the sum of seven terms r [ q1 - (-v 1 ±Xl i:ty 1j±Zl k) J -1 eq (3,4.11) has the form of '" (q) = 0 (3,2,2). If q 1 were also outside the Jensen surfaces of the points qj> 2~i~m, then we couldapplythe reasoningused toproveTheorem (3,2,1). Thus for all terms in (3,1.11). except possibly r [ql -(w1±x~i:tyd

±Zlk)]"l. the sign of the i-part would be that of sg(-X 1) etc. But since r [q1-{~ _{±x 1i:ty1 j±zl k )]-1}

r r r r r r

= - - +. + + - - + - - + - - - +

2x1i 2x₁i+2Ylj 2x1+2Ylj+2z1k 2Y1j 2z1k 2Ylj+2z1k

r

the sign of the i-part of all terms would be that of sg(-xd etc. That is, if ql were outside the Jensen surfaces for all ai (1.$ i ~p), and all the qd2~iS.m) then it would not satisfy eq (3,4,11). Hence ql lies within at least one surface VI (qi) of each set.

Theorem (3,4,2). If the coefficients ai and r of (3.4,4) are positive real numbers and if the system has a 8mnon-real zeros. then there are at least three Jensen surfaces J,tVm(a), J,t =x, y, z containing a zero.

To prove this theorem let us ~ssume that point q1 is exterior to all the Jensen surfaces ,Fl (1 a), i=l, 2, .. , p; J,t =x, y, z . By Lemma (3, 4, 2b) point q is interior to, say x V (q ). If then q2 is also exterior to

ad

the Jensen surfaces xV1{fa), it lies interior to, say x VI (q 3)' and so on. Eventually we must come to a value of k, k S m. such that a~though the point q k-1 lies exterior to all the surfaces xVI (1 a), i=l, 2, ..• p, and thus interior to the surface x VI (qk)' the point qk lies interior to at least one surface xV dia), say x VI (la). If there was no k with the above mentioned property the last zero would lie outside allJensen surfaces of the x-set. Now applying Lemma (3.4. 2a" we see that xVI (qk) lies in x V2 (1 a). that therefore x VI (q kol ) liesin xV3 (la) etc, finally that xVI (q2) lies in xVk(la). Since.

There are systems of the type (3,4,4) whichhave solutions. We give two example.s.

(3,4,1) Let us choose p points al' a2' .. ,

a"

and n points ql' q , .•• q in _{2 n} R4' Then (3,4,12 1 1 1 r r 0'1--+0'2--+ +0',, _ _ + _ _

+

+--=0 ql- a l ql- a 2 ... _{ql-ap 'l1- q 2 .... qfqn} ·1 1 1 r r 0'1--+ 0'2-- + •..•• +O'p _ _ + _ _ + _ _ + ..••. CJ._{2-a 1} q2- a 2 'q2- ap q2- q l q2- q 3 r - - = 0 ~-qn 1 1 1 r .r 0 ' 1 - - +(l2--+ _{••• •} +O'p _ _ + _ _ + _{a ••} =0 qn-al qn-a_{2 qn~ p q n- q1 qn- q n-l}

where r is a given rea! number, is a system of k equations in unknowns 0'1' 0'2' •.••• apo Such systems of linear unilateral equations, the coefficients of which are quaternions, are studied extensively by Study (8) pp 62,71 and Wachs (10), chapter 2.

2. We take al=(O. O. 0.0); a2 =(2, O. O. 0); a3=(1. ±1, ±l,O) ql = (1.0.±y), 0'1= a_{2=r=l, 0'3 =} a.

Then we have

1 1 a a 0' Cl l '

. - - + - + + + + - --1---=0

l+yj -l+yj -i+(y-1)j -i+(y+1)j i+(y-1)ji+(y+1)j2yj (3,4,13)

1 1 a a Cl 0' 1

- 1· _{-y - -YJ-l+ -y-}j+-1-'~' ( 1)*' _J-l+\;,yL +1)j+' _{, l+\~y-}L 1\'+'+( +1)_{IJ 1 -y J - YJ} '+-2 ,==0

or

-2yj -2a(y-1)j -2a(y+1)j 1

- - + + - - - =0

1+y2 l+(y-l) 2 1+(y+1) 2 2yj (3,4,14)

2yj -20'(-y-1)j -2a(-y+l)j 1

- - + I + - - =0

1+y2 1+(-y-1)2 1+(-y+1)2 2yj Then we have

2 Cl =

_(5y 2+1) 2y(1+y2)

[(y-l)2+1]

x

[(y+l) ~1] 2<T +2y 2_2).

Now we choose ay such that Cl ispositive. Forinstance take y =

t.

Then we have constructed a system of type (3,4,4) with rea! positive coefficients.

CHAPTER IV

4,1. Inconnectionwithapaperby Study (see [8] pp 56,57 and 60) we make the following remark. The continuum of the real points (xo' •••• , x n-1 ) of an n dimensional euclidian space Rn

can be closed by introducing an improper point. By means of the so called "stereographic projection", the points of the closed euclidian space are mapped one-to-one, contin~ously

and conformally onto the real branch of a sp~erical M_n of a Rn+1 (or in general onto the real branch of a M_n of the inertion index 1). Spherical surfaces of all possible dimensions in the Möbius continuum (to which the pl~ne also belongs) correspond to spherical surfaces on the M , or more general, plane intersections of the M~ corresponJto spherical surfaces in the Möbius continuurn. Generalization of the Möbius transforma-tions however causes difficulties, on account of complicate singularitieswhich areproduced bydivisors of zero in the set of the Clifford numbers. We need those numbers generalizing the Möbius transformations. Anexceptionhereuponis the case n=4. Infact, it is possible in the real Möbius continuum offour dimensions to describe the 2.0015 conformal transformations by re al quaternions in such a manner that each transformation corresponds to a formula which is analogous to

So far Study. az + b

----z'

_{c.z + d} and

12: + m - - - z' _{nz +r .}

In the previous chapter wesaw that a hypersphere H enclosing the points qi(i=l, 2, .. , p)also containsthe zeros off(q)=

Pf

md (q-qi)' ~ positive. We transform the points CJi by a non sin-gular linear transformation

(4,1,1) q = (aQ+b )(cQ+dr1 or q = (Q'Y+6 j1 (QCl'+f3). with and

'iJ

={

Cl'

'Y}+

0,

r f3 6 where, by definition,

'iJ

=

{~~} =

aadd- acdb- bdca+bbcc, see [8 J

which implies that the closed interior of a hypersphere H can be mapped conformally upon the closed exterior of a

hyper-sphere

H*.

We ask now the following problem: when r*(q) is the transform of f(q) by (4,1,1), are the zeros of r'(q) the