Download Complex Numbers PDF

TitleComplex Numbers
TagsAlgebra Euclidean Geometry Sine Mathematical Concepts
File Size33.8 MB
Total Pages58
Table of Contents
                            IMG
IMG_0001
                        
Document Text Contents
Page 2

lGonmilltuunm

A number in the form of a + ib, where a, b are real numbers and i = J-t is called a complex number.
A complex number can also be defined as an ordered pair of real numbers a and b and may be
written as (a, b), where the first number denotes the real part and the second number denotes the
imaginary part. lf z = a + ib, then the real part of z is denoted by Re (z) and the imaginary part by
lm(z). A complex number is said to be purely real if lm(z) = 0, and is said to be purely imaginary if
Re(z) = 0. The complex number 0 = 0 + i0 is both purely real and purely imaginary.
Two complex numbers are said to be equal if and only if their real parts and imaginary parts are
separatelyequal i .e.a+ib=c+idimpl iesa=candb=d.However, thereisnootherrelat ion
between complex numbers that is of the type a + ib < (or >) c + id.

Remark:
r Clear ly i2 =-1, i3 =i2. i =- i , ia =1.In general , i4n =1, 14n+1 - i , i4n*2 =-1 , i4n+3 -- i foran

integer n.

Algebra of Gomplex Numbers, modulus and argument, triangular inequality,
cube roots of unity.

A complex number z = x + iy, written as an ordered pair (x, y),
can be represented by a point P whose Cartesian coordinates
are (x, y) referred to axes OX and OY, usually called the real
and the imaginary axes. The plane of OX and Oy is called the
Argand diagram or the complex plane. Since the origin O lies on
both OX and OY, the corresponding complex number z = 0 is
both purely realand purely imaginary.

Modulus and Argument of a Complex Number:

we define modulus of the complex number z = x + iy to be the real number and denote it
by lzl. lt may be noted that lzl > 0 and lzl = 0 would imply that z = 0.

lf z = x + iy, then angle 0 given by tan 0 = I is said to be the argument or amplitude of the complex
X

numberzandisdenotedbyarg(z)oramp(z). lncaseofx=0(wherey+o),arg(z)=+nl2or-nl2

^ '*y '

']llltCc
Ltd., FuDEE House,2g-+ KaIa aarai, Sarvapflya WhaD tYew Delhi -tAo ot6, ph 265t5g4g, 26569493, Fax 26513942

Page 29

E=

General equation of the line:

From equation (1) we get za+-za+b
= 0 '

where e=i(4- l ) andib=Zz2-2.2r, b
eR

This is the general equation of a line in the complex
plane'

Slope of a given line:
ne(a)

lI za +-za+ a = 0 is the given line then its slope
= -

t("')

Equation of a line parallel to the line za +-za+ b
= 0 is zd +-za + l' = 0

(where l, is a real number)' - _7a+ i1. = 0 .
Equation of a line perpendicular to the line za

+-za+b = 0 is za

(where l, is a real number)'
Equation of Perpendicular Bisector:

consider a line segment joining A(21) and B(22)'
Let the line 'L', be it's perpendicular bisector'

lf P(z) be any point on the'L', we have PA
= PB'

=lz-z l = lz-z2l

+ z(2, - a\ +z(2, - zt) + 2.4 - zr2,
= o

Distance of a given point from a given line:

Let the given line be za+-za+b= 0 , and the
given point be z" then distance of z" from this

. l

lz^a+z^a+ol
l ineis--r :# '

zPl

Equation of a Circle

Equation of a circle of radius r and having its
centre at zs is lz-zsl= r'

> |z _zo|, = f * ( z - zi (z
_^) = f2 > f i . + az +62 +b = 0, where

_ ? = 7o and b = zo4 _ 12'

It represents the general equation of a circle
in the complex plane'

oEquat ionofacirc ledescr ibedona| inesegmentAB,(A(21),8(22))asdiameter is
(z-z)(2-2r)*( t - zr) (7-4)=o '

oLetzlAf idzzbetwogivencomp|exnumbersandzbeaanycomp|exnumbersuchthat,
(-

- \
arol ' - " l=a,wherea-

\z-zz )
e (0, n). Then'z'will lie on the arc of a circle'

o Let ABCD be a cyclic quadrilateral such that
A(21)' B(22)' c(4) and D(a) lie on a circle'

Then

(zo -2.) .(zr-zt) is purely reat.
(zr-2.)(zo-2")

Some lmportant Resultto Remember -^--,^-. ^..*hara z.
r The triangle whose vertices are the

points represented by complex numbers z1' 22' zs
ts

equi lateral i f l++-.=+=0i '€ ' ' i tz l+z?'+z! :z lzz+2223+zsz1'
zz-zg zs-zt 2 ' t -72

C|z-21|+|z_: ,12|=l , , representsane| | ipsei f |21_22|<x,havingthepoints4andzzasi ts

foci. And iI lz1-z2l= 1" then z lies
on a line segment connectin Q 21 and z2'

. |z_2. | - |z_zz|=^, , representsahyperbo|ai f |2. ,_2, |>}, ,havingthepointsz, tAndz2aSits

foci. And if lz1-z2l= 1., the z lies on
in" tin" passing through zt and z2 excluding

the points

between 21 artd 72'

f||?lcc
ttt"ot

Page 30

RSM-91 1-P3-MA-Complex Number 29

s01$D Pn0Bfirs

I

I

i

,
I

{

Subjective:

Problem 1.
lf iz3 + / - z + i = 0, then show thatlzl =1.

Solution:
iz3 + 22 -z+ i= 0
By substituting z = i in the equation, we get 0 = 0
Hence z- i is a factor of iz3 +22-z+i
+ iz2 1z-i) - 1(z-i) = 0 = (t - 1) (z - i1 = g

Either izz -1 = 0 or z- i= 0

When z- i= 0,2=i ; . lz l = | 0 +i .11 =JO' * 12 =1.

When iz2-1=0,22=1=-i

. ' l t l= lo-1. i1= !6 '*( lF =1+ l= ' l=1 or lz l=1
.'. In anY case we have lzl = 1.

Problem 2.
tf lzl <1,lwl <1, showthatlz-vn1' <Ml-lwl)t + ( argz-argw)2.

Solution:
Let us consider a unit circle with its centre as the
origin Let ZAOX =0r dnd IBOX= 0z
.'. arg(z) = 0r ?nd arg(w) = 02
z =OA, w=OB

Now in A oAB; cos o = t
on f t]_oe t1: tBA l'z

2ton t toe l '
t_t2 t_t2 t_t2

= lenl =loAl +loal -2loAl loBl coso
+lz-w l2 =lzl2 * lwl '- 2 lzl lwlcosO =(lzl - lwl ) ' * qEl lwl sin2 (e/2)

we know tnat sing = 9 =[9)' > sin2 9
22 \2) 2

= ( : i > lz l lwl s in29 [s incelz l , lwl<l , i t isauni t c i rc le]

Hence lz-wl ' < ( lz l - lwl) ' * (argz- arg w)2

Problem 3.
(i) Prove that the polynomial x3n + i:m+1 + v3k+2 is exactly divisibte by f + x+ 1 if m, n, k are
non negative integers.
(ii) Show that the polynomial f sin0- p''1x sinne +pn sin(n-l)0 is divisibte by f-2px cosl+ p2.

I

'l']ntcc
Ltd., EIITJEE House,2g-+ Kala hnl, sarraprita whaq ilew Delhl -tto olo Ph 26515949, 26569493, Fax 26513942

Page 57

RSM-9 1 1 -P3-MA-ComPIex

Subjective:

Level- |

3(i i).

4(i).

6(i i) .

B(i).
( i i ) .
( i i i ) .
(iv).
(v).
(vi).

L

| ,.1
= 2l sin %1,^rg 1zi

=
#

or lz,l = 0' arg
(21) is not defined'

lz-7-ai1=tfag+a2,aeR

Fora> s,r=({za. , j (# ,u) ;For
a 1o'z=l tz^- t I [ -#.#)

A|Ipointstowardsther ightofx=lexceptthepoint(2 '0)
All points inside the circle lzl

= S

Allreal and purely imaginary number

Exterior to a circle of radius 10 with centre
(1' 0)

Regionbetweenthetwoconcentr iccirc|esofradiusland4withcentre(2,0).
n"iion outside the circle of radius ll2wilh

centre (0' 0)
1 r

_nt2 11.
_L(rriJd)2,.1(.t $),,

13. 4y2+4x+sec2 2
=o'

15. lz2l> 1
12.

14.

3.

4(i).

5.

$i.nTa
a+b+c=0

Level ' l l

14.
15( i ) .

11.

r: -"'x":12:!-i ; no sotution for c, J2
l f c = 1,2= -1- i ; l f 1 < c<' !2 ' r= --G, _1)

2n
(i i) ' n

,_\_ t tJs (b) +-4i , t i
\a)z= ---;-) / \

ft * *u .= r(E *-! *^'li "no ,=r^,=4=,2=
!l!I+=J,, ,n''

lz l^ ' "=---2 ' ' - - [ z )-

true for all real a > 0; z= !2i '

1, -1(ii).
5
1

Ju'lcc Ltd,, HtrtEe aour.,zffif,En"priv"
vihar, New Dethi -770 oto Ph 2

I

Page 58

57RSM-91 1 -P3-MA*Complex Number

Ob"iective :

1.
4.
7.
10.
13.
16.
19.
22.
25.
28.
31.
34.
37.
40.
43.
46.
49.

Level- ll

1.
4.
7.
10.
13.
16.
19.

A
B
A
D
c
B
B
B
B
D
D
A
B
c
D
c

c
D
A
A
c
D
A
c
B
c
A
B
A
c
D
B
A

A
B
c
B
c
D
B
A
c
A
c
A
A
A
c
c
A

2.
5.
B.
11.
14.
17.
20.
23.
26.
29.
32.
35.
38.
41.
44.
47.
50.

3.
6.
9.
12.
15.
18.
21.
24.
27.
30.
33.
36.
39.
42.
45.
48.

B,C,D
A,B
A,B,C,D
A,D
A,C
A,C,D
A,D

A,B
A,B,C,D
A,C
A,B,C
A,B,D
A,B,C
A,D

B,C
A,B,G,D
A,B,C
A,B,C
A,B,C
A,C

2.
5.
8.
11
14
17
20

3.
6.
9.
12.
15.
18.

Numerical Based

1. 0 2.

Comprehension :

1. D
4.C
7. A

['latch ttre Calumn

2.
5.
8.

( i )
(iv)

B
A
c

D
B
D

(D)
(D)

3.
6.
9.

1.
2.

(A)
(A)

( i i i )
( i )

(c)
(c)

( i i )
( i i )

(iv)
( i i i )

(B)
(B)

Ja,lrCC Ltd., FIfiJEE House,29-A, Kalu Sani, Sarvapilya Vihar, New Delhi -tto OIO Ph 265t5949, 26569493, Fax 26573942

Similer Documents