CHAPTER V
COMPLEX NUMBERS
Preliminary. In order that the operation of taking the square
root of a number may be always possible, we extend the number system
so as to include numbers of a new class known as imaginary numbers.
1.
Consider the equation
x*-2x 4-5 = 0.
Proceeding in the usual way,
we obtain the formal
a?=l%/^T,
x=
or
solution,
l2j~^l,
which at present has no meaning.
Suppose
(i)
the
symbol
possess the following properties
to
i
It combines with itself
and with
real
:
numbers according
to the
laws of
algebra.
(ii)
We may substitute
(
-I) for
t
2
,
wherever this occurs.
The reader can
verify for himself that (at any rate so far as addition,
subtraction, multiplication and division are concerned) reckoning. under
these rules will not lead to results which are mutually inconsistent.
Ex.
We
1.
Show
that if
x = l+2i, then x* x a = l + 4t + 4i a = l + 4i-4 = -3+4i;
have
.*.
x 2 -2z + 5=
Immediate consequences
-3 + 4t-2(l+2t)+5=0.
of the supposition are
(i)
(x
+ ly) + (x + iy') = (x + x
(ii)
(x
+
1
iy)-(x'
f
(x
(iii)
1
+ itf
f
(x
4-
+
iy) (x
f
iy') (x
f
)
+ iy') = (x-x') + i(y-y ).
+ iy') = xx' -yy' + i (xy + x'y).
f
f
-
f
iy
)
x'*
+ y'*
x'*
+ y'*
'
This reckoning is at present meaningless, and can only be justified by
an extension of the number system.
THE FUNDAMENTAL OPERATIONS
A Complex Number
2.
x+
iy,
where
y are
x,
represented by an expression of the form
is
The sign + does not
numbers.
real
53
indicate addition
as hitherto understood, nor does the symbol i (at present) denote a number.
These things are parts of the scheme used to express numbers of a new
class,
a
and they
signify that the pair of real
single complex
(x, y)
are united to form
number.
The complex number x +
number x, and we write
and
numbers
be regarded as identical with the real
lO is to
iQ = x
x+
in particular
=
tQ
-f
(A.)
(B)
Thus the system of complex numbers includes all the
together with new numbers which are said to be imaginary.
For shortness we write
_
Q
real
numbers,
+
and
+
in particular
tl
/QX
= *,
(D)
an abbreviation for + il, the symbol i denotes a complex number.
Thus two real numbers are required to express a single complex number.
This may be compared with the fact that two whole numbers are required
so that as
to express a fraction.
In order that expressions of the form x
numbers,
we have
to say
what
is
+
ly
meant by
be regarded as denoting
equality and to define the
may
*
'
fundamental operations.
We
3. Definition of Equality.
x+
ly
= x' + iy'
if
say that
and only
if
x = x' and y = y'.
In applying this definition, we are said to equate real and imaginary parts.
It should be noticed that the terms greater than and
less than -have
no significance in connection with imaginary numbers.
'
4.
'
The Fundamental Operations.
by the formal reckoning
of Art.
1,
In defining these we are guided
and the equations (i)-(iv) are taken
as defining addition, subtraction, multiplication
5.
'
'
The sum
Addition.
of the
and
division.
complex numbers x-f
iy
and
x'
+ iy'
is
defined as
(x
and we write
(x
+
iy)
It follows that
so that x
+ iy
is
the
+
')
+
i(y
+ y')
+ (v' + iy') = (x+x') + i(y+y')
(E)
x+ iy = (x + 40) -f (0+ iy),
sum
of the
numbers denoted by x and
(P)
iy.
MULTIPLICATION AND DIVISION
54
Subtraction
6.
is
defined by the equation
Hence subtraction is the inverse of addition, for x-x' + i(y-y')
number to which, if x' + iy is added, the result is x + ly.
The meaning of - (x + ty) is defined by the equation
is
the
r
Hence
= (0 - a + i (0-y) aB (- a? + 4(-y) ................ (H)
= 0, we have
.
)
In particular, putting
)
.)
-y=*(-y)
7. Multiplication
is
by the equation
defined
x'y) .................. (J)
It follows that multiplication
Putting
#'
= &,
2/'
=
is
commutative.
in the last equation,
number,
we
see that
if
k
is
a real
+ iy)k = xk + iyk = k(x + iy) ......................... (K)
(x
we put x=x = and y = y' = l, we have
2
or i 2 =-l .......................... (L)
(0 + il) =-l
f
if
Again,
8.
Division
is
the inverse of multiplication, that
is
to say, the
equations
x +
mean
and
ly
From
the same thing.
Equating
real
and imaginary
the last equation
parts,
Xx'-Yy' = x
mi
and
we have
and
Xy'+Yx' = y.
v x
*-
f
Therefore
unless x'
we have
y'
are both zero.
Thus
the equation defining division is
x + iy _xx' + yy'
x'
+ iy'
excepting the case in which x'
Putting x'=k,
y'
=0, we
x'*
yx'-xy'
+ y'*
x'*
+ y'*
'
...................... (
'
+ iy' =0.
see that
if A; is
,)=
a real number different from zero,
+i
................................. (N)
ZERO PRODUCTS
55
It is easy to see that addition, subtraction, multiplication
and
division
above) are subject to the same laws as the corresponding operations for real numbers.
(as defined
We have therefore justified, and attached a definite meaning
of reckoning described in Art.
Zero Products.
9.
is zero, then
where
the process
1.
For complex, as for
one of the factors of the product
1
xx -yy'
real
is zero.
numbers, if a product
For let?
Then
numbers.
x, y, x' , y' are real
to,
+ i (xy + x'y) = 0.
f
Hence, by the rule of equality,
.\
xx'-yy'~Q and
2
2
and
z(*' + 2/' ) =
zy'
y(x'*
+ x'y
Q;
+ y'*)=Q.
+ y' 2 is not zero, it follows that x = 0, y = 0, and therefore x + ty = 0.
If x' 2 + ?/' 2 = 0, then, since x', y' are real they must both vanish, and
therefore z' + 4j/' = 0.
If
a;'
2
10.
Ex.
1.
Examples.
Express
(1
+ 24) 2 /(2 +
(2+4)
2
2
in the farm
=
=
~4+4i-l""
X + iY.
3+4i~~ (3+40(3~4t)
24
7
-9 + 16 + 244
Ex.
2.
Solve the equation
x3 = 1
.
The equation can be written (#-l)(a; 2 + a; + l)=0, giving # = 1 or
If a;2 + a; + 1 =0, we have in the usual way
x=
Hence
Ex.
3.
the
numbers
Fen/y
that
1,
J(
J(
-
1
11.
2
V3)
- 1 + K/3)
are ca//ed JAe ^reg cw6e roofe of unity.
M a roo
CD,
3
o/ x
= 1.
we have
= |{ - 1 +34^3 -3(tN/3) 2 + (^3) 3} = |( - 1+34^3 + 9-34^3) =
Geometrical
+ a: + 1=0.
-
Denoting the given number by
to 8
a;
Representation
of
1.
Complex Numbers.
In naming line-segments and angles the order of the letters will
denote the direction of measurement. Thus AB denotes the distance
travelled by a point in moving from A to B, and we write
(1)
AB + BA-0,
and
EA--AB.
ARGAND DIAGRAMS
56
Again, we use L BAG
must turn in order that it
'
'
to denote the least angle through which
may
AB
along AC.
lie
Thus L CAB = - L BACy and we
shall
suppose that
-7T<
= x + iy, we
say that the number z is represented by the point
whose coordinates referred to rectangular axes are x, y.
(2) If z
When
there
is
no
risk of confusion, this will be called the point
z.
A
diagram showing points which represent
complex numbers is called an Argand
diagram.
Let
point
be the polar coordinates of the
6)
(r,
then
z,
= x + iy = r(cos + 1 sin 5),
z = rcos#,
y = rsin0,
tan = y/x.
r = J (x 2 + y 2)
z
where
and hence
Here
r,
which
denoted by
z
|
is
or
\
It follows that if
The angle 6
is
is
essentially positive,
called the
modulus of
z
and
is
by mod z. Thus
modz =
=
z
\
\
then z
|
z\
= 0,
= r = J(x 2 + y 2
many
).
y = 0, and consequently
called the amplitude of
angle has infinitely
of 6 such that
is
FIG. 2.
,
z,
and
is
z
= 0.
denoted by
values differing by multiples of
2rr.
am z.
This
The value
called the principal value of the amplitude.
With the convention
am z is L XOz.
of
Unless otherwise stated,
amplitude of
(1), it will
amz
will
be seen that the principal value of
mean
the principal value of the
z.
and Subtraction.
z = x + ty, z' = x' + iy'
Draw
12. Addition
(1)
Let
'.
parallelogram Ozsz', then s represents z
For
then
if
(X, Y) are the coordinates of
the
+ z'.
5,
X ==sum of projections of Oz, zs on OX
= sum of
= + X'.
05
projections of Oz, Oz' on
OX
FIG. 3.
ADDITION AND SUBTRACTION
Y = y + y',
Similarly
that
therefore s represents z
+ z' = the
z
length Os,
|
|
Now Os<0z + 0z', the sign
Therefore
i.e. if am z = am z'.
To prove
this algebraically,
where each root
(x
is
and
+ z'.
am (z +
')
57
It is to be observed
= L XOs.
of equality occurring
we have
to
if
LXOz =
L.XOz'
show that
to have its positive value.
This will be the case
+ z') 2 + (y + y')*<x* + y 2 + x' 2 + y' 2 + 2>/(x2 + y2
)
(x'
2
if
+ j/ /2 ),
i.e. if
or
if
(x
or
if
(xy'
(2)
^en
Draw
the
- x'j/) 2 >0, which
parallelogram
is
the case.
Oz'zd
d represents z-z'
(Fig.
4),
Y
.
by the last construction, the number
=
represented by d + z' z.
For,
Observe that
= the length
am (z - z') = z. Jf Od.
|
and
(3)
it is
=
// s n Zt
- 2'
2
|
+ z2 -f ...
4-z w ,
FIG. 4.
wAere zl5
required to find the point s n
~O
z'z,
2J
2>
z n are given complex numbers,
.
FIG. 5.
Draw in succession z^,
sense as 022 Oz3 Oz4
,
2j
+ 22 + 23,
This
is
,
,
...
,
s 2s3 , 5 35 4>
then
s 2 , s3 ,
equal to, parallel to, and in the same
... are the
points representing 2 1 + 2a ,
etc.
a continued application of the
first
construction of this article.
PRODUCTS AND QUOTIENTS
58
The modulus of the sum of any number of complex numbers
(4)
or equal to the
sum
For in the
Ozl9 21$2> S2S29
of the moduli of the numbers.
the moduli of z v z 2
last figure,
>
an(^
^at
Osn
Therefore
|
f sn
< Oz
*n
The sign of equality is
the same straight line and
Os n
z2
+|
to be taken
l
than
eng^s
Also
.
+ 2^2 + s2s3 +
l
|<K|
XOTE.
are in
1s
^e
are
2s
>
is less
|
. . .
4-...
v
+ sn _ r
+| * n
and only
|.
tS
che points O, z l9 * 2
occur in this order, that is, if z l9 z t ... z n have the
if,
if,
,
same amplitude.
To prove
this algebraically,
|
and
sn
|
^
|
sn
^
|
+
we have by
\z n
I
<
I
(1) of this article,
s n _2
z n _i
-h
I
1
-h
I
\z n
I
,
so on.
13.
(1)
Products and Quotients.
Let
z = x + iy =r(cos^H-tsin0),
z'
By
= x' + iy' =r'(cos 0' + 1 sin 0').
the definition of multiplication,
==
rr'{cos
cos
0'
- sin
sin
& + t(cos
sin 0'
+ cos 0' sin 0)}
;
^^rr'tcos^ + ^-ftsin^ + fl')} ........................................ (A)
/.
Since division
is
the inverse of multiplication,
4 = ^{cos(0-0') + 'sin(0-0')} ........................ (B)
For
this last is the
number which when
multiplied
by
z'
produces
z.
Hence
also
)
z l *=x l
If
with similar notation for z2
(A)
,
+ iy^ = fj (cos
23 ,
...
zn ,
l -h i
.............. (C)
sin 0j),
by continued application
of equation
we have
^
2
---n = ^2---M COS (^l^^+---+^n)
In (D) put
Z1
...(D)
=z2 ==...~2 w = z = r(cos0 + tsin0), then
zn
where n
+ ^sin(01 4-02 +...+0n )}.
rn (cos
n0 +
1
sin n0),
........................... (E)
any positive integer.
Similarly, from (C) it follows that
is
) ........................... (P)
DE MOIVRE'S THEOREM
Show
Ex.
1.
We
have
few 1 -
that
is
nearly equal to
.
+ i)=*/26(cos0 + i sin 0), where tan = 1/5,
4=
676(cos 40 + 1 sin 40), by (E).
(5 + i)
2 =
4 =
476 + 480* hence we have
(5 + 1)
(24 + 10t)
(5
and therefore
But
;
sin 40 =
cos 40 = 476/676,
tan 40 = 1, nearly.
and
480/676,
40=rr/4 approximately.
.*.
Modulus and Amplitude. From (D) we
the complex numbers zl9 z2 ... z w
(2)
of
59
|Z|=V2 ...rw = KN*2 |...|*n|,
axnZ
and
but note that the
---
last
if
Z
the product
is
,
,
,
see that
..................... (0)
+ amz n
(H)
equation does not give the principal value of
unless
am Z
-7T<01 +02 +. ..+0n < 77.
Again, from (B)
we have
M
am =
and
am
0'
z
21
.(I)
am z'
(J)
,
although the last equation does not give the principal value of amz/s'
unless
-7r<0-0'<7T.
important to notice that the amplitude of z/z'
through which Oz' must be turned in order that it may
and that, with the convention of Art. 11, (1),
It
is
is
the angle
lie
along Oz;
ft
the principal value of
Again,
if z,
a, a' are
am - = L z'Oz.
any numbers,
z-a
am
and
(3)
the length az
the length a'z
:
z-a =
L aza.
z-a
,
.
De Moivre's Theorem.
FIQ. 6.
From
(E)
and
(F) it follows that if
n
is
a
positive or negative integer
(cos
This, with
+ i sin
0)
n
an extension to be given
= cos n9 + 1 sin n0.
later (Art. 16), is
known
as
De
Moivre's
Theorem.
B.C.A.
COMPLEX ROOTS OF EQUATIONS
60
14.
Conjugate Numbers.
z=x + iy and
//
(1)
= x-iy
z'
9
then
are called conjugate numbers.
z, z'
Their product is a? 2 + y2 and
polar coordinates of the point
,
point
z'
are r
- 0.
,
6 are the
if
r,
z,
those of the
Hence
O
am(x + ty) = = -am(x-iy).
(2)
a polynomial in
is
// /(z)
with
z
real
= X + 1 F, where
coefficients, and f(z) =f(x + iy)
X, Y are real, then f(x - iy) = X - 4 Y.
We have /(# 4- iy) = a + a t (x + iy) + a 2 (x + iy) 2 -f
FIG.
. . .
where a
7.
ax
,
,
are
. . .
real.
We
can therefore obtain expansions of the form
f(x +
iy)
= 2axm (iy) n
f(x
)
-
ty)
= Zaxm - iy) n
(
,
and m, n are positive integers or zero.
of f(x + iy) in which n is even or zero is real, and occurs with
the same sign in/(x - iy).
Every term of f(x + iy) in which n is odd is imaginary, and occurs with
where every a
Every term
is
real
the opposite sign inf(x-ty).
Hence the result follows.
(3)
// an equation with real
coefficients
has complex
= 0, where
a,
roots, then these occur
in conjugate pairs.
Let a +
tjS
be a root of f(x)
the polynomial /(x), are
Let /(a +
ij8)
A = 0, B = 0.
j8,
as well as the coefficients of
real.
= A + iB,
where
A B
9
are real.
Since /(a +
1/?)
==
0,
we have
Hence
Therefore a -
ijS
is
also a root of f(x)
= 0.
EXERCISE VII
COMPLEX NUMBERS
1.
Find the modulus and amplitude of
2.
The condition that
3.
Find
4.
Express in the form JC
2 so that 2(3
(i)
2~+3i
(a + t6)/(a'
-f
f
i4)
4- ib')
s/3
may
=2 -f 3*.
+ 1 7,
iJ
+ 1, V3 - 1, - <s/3 -f *, - \/3 ~
(iii)
be real
is
ofe'-a'6=0.
.
COMPLEX NUMBEEb
x 4- iy> express
If 2
5.
form
in the
X + iY
Z
6.
Find the moduli of
7.
What
(i)
If
(ii)
If
8.
Let
9.
|
z
(i)
if
(ii)
if
(2
+ 3i) 2 (3-40 3
2
2
|
|
=
=
|
|
z'
|
z'
= 1 and am z =
2
x + ty where
differ
by
2|z-l| = |z-2|, then 3 (z -f y ) = 4z
22- 1 = 2-2 then # 2 4- y 2 = 1
|
a
z describes
Prove that
12.
22 cos a 4-2*
|<1
<2
2
cos a
Ifz=x + iy and Z = Jf + 1 7, show
that
|
22 cos a
4-
2
1
/
|
|
4-
1
Given that
1
+ 24
is
find the other three roots.
From
cos w0 = cosn
sin
if
___
one root of the equation, x 4 - 3a;3 -h 8x 2 -7a? + 5=0,
.^
+1
the formula cos nO
that
deduce
integer,
14.
2
|
..
13.
.
< 2r 4- r < 1
2
z
then ,,,
,
2
|
|2|<O41.
if
z
=r, then
|
'
circle.
11.
2
-C\ cos
n0=(7?
cos*- 1
1
sin
n^ = (cos 9 +
n ~ 2 ^ sin 2
sin
sin 0) n ,
cos n
+ <7
-C?" cosn ~ 3
1
tf
~4
where
^ sin 4
+ ...
sin 3 ^
-
a positive
ri is
...
,
,
(
tannf.
15.
By
putting
2=r (cos
=
-f i
sin 0) in the identity
prove that
1 -f t
coM+r
2
cos 26 +
2
r sin 04- r sin 26
?
;
Prove that |24-2'| 2 4-|2-2'| 2 =2|2| 2 4-2|2
|
6
.
|,
|
50)
.
then z'= -z.
TT,
sin
.
2
2
|
=-
1
4-
+ 4t) 3
Prove that
y are variables.
x,
z'
10.
|
(2-f 3t) /(3
(ii)
;
- am 2', then
am 2, am z'
and
|
Thus, in either case, the point
[Let
*
2~ 1
9
the principal value of the amplitude of (cos 50
is
|
(i)
61
...
+rn ~ l cos (n - 1)0
I-rco&O - rn cos n6 -f rn+1
+ ... -t-r - 1 sin
11
(rc
~ 1)0
cos
'
rsin0-rn 8inn0-f rn
" "
t
1
sin
(n~
(TI
-
1
)
ROOTS OF COMPLEX NUMBERS
62
= cos
16. If z
and n
sin
4- 1
zn
is
+ z -n = 2
a positive integer, prove that
- z~n = 2i
zn
cos nd,
sin n0..
Hence show that
~
2n 1 cosn 6 = cos n0 + C%coa(n -2)6 + 0% cos(?i-4)04-
and
if
n
is
and
if
n
-
2"- 1 sinn
1)2
odd
n-l
n-l
(-1) 2 2
= cos ?i0 -Of cos (n -2)0 + 0% cos (n -4)0 -
Binn
0==8mn0-C%8w(n
By
similarly for
(*-ar
n
l
)
-l
the method of Ex. 16, prove that
16 cos 8
18. If cos a
+ cos
0= - cos
sin*
50 - cos 30 4- 2 cos
0.
+ cos y=0 and sin a 4- sin + sin y=0, prove
cos 3 + cos 3jS 4- cos 3y = 3 cos (a 4- ]8 4- y)
/?
that
8in34-sin3)54-8in3y = 3sin(aH-)S4-y).
and
[Put a = cosa4-t sin
and a 8 4- 6 8 + c3 = 3a6c,
(19)
.
is
...;
17.
,
even
n
(
...
From
a,
b
= cos j3 4-
1
sin
c
j5,
= cos y4-t sin y,
then
etc.]
the identity
(x-b)(x-c)
t
(x-c)(x-a)
"
(a-6)(a~c)"
t
(ft-cXft-a)"*"
(x-a)(x-b)
(c-a)(c-6)
deduce the identities
-
sin (0
sin (a
B) sin (0
j5)
-
sin (a
y)
y)
[Put x = cos 20 4- i sin 20, a = cos 2a 4- 1 sin 2a,
Roots of Complex Numbers.
15.
any number whose nth power
is
r>0 and
root of
r.
is
4- 1
n
is
a positive integer,
called afc n-th root of
is
sin 0),
Consider the values of
s
any integer or zero.
un = r {cos (0
from which
it
follows that
z.
Let ^r denote the positive arithmetical nth
-7r<0<7r.
>
where &
(1) If
equal to z
z = r (cos
Let
where
etc.]
u
By
-I-
is
+ L sin
n
Art. 13
2for)
+ 1 sin
I
n
)
we have
(0 4- Zkn)}
an nth root of
z.
z
;
PRINCIPAL VALUES
63
and only if, the corresponding
That this may be the case, the
angles must differ by a multiple of 2ir and the corresponding values of k
must differ by a multiple of n. Therefore u has n distinct values, namely
Again, two values of u will be equal
angles have the same sine and cosine.
if,
y
those given
Thus
by
& = 0, 1,2,
the n-th roots
cos-
ofz = r(cos 6 +
-
n
lvalue of
(2)
,
n
.
cos-
_
1, 2, ...,
,
where
&=A
0,
.
0\
i
sin
~i
is
,
n-1.
,
called
the
principal
^z,
To find
the points representing the
find the length represented
Find
1.
points
values of
the
representing
where
%Jz,
we must be
able to
V#
into
Q lt Q2
the
n values of
by $r
and to divide the angle
n equal parts.
Ex.
J
/n +
|
n-1.
sin 6) are the values of
\
+ 1 sin
v^. v
,,^ v ^,,,
i
...,
Q^
,
/z,
z
Here
and
Oz=r=\/5 + 3~=
^Oz =tan 8 ^=
tan
Also fyr= */2* an(* so Ci ^2 ^3 are
and
points on the circle with centre
radius */2, such that
8.
General Form of De Moivre's Theorem. If x
w one o/ the values of (cos 6 + isin 6) x
cos x0 + M*n
16.
then
FIG.
^
The case
in
is rational,
.
which 2
is
a positive or negative integer has been considered
in Art. 13.
Let x~p/q where p, q are integers and q
(cos 0/q
is
positive, then since
+ 1 sin 0/q) q = cosO + t sin 0,
therefore
i
cos 0/q
also, since
j? is
an
+ 1 sin 0fq
is
a value of
(cos
+
t
sin 0)*
;
integer,
(cos 0/q +
1
sin 0/q) p
= cos pO/q + 1 sin j?0/#
therefore
;
,,
cos p0/q 4-
1
sin jo0/j
is
a value of (cos
+ 1 sin 0)
.
ROOTS AND FACTORS
64
The n-th Roots
17.
+1
then since cos
of Unity.
= 1,
sin
2k-rr
r-1 and 0=0,
let
the n-th roots of unity are the values of
2kn
-hsm~
n
n
cos
In Art. 15,
where #7 = 0,
.
,
n
1, 2, ...
1
1.
The principal n-th root, given by A = 0, is 1.
1 is given by k
If w is even, the root
\n.
~2
-1
n
n
=
1 (x
1
Since x
+x
+ + 1 ), it follows that
(x
71
. . .
)
other than
of x"~-
1,
71
The
-
+1)77
(2*
VL-
.
sm
-ft
.n
-
(
If
n
19.
(1)
n-th root of
odd, the root
is
root,
(2A
These are the values of
1 ).
+ 1)7T
T
where k = n
Q,
n
,
This follows from Art. 15 by putting r =
The principal
any
is
+an ~ 2 +...+<* + 1=0.
n-th Roots of
cos
a
1=0, then
a -1
18.
if
(
-
(
-
1) is
1) is
cos
Factors of x n -1 and x n +
O
1, 2,
...
1
n~l.
6 = 77.
l,
to k = 0.
n corresponding
& = -|(n - 1).
-+
n
given by
1
1
sin
,
1.
n
The n factors ofx -l are
x - {cos 2r7r/n + 1 sin 2rrr/n} where
= 0,
r
1, 2, ...
n-1.
This follows from Art. 15.
are given by r = 0, r = n/2.
The remaining (n - 2) factors can be grouped in pairs as follows
Since 2r7r/n + 2(n-r)7r/n = 27r, the factors corresponding to r and
(i)
If n
is even,
the factors
x-1
J.nd x
+1
:
n - r are
x-
f
{
cos
2rn
--
[
And
n
f-
sm 2f7rl
.
*
the product of these
j
x-
f
{
I
cos
2rn
-n
1
--
2f7T
- cos 2f7T\
= x^ - 2x cos 2f7T
+ sm^2
n
n /
n
.
a;
.
and
sm 2r<jr}/
n
.
]
is
2
/
\
n
'/
)
h 1
;
therefore
n
/
where 77 denotes the product of the factors as indicated.
- 1 is
=
(ii) // n is oddy the factor x
given by r 0. The remaining
factors can be grouped in pairs, and
(A)
n-1
(B)
CUBE ROOTS OF UNITY
x-{cos
// n
(i)
15
(2r
//"
n
+
l)77/n
l
=
+ ism
+1
(2r
are
+ l)7r/n},
where
'r
= 0,
1, 2,
...(n-1).
even
x +
(ii)
factors of xn
n
the
(2) Similarly,
65
is
n^
n - 2)
C
(x*-2x
\
os&-^
+ l) .............
n
(0)
/
ocW,
(D)
20. The Imaginary Cube Roots of Unity.
of x 2 4- x -f 1 = 0. Their actual values are
cos
.277
277
i
sin
j
Denoting either of these by
Moreover,
l
when
Again, except
cube roots of
1,
and
o
co,
(
+w+
co
2
=
/Q
-1,
the other
.
^3).
for (a> 2 ) 3
o> 2 ,
is
= (6t> 3 2 = l.
)
.................................. (A)
r is a multiple of 3, co r
and
o>
2r
are the imaginary
so
l-f-o/4-
The following
-,
or
These are the roots
o> 2r
=
.................................. (B)
identities are important.
x*-y* = (x-y)(x-o)y)(x-aj
2
y) ....................... (C)
+ y* = (x + y)(x + ajy)(x + a> y) ....................... (D)
x2 + y2 + z 2 - yz - zx - xy = (x + coy 4- o> 2 z) (x + o> 2 y 4- ojz) ......... (E)
2
z?
x 3 + y3 + z3 - Sxyz =
Ex.
Prove *Aa*
1.
(x*
(x
-f-
y + z) (a; +
coy
+
+y* + z* -3xyz)* = X*+Y* + Z*
X=x
z
+2yz,
2
co z)
(x
+
co
-3XYZ,
2
y -h
coz) ....... (F)
where
Y=
This follows from (F) on observing that
and
Ex.
If (l+x)
2.
n =c
Q
+ c 1x + ctf(;* + ...+cnxn and
o/ the series being continued as far as possible, show that the values of
are
|(2
\
n
+ 2 cos ~
Putting
*
1, o>, o>
2
) ,
where r = n,
n - 2, n + 2,
Slt #2
,
3
respectively.
/
for #, in succession,
(l-fl)
n -c
+ Cl -f...-fc r 4-...-fcn ..................................... (A)
n
...-hc n a>
,
4-...+c n cu
........................... (B)
2n ..........................
(C)
COMPLEX ROOTS
66
r
3, 1-f co
not a multiple of
If r is
we have
Now
-i
1
hence, adding (A), (B), (C) together,
n
2 n
+o>) + (l+o> )
n
+c3 +c 6 4-..0=2 + (l
3(c
XT
+ o>2r =0;
+co = l
i
2rr.27r
.
7r/7r.
TT\
+ cos -^- + 1 sin -5- =2 cos- cos^+tsmO /),
O
O
O
O
rt
\
and, since
AI
i
Also
1
cos ~
=i, we have
01l 4-cos 27T -i sin
+aj 2 =
.
O
Again, multiply (B) by
co
a
,
2-7T
,
,
O
by
(C)
1 -f o>
= cos - + 1 sin -
,,
and
+o> 2 ) n
,.,
(1
cos
and add to (A)
co,
Ttrr
o
;
.
-tsui
Al
a>
277
mn /
(l+eo) =( cos -^--4
2 /i
.
sm
27T\/
(n-2)ir
-
^
If
.
tJ
then
;
n
2 n
n
2
8(^+04+07 + ...)=2 +co (l4-a>) +o)(l+a> )
Also
nrr
TlTT
cos-
-
W7T\
+tsm-~
.
1
(n-2)ir
-^~
.
^
-
,
and
3
Finally, multiply (B)
by
o>,
(C)
by
o>
2
,
and add
EXERCISE
to (A)
;
it will
be found that
VIII
COMPLEX ROOTS
^-1
1.
Prove that the values of
2.
Find the values of ^(1 + 1) and *J(l-i).
3.
Use
Art. 19 to
are
~^(1
v^
t).
show that
1),
)
4.
Prove that
=a; 2 (a;
where
a,
j3
are the roots of z* + z- 1=0.]
'
(#-
2x
cos^-H
l).
COMPLEX ROOTS
67
Give a geometrical construction to find the points
5.
the values of
z l9 z 2 corresponding to
*Jz.
[If 01 is the unit of length, Oz l bisects the angle XOz, and is a mean proportional
to 01 and Oz ; and 2 2 is on z-f) produced, so that Oz a =2 1 O.]
If a, b are complex numbers, show that
6.
a
[Let 2 1 =a + */a
I
*i
I
a-*Ja*-b 2 then by Exercise VII, Ex. 10,
2
+ *$J 2 =i *!+** 2 + i *i -*2 2 =2 a )t + 2 a*-6>
-6 2
,
z2
,
I
!
I
I
I
I
.
|
|
|
a -6
|
-f
|
a -6
;
2
|
}
= {|a + 6| + |a-6|}M
Solve the equation
7.
f
+ 2(l+2i)z -(11 + 20=0.
- (11 + 20.
Verify that the sum of the roots is -2(1 + 20 and the product
[Put z~x + iy, equate real and imaginary parts to zero and solve for x, y.]
Prove that, with regard to the quadratic
8.
z2
(i) if
(ii) if
+ (p + ip')z
the equation has one real root, then
the equation has two equal roots, then
and
p*-p'*=:4q
If a
[(i)
Eliminate
In
(ii)
is
a
a.
this case (p
+ ip')*=4:(q + i,q'),
If z ~x + ly =r (cos 6 +
9.
rule of equality a 2
by the
real root,
1
etc.]
sin 0), prove that
Jz=z--={'Jr + x + i*/r-x}
according as y
Also
them
if
is
y>Q
-
-f l
sin
J
then 0<^<?r, and
and
if
[The roots are -J(
lie
on the
line
l
+i
cot
x + i=0.]
cos 6 = -
y<0 we
the roots of zn = (z + l) n ,
are collinear.
points
{Vr -f x
or
- i\lr - x}
9
positive or negative.
*/r ( cos
[\/z
pp'ty'*
J,
;
have -7r<8<0
9
etc.]
and show that the points which represent
where r=0,
1, 2, ...
n-1. The corresponding
COMPLEX FACTORS
68
r = 0,
that the roots of (1-f 2) n = (l -z) n are the values of
... n - \
but omitting n/2 if n is even.
Show
1, 2,
12.
i
tan-^, where
n
,
Prove that
* 2W -2xn cos
(i)
194- 1
*n + ar* - 2 cos
(ii)
cos
(iii)
rz<
=
1
2
/Trio" (* - 2x cos
=
/T^I JJ
- cos nS = 2n - 1
77^o
"*
(a?
~*
x m - 2#n cos 6 + 1 ~~ {xn - (cos 4 t sin
Art. 15. (iii) Put x =cos <f> + 1 sin ^, and for
n
is
odd and not a multiple of
+ l) n -xn -l.
- 1, co, where
[Put x=Q,
cos
put
3,
cos
- cos
<
n-
0)} {z
[(i)
13. If
+ ar * - 2
+
-
(cos
1
~
sin 0)}.
Now
use
n0.]
prove that x(x + l)(x* + x +
1) is
a
factor of (x
14. If
(
1
+ x + x2
)
n=a
-f
a; is
an imaginary cube root of unity.]
a^ -f a 2 +
a:
2
. . .
-f
2
n^
2n
>
prove that
aQ + a 3 + a 6 +
u
15. If
v=x+y
+ z + a(z
w=x + y + z + a(x + y-2z),
2
3
3 3
Sa^z)
prove thUt 21 a (x 4- y -f z
(a
where
a>
= w 3 -f v 3
- Suvw.
,
r
4-
2 2 2 4- z 3 2
[For
o>=J(
then
where
3
and w are the imaginary cube roots of unity, prove that
[IT] If 2^
real,
1^
+ w')- 1 -f (6 -f co')- 1 + (c + co')" 1 + (d + w')- 1 - 2ft/- 1
1
[Consider the equation (a -f a)"" -f (6
18. If
-f-
-
2;
2 2;3
1 4-t\/3),
*-
^^s = 0,
l
-f (c 4-
z)-
1
4-
(d
4 x)- 1 = 2ar 1 .]
/.
ZB
- zt =
being an imaginary cube root of unity, and
+ oA+ <o c)=
2
V(a 4- co 6 + we) =
</(
x)~
- Zfa - 2^2 = 0,
prove that
x 4- a>2:2
-
4-
2
a, 6, c
are
J{>
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
21. Points representing the Product
Let z = r(cos0 + *sin0),
Given Numbers.
Construction for
(1)
the
point
69
and Quotient of
z' = r'
(cos 0'
Two
+ 1 sin 0').
representing
the product zz'.
be the point on OX which represents
Draw the triangle Oz'P
unity, so that 01 =1.
directly similar to the triangle Olz.
1
L^t
P
Then
represents the product zz'
z
.
For by similar triangles
OP
Oz
-,
OP
r
IB,
Lz'OP^ LlOz = 0,
also
But
zz'
Therefore
P
= rr
f
{cos (0
represents
Construction for
(2)
..OP = rr
r-,
.
that
i
FIG. 9.
-f i
;
LlOP =
:.
+ 0')
,
sin (0
+ 0')}.
zz'.
the
point
representing
the quotient z/z'.
Draw
the triangle
the triangle Oz'l
Then
Q
directly similar to
.
i
represents the quotient
by the
For,
OzQ
FIG. 10.
z/z'.
last construction,
(number represented by Q)
// k
(3)
is constant,
z-a
z-a
and
.z'
z.
z varies so that
"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
For
at d,
let
d
f
.
aa
f
.
FIG. 11.
the bisectors of L aza' meet aa
Then, by Art. 13,
Therefore
ri,
(2),
az
:
d' divide aa' internally
r
a'z
=k
:
1.
and externally
and are fixed points.
Therefore z describes the circle on dd' as diameter.
>
in the ratio
k
:
1,
DISPLACEMENTS AND VECTORS
70
Also
f
,
we have
ca.ca'
= cd2
;
hence
are inverse points with regard to
a, a'
the
the mid-point of dd
c is
if
circle.
If
&=
1,
then az = a'z\ and
on
z lies
the perpendicular bisector of aa'.
if
Conversely,
which
circle of
the point z describes a
a, a' are inverse points,
Z ~Q>
then
z-a
For, since ca
.
ca'
the same ratio, k to
Hence,
:
=cd 2 then
,
where
<f>
on
is
dd'
is
divided internally and externally in
say.
1,
a'z = k
:
1 .*
If z varies so that
(4)
circle
az
FIG. 11.
K.
,
am
z-a
a constant angle, then the point
aa', containing
an angle
z describes
an arc of a segment of a
<f>.
\a
FIG. 12.
FIG. 13.
For by Art. 13, La'za =
on which the segment lies.
<f>.
The
Thus
sign of
<f>
is
<f>
determines the side of aa'
positive in Fig. 12
and negative
in Fig. 13.
22. Displacements
and Vectors.
In connection with the geometrical representation of complex
numbers, we introduce the notions of displacement and directed length or
(1)
vector.
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.
(2)
* See Elements
of Geometry,
Barnard and Child,
p.
316
'
:
Circle of Apollonius.'
ADDITION OP VECTORS
If
any
sense as
71
drawn equal to, parallel
PQ, the displacements PQ, P'Q' are
straight line P'Q'
is
and
to,
in the
same
said to be equal.
To specify completely a displacement
we must know
PQ
:
(i)
its
magnitude,
(ii)
its
direction
(iii)
its
i.e.
denoted by the
sense,
we draw OL equal and
that L
;
;
the letters, and
arrow.
If
PQ
the length
if
order of
-
necessary by an
parallel to
XOL is the angle which PQ
PQ
and
FIG. 14.
in the
same
sense,
we say
makes with OX.
This angle determines the direction and sense of
An
(3) Vectors.
with reference to
line
PQ) used to denote a line-segment
and sense, the actual position of the
expression (such as
length, direction
its
being indifferent,
is
called a vector.
Quantities which can be represented by lines used in this way are called
Velocities and accelerations are vector quantities.
vector quantities.
A
force can be represented
action of the force.
by a vector
'
localised
'
to
lie
in the line of
Quantities (such as mass) which do not involve the idea of direction are
called scalar.
Connection with Complex Number.
(4)
If
z=x + t,y and P
the point
is
a one-to-one correspondence exists between the number z
of the following
(ii) the displacement OP
(i) the point P
(x, y),
:
and any
;
;
(iii)
the
vector (or directed length) OP.
Any
to be
one of these three things
represented by
may
therefore be said to
represent
%*
z.
23. Addition of Displacements
and of Vectors.
Q R
Let P,
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
(1)
same as
to
R.
if
We
9
it
had moved
therefore
displacements as follows
*
Some
from
P
addition
of
directly
define
the
Flo
:
writers use
an underline instead
of
an
overline.
15
z,
or
ZERO AND NEGATIVE VECTORS
72
The
result of
adding
QR
expressed by writing
This equation
PQ, QR,
If
to
PQ
is
defined to be
PR\ and
this is
PQ+QR=PR .................................. (A)
also taken as defining the addition of vectors.
is
RS
are
any three displacements or vectors
(Fig. 17),
Q
FIG.
FIG. 16.
17.
The Commutative and Associative Laws hold for
placements and vectors.
(2)
In Fig. 16, complete the parallelogram
(i)
and
PQRS\
the addition of dis-
then by Arts. 23,
(1),
22,
Hence the commutative law
(ii)
holds,
and
PR is called the sum of PQ and
QR.
In Fig. 17,
therefore
and the
NOTE.
associative law holds.
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
the meaning of
is
also written in the
form
-PQ = QP,
which
defines
- PQ.
These equations are also taken as defining the meaning of zero and
negative vectors.
DISTRIBUTIVE
LAW
73
For displacements and vectors the meaning of
25. Subtraction.
PQ-QR is defined by
PQ-QR - PQ + (-QR) = PQ + RQThus
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
The
resulting displacement
its
length
or vector
by
is
A,
its
direction unaltered.
denoted by JcPQ or by
we
Further,
leaving
define
PQ
.
k.
- k)PQ by the
equation
(
In particular,
(
So that
to
multiply a vector by
27. The Distributive
number then
(
-
1) is to
Law.
turn
We
it
through two right angles.
shall prove
that if
k
is
a real
,
Let k be positive. Along PQ set off PQ' = kPQ.
in R'. By similar triangles,
to QR to meet
(i)
Draw
PR
and
and
Q'R'
Q'R' parallel
= kQR\
kPQ + kQR^PQ'
Hence the theorem holds
for
positive
-k),
we have
numbers.
(ii)
For a negative number
and
(
IG *
(
hence the theorem holds for negative numbers.
Thus
the distributive
vectors by real
law holds for the multiplication of displacements and
numbers.
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 less than
NOTE.
unity.
The diagram of
VECTORS AND COMPLEX NUMBERS
74
Complex Numbers represented by Vectors.
28.
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
AB
and am z
the length
makes with the directed line Ox.
then
|
z
|
is
Theorem
1.
If
C
divides
is
AB in
z is represented
by a vector AB,
the angle which the directed line
the ratio
n
:
m and
is
AB
any point, then
(m + n)OC - mOA + nOB.
For
mOC^mOA+mAC,
Also
mAC = nCB = - nBC
whence the
Theorem
in the ratio
result follows
2.
If
n:m,
by
;
O
addition.
FIG. 19.
z z
point which divides the straight line joining v 2
then the corresponding numbers are connected by the
z is the
relation
This follows from Theorem
1, in
accordance with the principle stated in
this article.
In particular, if z
Ex.
is the
mid-point of
then
z
= %(z l + z 2 ).
that
// a, b are complex numbers, prove geometrically
1.
A B
be the points which represent
Bisect AB at C, then
Let
z^
9
a, 6.
OA+OB=20C
OA-OB^'BA^Z'CA.
and
Therefore
a + 6 and a -b are represented by
200"and2CJ;
Now, since C
hence
is
mid-point of base
A B,
SYMBOLS OF OPERATION
Ex.
2.
//
OA, OB, OC
are connected by the relation
= 0,
?
Men A, B, C
[This
a -f b
where
-f
c
= 0,
are collinear.
the converse of
is
75
Theorem
We
of this article.
1
+ c)(W==a.OA+c.OC'; hence
(a
The Symbol
a
have
AB^c
.
.
BC.]
as an Operator.*
Along two straight lines at
consecutively, equal lengths OP, OQ, OP', OQ' in the
29.
i
right angles set off,
positive direction of rotation.
Let the symbol
i
applied
through a
it
of turning
operation
a vector denote the
to
angle
right
in
the positive direction of rotation.
To bring our language into conformity with
that of algebra, we say that to multiply a vector
by
is
i
turn
to
it
through
a right angle in the
FIG. 21.
positive sense.
Thus
in Fig. 21
OQ=i OP
,
OP' =
Therefore
where i*OP
Thus
i
2
an abbreviation
is
1(1
OP) -
1
2
OP,
Hence
for i(iOP).
t
and -1 denote the same operation, and
OQ' =
OQ' = iOP' = i(-l)OP and
Again,
Either of these results
to
OP' = i OQ.
and
multiply a vector by (-
is
i)
turn
it
OP -
(
(
- 1)OP.
in this sense
-OQ =
written in the form
is to
2
-
1)
(
we
write
-l)iOP.
OP^OQ',
so that
through a right angle in the negative
sense.
Again,
3
.
i
2
OP
is
taken to mean
t
3
(i
2
OP),
it is
obvious that
b are complex numbers, find numbers
so that the points z, z' and a, 6 may be opposite
Ex.
z, z'
if i
1
.
//a,
corners of a square.
Let
c
be the mid-point of a6, then
Oc + icb
Similarly,
*
2'=!- (a
;
+ &)+* (a- 6).
For the moment, the reader should forget
FIG. 22.
his conception of
I
as denoting a number.
B.C.A.
PRODUCT OF COMPLEX NUMBERS
76
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.
ON = ^ n OP - cos
Also
OF - cos
/.
.
OP +
*
.
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.
and Division of a Vector by a Complex
31. Multiplication
Number.
say that
to
to
multiply
Here
In accordance with Arts. 26, 29 and 30 of this chapter,
multiply a vector OP by the complex number r (cos -h i sin 0)
length by r
its
r is
and turn
applied
If
z,
OQ
we
and cos -ft
each other, and the
is
the vector obtained by multiplying
OP
by the complex number
write
also say that the ratio of
Division
is
the turning factor.
order in which they are
sin
indifferent.
is
OQ^zOP
we
is
the resulting vector through the angle 0.
the stretching factor
These are independent of
we
OQ
arid
to
resulting vector through the angle
result
is
the
number
the inverse of multiplication, so that
is
the result of dividing OQ by z.
Therefore to divide a vector OQ by
The
OP
OQ/OP^zi
is
(
-
if
OQ = zOP
z is to divide its length
y
then
OP
by r and tnr^ the
0).
the same as that obtained
32. Product of
z.
by multiplying
OQ by
I/z.
Complex Numbers.
OQ^z'OP and OR^zOQ,
Let
then we write
where
zz'
applied to
OP
denotes that the operators
in succession, in this order.
z',
z are to
be applied
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 turning the resulting vector through the angle (6 + 6').
Hence the operations denoted by
and rr' {cos
z'z
zz',
(0
+ 0') +
i
sin (0 + 0')}
are equivalent.
Again,
if
we take
(cos 0-h
i
sin 0) n to
+1
(cos
is
to be applied
n times, the
result
is
mean that
the operation
sin 0)
the same as that given by the operator
cos n0
-f i
sin 0) n
= cos n0 +
sin n0.
In this sense then
(cos
+
i
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
:
If z = 3
+ 2i, z' 1 -f i, mark the points z, z' in an Argand diagram, and
by geometrical construction, the points representing
1.
Z
+ Z',
Z-Z',
ZZ\
find
Z/z'.
z, a, b be complex numbers of which a, b are constant and z varies.
is given in terms of z by one of the following equations, it is required to find
2.
Let
If Z
the point Z corresponding to a given point z. Explain the constructions indicated
in the diagrams, 01 being the unit of length.
FIG. 24.
(ii)
(iii)
Z=
(iv)
Z=tz where
Z~az+b.
t is
real,
CENTROIDS
78
find tho point Z corresponding to a given point z t
the point Z is on the y-axis.
Z=(l + z)/(l-z),
3. If
and show that
if
|
= 1,
numbers and
are given complex
b
4. If a,
z
|
J^
6^,
find the point z corresponding to
value
of
and prove
t,
-oo to + 00,
z
that
varies
i
the entire
describes
real
any given
as
from
which
line
passes through a, b, the segment ah corresponding
to 1 of t.
to the values from
[Along the
ab set off the length az = t.ab,
line
Oz=0a+az~0a + t
then
ab;
.
:.
z=a + (b-a)t,
etc.]
5. If z=a(l + it) where t is a real number, prove that as
tho line through the point a perpendicular to Oa.
6. If
c,
a are given numbers, a being
z
(i) if
real,
= c+a(cos< + isin<),
show that as t varies from with centre c and radius a.
is real,
t
once the
circle
7. If
A, B, C,
that this theorem
8. If
and
O
To
G
is
G
is
any
GO to
is
[Put
.
an immediate consequence of the identity
m m w 3 ... at A A A& ...
=
(m + m z + m 3 + ...)OG m OA l + msOA 2 + maOA 3 4- ...
point, then
l9
29
. . .
are not
9. If z is the centroid of particles of
all
of the
three coplanar
Moreover,
same
m m
29
2,
l9
2
and non-parallel
pOA +qOB + rOC = Q,
OA,
vectors
where
s , ...
p
9
:
.
at z l9 z 29 z a ,
. . .
OB OC
9
...
,
then
.
are connected by a
q 9 r are real numbers.
p:q:r=& OBC A OCA A OAB
:
+
3 3
2
we may suppose
acting at A l9 A% ...
sign,
2,
l9
m w m
m^ + m z + w z
mass
...)s
relation of the form
19
,
1
l
m m
z,
19
to be the centre of parallel forces, proportional to
Any
the point z describes
,
= tan J<^.]
t
.
the centroid of particles of mass
include cases where
10.
+ oo
D are any four points in a plane, then
AD BC^BD AC + CD AB.
.
Show
;
z=c + a(l + it) I (I - it),
if
where
varies z describes
and,
find the point z corresponding to a given value of cf>
(ii)
t
9
where the signs of the area# are determined by the usual rule.
Also the points A, B C are collinear if p + q +
and conversely.
rQ
9
[For
and
p
9
q,
let
G
p:q:r= &OBC AOCA
:
:
9
&OAB,
be the centre of parallel forces, acting at A, B,
r then G coincides with O, and
pOA
let
9
(7,
and proportional to
.
VECTORS AND COMPLEX NUMBERS
11.
(i)
of the
//a,
j8
are non-parallel vectors
andp,
q,
79
p' 9 q' are real numbers, adequation
form
involves the two equations
If a,
(ii)
,
p=p'
q=q'.
9
y are non-parallel coplanar vectors connected by the equations
= Q and p'a + q'fi r'y = 0,
p<x + qfi + ry
p' q', r' are real, then these are one and the same equation, that
-fc-
where p
9
q, r,
9
is
[These theorems follow at once from Ex. 10.]
12.
three complex
Any
form
where p,
q, r
numbers
z l9 z 2 , z 3
are connected
p:q:r~ A0z 2z 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
Prove this algebraically, and deduce Ex. 11.
Q,
,
=x + iy
l
whence the
l9
z2
=z 2 + iyl9
zs
=z 3 +
[In Art. 28,
(i)
we have
B C are collinear, and is any point,
OA EC + OB CA + OC AB^O.
Theorem 1, m n m + n=CB AC AB.]
A,
.
:
Zj, z 2 , z a
then
9
.
Let
it/ 3 ,
and conversely,
results follow immediately.]
13. If the points
14.
relation of the
are real numbers.
Moreover,
[If z l
by a
.
:
:
:
be complex numbers, no two of which are equal, then
If the points z l9 z 2 z 3 are collinear,
,
l
\
2|2~2l
Z 2~*3\
!
*8hl-2l|
=-
the point z l lies between z 2 and z 3 , the ambiguous signs are both minus.
(ii) If the above equation holds, then either z l9 z 2 , z 3 are collinear, or else O is
the centre of a circle which touches the sides of the triangle Z 1 z 8z 3 .
Also, if
15. If A i^4 2^4 a is an equilateral triangle, the vertices occurring in the positive
direction of rotation, prove that
s
~
4- 1
sin
an imaginary cube root of unity.
numbers corresponding to A l9
3 X -f wz 2 + cj*z 3 = 0, and consequently
2
2
z^ 4- z a + z 3 - z^ - Z& - Z& = 0.
where
o> is
Also, if z l9 z 2 , z 3 are the
16. If
numbers
A X YA'X'Y'
a, a' 9
is a regular hexagon and
then the numbers represented by
where 6 has the values
~,
3
~
3
-
.
A
X
9
9
A t9 A
B9
prove that
A' represent given complex
X' Y Y' are given by
9
9
TRANSFORMATIONS
80
The
17.
triads of points
A B C and X,
9
9
triangles if the corresponding
by the relation
x(b
- c) +
y(c
[The triangles are directly similar
18. If
CA, AB,
ABC
Y,
Z are the vertices of directly
complex numbers
and
a, b, c
- a) + z(a - b) =0.
if
=
AC XZ
i.e*.
,
if
a triangle and triangles BCX, CA Y,
similar
to one another, the centroids of
directly
[By
1
J Ex.
is
x
-
a
x, y, z are
-
c-a
ABZ
similar
connected
=.]
z-x
are
drawn on BC,
XYZ &ndABC coincide.
= y--- = z br. Hence show that# + yv + z=a + & + c.lJ
b-c c-a a-b
c
17. i
drawn on the sides of a given triangle ABC, all
inwards. Prove that their centroids form an equilateral triangle.
be the centroids of the triangles drawn outwards on BC, CA, AB.
19. Equilateral triangles are
outwards or
[Let P, Q,
all
R
Prove that
_ _
I
QA = CA
\/O
(cos
30 + L sin 30),
__ __
A R =AB
i
.
*Jd
30 -
(cos
L
QR \CB + ^ (CA -AB), and that RP has
show that EQ = EP (cos 60 + sin 60), and use Ex. 15.]
Hence show that
Hence
sin 30).
a similar value.
L
'
TRANSFORMATIONS
'
and
20. If Z, z are connected by any of the relations in Ex. 2
given curve s, then Z will describe a curve S.
if z
describes a
Explain the following, where a, b are given complex numbers.
= z -f a, S can be obtained from 8 by a translation.
(i) If Z
Z=tz where t is real, the curves s, S are similar and similarly situated,
the
centre of similitude. In this case we say that S is a magnification
being
(ii)
of
If
s.
(iii)
If
Z = (cos a-f
tsin a) z,
Zaz
S
through L a.
(iv) If
can be obtained from s by a rotation about
-f 6,
S
can be obtained from s by a
Z = l/z, S
is
the reflection in
a magnification and a
rotation,
translation.
(v) If
OX
of the inverse of
s,
O
being the centre
of inversion.
21.
Show
except that in
circle C,
22.
that each of the substitutions in Ex. 20 converts a circle c into a
Show
(v), if c
passes through O, then
[Z
L
or
a
circle into
= ~> H--a
is
a straight
line.
that the substitution
a'z
converts
C
a
more of those
?
a
circle or, in
-;
az
+ br?
in Ex. 20.]
9
an
+ b'
exceptional case, into a straight line.
therefore the transformation
is
to one
equivalent
^