, and
- 3
Finally, multiply (B) by o>, (C) by o> 2 , and add to (A) ; it will be found tha t
EXERCISE VIII COMPLEX ROOTS
1. Prove that the values of ^ - 1 are ~^(1 t).
v^
2. Find the values of ^(1 + 1) and *J(l-i).
3. Use Art. 19 to show that 1),
) (#- 2x cos^-H l).
4. Prove that
=a; 2 (a; '
where a, j3 are the roots of z* + z- 1=0.]
COMPLEX ROOTS 67
5. Give a geometrical construction to find the points z l9 z 2 corresponding to the values of *Jz.
[If 01 is the unit of length, Oz l bisects the angle XOz, and is a mean proporti onal
to 01 and Oz ; and 2 2 is on z-f) produced, so that Oz a =2 1 O.]
6. If a, b are complex numbers, show that
[Let 2 1 =a + */a a -6 2 , z 2 a-*Ja*-b 2 , then by Exercise VII, Ex. 10, I *i I 2 + I *$J 2 =i I *!+** ! 2 + i I *i -*2 I 2 =2 I a )t + 2 | a*-6> | ;
. | a -6 | -f | a -6 | 2 }
= {|a + 6| + |a-6|}M 7. Solve the equation f + 2(l+2i)z -(11 + 20=0.
Verify that the sum of the roots is -2(1 + 20 and the product - (11 + 20.
[Put z~x + iy, equate real and imaginary parts to zero and solve for x, y.]
8. Prove that, with regard to the quadratic z 2 + (p + ip')z
(i) if the equation has one real root, then
(ii) if the equation has two equal roots, then p*-p'*=:4q and pp'ty'*
[(i) If a is a real root, by the rule of equality a 2 Eliminate a.
(ii) In this case (p + ip')*=4:(q + i,q'), etc.]
9. If z ~x + ly =r (cos 6 + 1 sin 0), prove that Jz=z--={'Jr + x + i*/r-x} or {Vr -f x - i\lr - x} 9 according as y is positive or negative.
[\/z */r ( cos - -f l sin - J and cos 6 = - ;
Also if y>Q then 0<^<?r, and if y<0 we have -7r<8<0 9 etc.]
the roots of z n = (z + l) n , and show that the points which represent them are collinear.
[The roots are -J( l + i cot J, where r=0, 1, 2, ... n-1. The corresponding points lie on the line x + i=0.]
68 COMPLEX FACTORS
Show that the roots of (1-f 2) n = (l -z) n are the values of i tan-^, where r = 0, 1, 2, ... n - \ , but omitting n/2 if n is even. n
12. Prove that
(i) * 2W -2x n cos 194- 1 = /Trio" 1 (* 2 - 2x cos
(ii) * n + ar* - 2 cos = /T^I JJ " * (a? + ar * - 2
cos
(iii) cos rz< - cos nS = 2 n - 1 77^o ~ * cos < - cos + ~
[(i) x m - 2# n cos 6 + 1 ~~ {x n - (cos 4 t sin 0)} {z n - (cos - 1 sin 0)}. No w use
Art. 15. (iii) Put x =cos <f> + 1 sin ^, and for put n0.]
13. If n is odd and not a multiple of 3, prove that x(x + l)(x* + x + 1) is a factor of (x + l) n -x n -l.
[Put x=Q, - 1, co, where a; is an imaginary cube root of unity.]
14. If ( 1 + x + x 2 ) n = a -f a^ -f a 2 a: 2 + . . . -f 2 n^ 2n > prove that a Q + a 3 + a 6 +
15. If u
v = x + y + z + a(z
w=x + y + z + a(x + y-2z),
prove thUt 21 a 2 (x 3 4- y 3 -f z 3 - Sa^z) = w 3 -f v 3 -f- 1^ 3 - Suvw.
(a + w')- 1 -f (6 -f co')- 1 + (c + co')" 1 + (d + w')- 1 - 2ft/- 1 , where a> and w r are the imaginary cube roots of unity, prove that
[Consider the equation (a -f a)"" 1 -f (6 4- x)~ l -f (c 4- z)- 1 4- (d 4 x)- 1
= 2ar 1 .]
[IT] If 2^ 4- 2 2 2 4- z 3 2 - 2; 2 2;3 - Zfa - 2^2 = 0, prove that
[For x 4- a>2:2 *- ^^s = 0, /. Z B - z t =
18. If o>=J( - 1 4-t\/3), being an imaginary cube root of unity, and a, 6, c are real, then </( + oA+ <o 2 c)= J{>
V(a 4- co 2 6 + we) = where D = s /(
provided that 6>c. If 6<c, the sign of A must be changed.
[Use Ex. 9. For a positive number x 9 >Jx denotes the positive square root.]
PRODUCTS AND QUOTIENTS
69
21. Points representing the Product and Quotient of Two
Given Numbers. Let z = r(cos0 + *sin0), z' = r' (cos 0' + 1 sin 0').
(1) Construction for the point representing the product zz'.
L^t 1 be the point on OX which represents unity, so that 01 =1. Draw the triangle Oz'P directly similar to the triangle Olz.
Then P represents the product zz' . For by similar triangles
OP Oz . OP r ,
-, that IB, r-, ..OP = rr ;
z
i
FIG. 9.
also Lz'OP^ LlOz = 0, :. LlOP =
But zz' = rr f {cos (0 + 0') -f i sin (0 + 0')}.
Therefore P represents zz'.
(2) Construction for the point representing the quotient z/z'.
Draw the triangle OzQ directly similar to the triangle Oz'l .
Then Q represents the quotient z/z'.
For, by the last construction, (number represented by Q) .z' z.
(3) // k is constant, and z varies so that z-a
i
FIG. 10.
z-a
"k,
then the point z describes a circle of which a, a! are inverse points; unless k~I, in which case z describes the perpendicular bisector of aa f .
FIG. 11.
For let the bisectors of L aza' meet aa r
at d, d f . Then, by Art. 13, (2), az : a'z = k : 1.
Therefore ri, d' divide aa' internally and externally in the ratio k : 1, and are fixed points.
Therefore z describes the circle on dd' as diameter. >
70
DISPLACEMENTS AND VECTORS
Also if c is the mid-point of dd f , we have ca.ca' = cd 2 ; hence a, a' are inverse points with regard to
the circle.
If & = 1, then az = a'z\ and z lies on the perpendicular bisector of aa'.
Conversely, if the point z describes a circle of which a, a' are inverse points, Z ~Q>
then , K. FIG. 11.
z-a
For, since ca . ca' =cd 2 , then dd' is divided internally and externally in the same ratio, k to 1, say.
Hence, az : a'z = k : 1 .*
(4) If z varies so that
am
z-a
where <f> is a constant angle, then the point z describes an arc of a segment of a
circle on aa', containing an angle <f>.
\a
FIG. 12.
FIG. 13.
For by Art. 13, La'za = <f>. The sign of <f> determines the side of aa' on which the segment lies. Thus <f> is positive in Fig. 12 and negative in Fig. 13.
22. Displacements and Vectors.
(1) In connection with the geometrical representation of complex
numbers, we introduce the notions of displacement and directed length or vector.
(2) Displacements in a given Plane. Let P, Q be two points in the plane OXY. The change of position which a point undergoes in moving from P to Q is called the displacement PQ.
* See Elements of Geometry, Barnard and Child, p. 316 : ' Circle of Apollonius.'
ADDITION OP VECTORS 71
If any straight line P'Q' is drawn equal to, parallel to, and in the same sense as PQ, the displacements PQ, P'Q' are
said to be equal.
To specify completely a displacement PQ we must know :
(i) its magnitude, i.e. the length PQ ; (ii) its direction ;
(iii) its sense, denoted by the order of - the letters, and if necessary by an
arrow.
FIG. 14.
If we draw OL equal and parallel to PQ and in the same sense, we say that L XOL is the angle which PQ makes with OX.
This angle determines the direction and sense of
(3) Vectors. An expression (such as PQ) used to denote a line-segment
with reference to its length, direction and sense, the actual position of the line being indifferent, is called a vector.
Quantities which can be represented by lines used in this way are called vector quantities. Velocities and accelerations are vector quantities.
A force can be represented by a vector ' localised ' to lie in the line of action of the force.
Quantities (such as mass) which do not involve the idea of direction are called scalar.
(4) Connection with Complex Number. If z=x + t,y and P is the point (x, y), a one-to-one correspondence exists between the number z and any of the following : (i) the point P ; (ii) the displacement OP ; (iii) the vector (or directed length) OP.
Any one of these three things may therefore be said to represent z, or to be represented by z. % *
23. Addition of Displacements and of Vectors.
(1) Let P, Q 9 R be any three points. If a point moves from P to Q and then from
Q to R, the resulting change of position is the same as if it had moved directly from P
to R. We therefore define the addition of Flo 15 displacements as follows :
* Some writers use an underline instead of an overline.
72
ZERO AND NEGATIVE VECTORS
The result of adding QR to PQ is defined to be PR\ and this is
expressed by writing PQ+QR=PR ... (A) This equation is also taken as defining the addition of vectors.
If PQ, QR, RS are any three displacements or vectors (Fig. 17),
Q
FIG. 16.
FIG. 17.
(2) The Commutative and Associative Laws hold for the addition of dis-
placements and vectors.
(i) In Fig. 16, complete the parallelogram PQRS\ then by Arts. 23, (1), and 22,
Hence the commutative law holds, and PR is called the sum of PQ and QR.
(ii) In Fig. 17,
therefore
and the associative law holds.
NOTE. In Fig. 16, PQ + PS = PR t a fact which is expressed by saying that dis- placements and vectors are added by the parallelogram law.
24. Zero and Negative Displacements and Vectors. If after
two or more displacements the moving point returns to its initial
position, we say that the resulting displacement is zero. Thus we write
This equation is also written in the form -PQ = QP, which defines the meaning of - PQ.
These equations are also taken as defining the meaning of zero and negative vectors.
DISTRIBUTIVE LAW 73
25. Subtraction. For displacements and vectors the meaning of PQ-QR is defined by
PQ-QR - PQ + (-QR) = PQ + RQ-
Thus OP-OQ -OP + QO = QO + OP = QP.
26. Definition of Multiplication by a Real Number. To
multiply a displacement or a vector PQ by a positive number k is to multiply its length by A, leaving its direction unaltered. The resulting displacement or vector is denoted by JcPQ or by PQ . k.
Further, we define ( - k)PQ by the equation
In particular, (
So that to multiply a vector by ( - 1) is to turn it through two right angles.
27. The Distributive Law. We shall prove that if k is a real
number , then
(i) Let k be positive. Along PQ set off PQ' = kPQ. Draw Q'R' parallel to QR to meet PR in R'. By similar triangles,
and Q'R' = kQR\
and kPQ + kQR^PQ'
Hence the theorem holds for positive numbers.
(ii) For a negative number ( -k), we have
IG *
and (
hence the theorem holds for negative numbers.
Thus the distributive law holds for the multiplication of displacements and vectors by real numbers.
NOTE. The diagram of Fig. 18 is drawn for a value of k greater than unity : the student should see that the same result follows from a diagram in which k is les s than
unity.
74 VECTORS AND COMPLEX NUMBERS
28. Complex Numbers represented by Vectors. It will now
be seen that, so far as addition, subtraction and multiplication by real numbers are concerned, complex numbers are subject to the same laws as the vectors which represent them. This fact is fundamental in theory and very useful in practice.
It should be noticed that if a number z is represented by a vector AB,
then | z | is the length AB and am z is the angle which the directed line AB makes with the directed line Ox.
Theorem 1. If C divides AB in the ratio n : m and is any point, then (m + n)OC - mOA + nOB.
For mOC^mOA+mAC,
Also mAC = nCB = - nBC ;
O
whence the result follows by addition. FIG. 19.
Theorem 2. If z is the point which divides the straight line joining z v z 2 in the ratio n:m, then the corresponding numbers are connected by the
relation
This follows from Theorem 1, in accordance with the principle stated in this article.
In particular, if z is the mid-point of z^ then z = %(z l + z 2 ).
Ex. 1. // a, b are complex numbers, prove geometrically that
Let A 9 B be the points which represent a, 6.
Bisect AB at C, then OA+OB=20C
and OA-OB^'BA^Z'CA.
Therefore a + 6 and a -b are represented by 200"and2CJ; hence
Now, since C is mid-point of base A B,
SYMBOLS OF OPERATION
Ex. 2. // OA, OB, OC are connected by the relation
75
? = 0, where a -f b -f c = 0, Men A, B, C are collinear.
[This is the converse of Theorem 1 of this article. We have (a + c)(W==a.OA+c.OC'; hence a . AB^c . BC.]
29. The Symbol i as an Operator.* Along two straight lines at
right angles set off, consecutively, equal lengths OP, OQ, OP', OQ' in the positive direction of rotation.
Let the symbol i applied to a vector denote the operation of turning it through a right angle in
the positive direction of rotation.
To bring our language into conformity with
that of algebra, we say that to multiply a vector by i is to turn it through a right angle in the positive sense.
Thus in Fig. 21 , OQ=i OP and OP' = i OQ.
Therefore OP' = 1(1 OP) - 1 2 OP,
where i*OP is an abbreviation for i(iOP). Hence t 2 OP - ( - 1)OP.
Thus i 2 and -1 denote the same operation, and in this sense we write Again, OQ' = iOP' = i(-l)OP and OQ' = -OQ = ( -l)iOP.
Either of these results is written in the form ( - 1) OP^OQ', so that
to multiply a vector by (- i) is to turn it through a right angle in the negativ e
sense.
Again, if i 3 . i 2 OP is taken to mean t 3 (i 2 OP), it is obvious that
FIG. 21.
Ex. 1 . //a, b are complex numbers, find numbers
z, z' so that the points z, z' and a, 6 may be opposite corners of a square.
Let c be the mid-point of a6, then Oc + icb ;
Similarly, 2'=!- (a + &)+* (a- 6). FIG. 22.
* For the moment, the reader should forget his conception of I as denoting a num ber.
B.C.A.
76 PRODUCT OF COMPLEX NUMBERS
30. The Operator cos0-MSin0. Draw two equal straight lines
OP, OP' inclined at an angle 0. Draw P'N perpendicular to OP. Along NP set off NQ equal to NP'. Then
OF = ON + NP' = ON + iNQ.
Also ON = ^ n OP - cos . OP,
/. OF - cos . OP + * sin . OP, which we write in the form OF-(cos0+ism0)OP,
and we say that to multiply a vector by cos + i sin is to turn it through the angle 0.
31. Multiplication and Division of a Vector by a Complex
Number. In accordance with Arts. 26, 29 and 30 of this chapter, we
say that to multiply a vector OP by the complex number r (cos -h i sin 0) is to multiply its length by r and turn the resulting vector through the angle 0.
Here r is the stretching factor and cos -ft sin the turning factor.
These are independent of each other, and the order in which they are applied is indifferent.
If OQ is the vector obtained by multiplying OP by the complex number z, we write
OQ^zOP arid OQ/OP^zi
we also say that the ratio of OQ to OP is the number z.
Division is the inverse of multiplication, so that if OQ = zOP y then OP is the result of dividing OQ by z.
Therefore to divide a vector OQ by z is to divide its length by r and tnr^ the resulting vector through the angle ( - 0).
The result is the same as that obtained by multiplying OQ by I/z.
32. Product of Complex Numbers.
Let OQ^z'OP and OR^zOQ, then we write
where zz' applied to OP denotes that the operators z', z are to be applied in succession, in this order.
AEGAND DIAGRAMS 77
Since the stretching and turning factors may be applied in any order, OP may be transformed into OR by multiplying its length by rr' and turn- ing the resulting vector through the angle (6 + 6').
Hence the operations denoted by
zz', z'z and rr' {cos (0 + 0') + i sin (0 + 0')}
are equivalent.
Again, if we take (cos 0-h i sin 0) n to mean that the operation
(cos + 1 sin 0)
is to be applied n times, the result is the same as that given by the operator cos n0 -f i sin n0.
In this sense then
(cos + i sin 0) n = cos n0 + i sin w#.
It will be seen that complex numbers used as operators on vectors conform to the laws of algebra.
EXERCISE IX
ARGAND DIAGRAMS : VECTORS
1. If z = 3 + 2i, z' 1 -f i, mark the points z, z' in an Argand diagram, and fin d
by geometrical construction, the points representing Z + Z', Z-Z', ZZ\ Z/z'.
2. Let z, a, b be complex numbers of which a, b are constant and z varies.
If Z is given in terms of z by one of the following equations, it is required to find
the point Z corresponding to a given point z. Explain the constructions indicate d
in the diagrams, 01 being the unit of length.
FIG. 24.
(iii) Z=
(ii) Z=tz where t is real, (iv) Z~az+b.
78 CENTROIDS
3. If Z=(l + z)/(l-z), find tho point Z corresponding to a given point z t and show that if | z | = 1, the point Z is on the y-axis.
4. If a, b are given complex numbers and J^
6^,
find the point z corresponding to any given real value of t, and prove that as i varies from
-oo to + 00, z describes the entire line which passes through a, b, the segment ah corresponding to the values from to 1 of t.
[Along the line ab set off the length az = t.ab, then Oz=0a+az~0a + t . ab; :. z=a + (b-a)t, etc.]
5. If z=a(l + it) where t is a real number, prove that as t varies z describes tho line through the point a perpendicular to Oa.
6. If c, a are given numbers, a being real, and, (i) if z = c+a(cos< + isin<),
find the point z corresponding to a given value of cf> ; (ii) if z=c + a(l + it) I (I - it),
where t is real, show that as t varies from - GO to + oo , the point z describes once the circle with centre c and radius a. [Put t = tan J<^.]
7. If A, B, C, D are any four points in a plane, then AD . BC^BD . AC + CD . AB.
Show that this theorem is an immediate consequence of the identity
8. If G is the centroid of particles of mass m 19 m z , w 3 , ... at A 19 A 29 A
G to be the centre of parallel forces, proportional to m l9 m 2 , acting at A l9 A% ... .
9. If z is the centroid of particles of mass m l9 w 2 , m s , ... at z l9 z 29 z a , ... , then
...)s m^ + m 2 z 2 + w 3 z 3 + . . . .
10. Any three coplanar and non-parallel vectors OA, OB 9 OC are connected by a relation of the form pOA +qOB + rOC = Q, where p 9 q 9 r are real numbers.
Moreover, p:q:r=& OBC : A OCA : A OAB 9
where the signs of the area# are determined by the usual rule.
Also the points A, B 9 C are collinear if p + q + rQ 9 and conversely.
[For let p:q:r= &OBC : AOCA : &OAB,
and let G be the centre of parallel forces, acting at A, B, (7, and proportional to
p 9 q, r 9 then G coincides with O, and pOA
VECTORS AND COMPLEX NUMBERS 79
11. (i) //a, j8 are non-parallel vectors andp, q, p' 9 q' are real numbers, adeq uation
of the form
involves the two equations p=p' 9 q=q'.
(ii) If a, , y are non-parallel coplanar vectors connected by the equations p<x + qfi + ry = Q and p'a + q'fi -fc- r'y = 0,
where p 9 q, r, p' 9 q', r' are real, then these are one and the same equation, that is
[These theorems follow at once from Ex. 10.]
12. Any three complex numbers z l9 z 2 , z 3 are connected by a relation of the form
where p, q, r are real numbers.
Moreover, p:q:r~ A0z 2 z 3 :
the signs of the areas being determined as usual.
Also the points z l9 z 2 , z a are collinear if p + q + r Q, and conversely, Prove this algebraically, and deduce Ex. 11.
[If z l =x l + iy l9 z 2 =z 2 + iy l9 z s =z 3 + it/ 3 , we have
whence the results follow immediately.]
13. If the points A, B 9 C are collinear, and is any point, then OA . EC + OB . CA + OC . AB^O.
[In Art. 28, Theorem 1, m : n : m + n=CB : AC : AB.]
14. Let Zj, z 2 , z a be complex numbers, no two of which are equal, then (i) If the points z l9 z 2 , z 3 are collinear,
l \ Z 2~*3\ 2|2~2l ! *8hl-2l| =-
Also, if the point z l lies between z 2 and z 3 , the ambiguous signs are both m inus.
(ii) If the above equation holds, then either z l9 z 2 , z 3 are collinear, or e lse O is
the centre of a circle which touches the sides of the triangle Z 1 z 8 z 3 . 15. If A i^4 2^4 a is an equilateral triangle, the vertices occurring in the pos itive
direction of rotation, prove that
s ~ 4- 1 sin
where o> is an imaginary cube root of unity.
Also, if z l9 z 2 , z 3 are the numbers corresponding to A l9 A t9 A B9 prove th at
3 X -f wz 2 + cj*z 3 = 0, and consequently z^ 4- z a 2 + z 3 2 - z^ - Z& - Z& = 0.
16. If A X YA'X'Y' is a regular hexagon and A 9 A' represent given complex numbers a, a' 9 then the numbers represented by X 9 X' 9 Y 9 Y' are given by
where 6 has the values ~, ~ - . 3 3
80 TRANSFORMATIONS
17. The triads of points A 9 B 9 C and X, Y, Z are the vertices of directly simi lar
triangles if the corresponding complex numbers a, b, c and x, y, z are connected by the relation
x(b - c) + y(c - a) + z(a - b) =0.
[The triangles are directly similar if = , i.e*. if - = - .]
AC XZ c-a z-x
18. If ABC is a triangle and triangles BCX, CA Y, ABZ are drawn on BC,
CA, AB, directly similar to one another, the centroids of XYZ &ndABC coincide.
x c y a z b
[By Ex. 17. i = - -- = - r. Hence show that# + v + z=a + & + c.l 1 J b-c c-a a-b y J
19. Equilateral triangles are drawn on the sides of a given triangle ABC, all outwards or all inwards. Prove that their centroids form an equilateral triangle .
[Let P, Q, R be the centroids of the triangles drawn outwards on BC, CA, AB.
Prove that _ _ I __ __ i
QA = CA (cos 30 + L sin 30), A R =AB . (cos 30 - L sin 30).
\/O *Jd
Hence show that QR \CB + ^ (CA -AB), and that RP has a similar value.
Hence show that EQ = EP (cos 60 + L sin 60), and use Ex. 15.]
' TRANSFORMATIONS '
20. If Z, z are connected by any of the relations in Ex. 2 and if z describes a given curve s, then Z will describe a curve S.
Explain the following, where a, b are given complex numbers.
(i) If Z = z -f a, S can be obtained from 8 by a translation.
(ii) If Z=tz where t is real, the curves s, S are similar and similarly situated ,
being the centre of similitude. In this case we say that S is a magnification of s.
(iii) If Z = (cos a-f tsin a) z, S can be obtained from s by a rotation about through L a.
(iv) If Zaz -f 6, S can be obtained from s by a rotation, a magnification and a translation.
(v) If Z = l/z, S is the reflection in OX of the inverse of s, O being the centr e
of inversion.
21. Show that each of the substitutions in Ex. 20 converts a circle c into a circle C, except that in (v), if c passes through O, then C is a straight line.
22. Show that the substitution
a'z + b'
converts a circle into a circle or, in an exceptional case, into a straight line .
[Z = ~> H --- ? -; r? 9 therefore the transformation is equivalent to one L a a az + b ^
or more of those in Ex. 20.]
CHAPTER VI
THEORY OF EQUATIONS (1)
1 . Roots of Equations. Under this heading we consider equations
of the type f(x) =0 where /(x) is a polynomial, and ( equation ' will mean an equation of this kind.
The general equation of the nth degree will be written in one of the forms :
x n +p l x n ~ l + pfl n - 2 + ... -f j0 n = 0, a x n + c^z"- 1 + a 2 x n ~ 2 + . . . + a n = ;
M (yt 1 \
or, a Q x n + na^- 1 + -~ - a z x n ~ 2 + . . . + a n = 0,
where binomial coefficients are introduced. The last equation is written briefly in the form
(a 0> a 1? a 2 , ... a n $x, l) n = 0.
For the present, we assume that every equation has one root. This is the fundamental theorem of the Theory of Equations, and will be proved
in another volume.
(1) It follows that every equation of Ike n-th degree has exactly n roots.
For let f(x) = x n +p l x n ~ l + ...+p n and let a be a root of/(x)=0. By the Remainder theorem, f(x) is divisible by x a; we may therefore
For let f(x) = x n +p l x n ~ l + ...+p n and let a be a root of/(x)=0. By the Remainder theorem, f(x) is divisible by x a; we may therefore