study material for iit-jee on complex number
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Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 1
COMPLEX NUMBERS
There is no real number x which satisfies the polynomial equation 2 1 0x . To permit solutions of this
and similar equations, the set of complex numbers is introduced.
We can consider a complex number as having the form a + bi where a and b are real number and i, which
is called the imaginary unit, has the property that 2 –1i . It is denoted by z i.e. z = a + ib. ‘a’ is called as
real part of z which is denoted by (Re z) and ‘b’ is called as imaginary part of z which is denoted by (Im
z).
Any complex number is:
(i) Purely real, if b = 0 (b) Purely imaginary, if a = 0 (c) Imaginary, if b 0
Note :
(a) The set R of real numbers is a proper subset of the Complex Numbers. Hence the complete
number system is N W I Q R C.
(b) Zero is purely real as well as purely imaginary but not imaginary.
(c) –1i is called the imaginary unit. Also 2 3 4–1; – ; 1 .i i i i etc
(d) a b ab only if atleast one of a or b is non–negative.
(e) is z = a + ib, then a – ib is called complex conjugate of z and written as –z a ib .
Fundamental operations with complex numbers
In performing operations with complex numbers we can proceed as in the algebra of real numbers,
replacing 2 –1i by when it occurs.
(1) Addition ( ) ( ) ( ) ( )a bi c di a bi c di a c b d i
(2) Subtraction ( ) – ( ) – – ( – ) ( – )a bi c di a bi c di a c b d i
(3) 2Multiplication ( ) ( ) ( – ) ( )a bi c di ac adi bci bdi ac bd ad bc i
1.00 The complex number system
1.01
111
1
Algebraic Operations
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 2
(4) 2
2 2 2 2 2 2 2 2 2
– – – ( – ) –Division = = = + i
– – –
a bi a bi c bi ac adi bci bdi ac bd bc ad i ac bd bc ad
c di c di c di c d i c d c d c d.
Inequalities in complex numbers are not defined. There is no validity if we say that complex number is
positive or negative. e.g. z > 0, 4 + 2i < 2 + 4i are meaningless.
In real numbers if 2 2 2 2
1 2 1 20 then 0 however in complex numbers, 0 does not imply 0.a b a b z z z z
Illustration 1 : Find multiplicative inverse of 3 + 2i, then
Solution:
Two complex numbers 1 1 1 2 2 2 & z a ib z a ib are equal if and only if their real and imaginary parts are
equal respectively. 1 2 2 1 m 2. . Re( ) and I ( ) I ( )mi e z z z z z .
Illustration 2 : Find the value of x and y for which 2(2 3 ) – (3 – 2 ) 2 – 3 5 where , .i x i y x y i x y R
Solution:
Let z be the multiplicative inverse of 3 2 , then
.(3 2 ) 1
1 3 – 2
3 2 3 2 3 – 2
3 2 –
13 13
3 2 – Ans.
13 13
i
z i
iz
i i i
z i
i
1.02
111
1
Equality In Complex Number
2
2
2
2
( 3 ) – (3 – 2 ) 2 – 3 5
2 – 3 2 – 3
– 0
0, 1 and 3 2 5
5 if 0, and if 1, 1
2
5 0, and 1, 1
2
z i x i y x y i
x y x y
x x
x x y
x y x y
x y x y
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 3
Illustration 3 : Find the value of expression 4 3 2– 4 3 – 2 1 when 1x x x x x i is a factor of expression.
Solution:
Illustration 4 : Solve for z if 2 | | 0z z .
Solution:
2 2 2 2
2
2
– 0 and 2 0
0 or 0
when 0 – | | 0
0, 1, –1 0, , –1
when 0 | | 0 0
0 Ans. 0, , –1
x y x y xy
x y
x y y
y z i
y x x x
z z z i z
Illustration 5 : Find square root of 7 + 40i.
Solution:
2
2 4 3 2
2 2
2
4 3 2
1
–1 ( –1) –1
– 2 2 0 Now – 4 3 – 2 1
( – 2 2)( – 3 – 3) – 4 7
when 1 . . – 2 2 0
– 4 3 – 2 1 0 – 4(1 ) 7
– 4 7 – 4
3 – 4 A
x i
x i x
x x x x x x
x x x x x
x i i e x x
x x x x i
i
i ns
2 2 2
Let
( ) 0
z x iy
x iy x y
2
2 2
4 4 2 2 2 2
Let ( ) 9 40
– 9 ........( )
and 20 ..........( )
squing ( ) and adding with 4times the square of ( )
we get – 2 4 81 1600
x iy i
x y i
xy ii
i ii
x y x y x y
2 2 2
2 2
( ) 168
4 ........( )
from ( ) ( ) we can see that & are of same sign
(5 4 ) or (5 4 )
Sq. roots of 40 (5 4 )
(5 4 ) Ans.
x y
x y iii
i iii x y
x iy i i
a i i
i
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 4
Illustration 6 : Express the complex number –1 2z i in polar form.
Solution:
Illustration 7 : If | – 5 – 7 | 9z i , then find the greatest and least values of |z – 2 – 3i.|
Solution: We have 9 = |z – (5 + 7i)| = distance between z and 5 + 7i. Thus locus of z is the circle
of radius 9 and centre at 5 + 7i. For such a z (on the circle), we have to find its greatest
`and least distance as from 2 + 3i, which obviously 14 and 4.
Illustration 8 : Find the minimum value of |1 + z| + |1 – z|.
Solution:
1.03
111
1
Representation Of A Complex Number
22
–1 –1
–1
–1 2
| | (–1) 2 1 2 3
2Arg – tan – tan 2 ( )
1
3(cos sin ) where – tan 2
z i
z
z say
z i
1.04
111
1
Modulus Of A Complex Number
|1 | 1– |1 1– | (triangle inequality)
1 1– 2
minimum value of (|1 | |1– |) 2
Geometrically | 1 | |1– 2 | | 1 | | –1| which represents sum of distances of from 1 and –1
it can be seen easily t
z z z z
z z
z z
z z z z
1 1/4 8
hat minimum ( ) 2
2 Ans.n
PA PB AB
e
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 5
Illustration 9 : 2
– 1zz
then find the maximum and minimum value of |z|.
Solution:
Illustration 10: Solve for z, which satisfy
2Arg ( – 3 – 2 ) and Arg ( – 3– 4 ) .
6 3z i z i
Solution: From the figure, it is clear that there is no z, which satisfy
both ray.
Illustration 11: Sketch the region given by
(i) ( –1– ) / 3Arg z i (ii) 5& ( – –1) / 3z Arg z i .
Solution: (i) (ii)
2 2 2 2– 1 – – –
2
Let
2 2 – 1
2 1 ..........( )
2 2 and – 1 –1 – 1
(1, 2) .
z z z zz z z
z r
r rr r
r r R ir
r rr r
r .........( )
from (i) and (ii) (1, 2)
(1, 2) Ans.
ii
r
r
1.05
111
1
Argument Of A Complex Number
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 6
Illustration 12: If –1
1
z
z is purely imaginary, then prove that 1z .
Solution:
Illustration 13: If –1
1 3
zArg
z then interrupter the locus.
Solution:
Here 1–
arg–1–
z
z represents the angle between lines
joining –1 and z and 1 + z. As this angle is constant, the locus of z will be a of a circle
segment.(angle in a segment is count). It can be seen that locus is not the complete side
as in the major are 1–
arg–1–
z
z will be equal to
2–
3. Now try to geometrically find
out radius and centre of this circle. 1 2
centre 0, Radius Ans.3 3
Illustration 14: If A(z + 3i) and B(3 + 4i) are two vertices of a square ABCD (take in anticlock wise
order) then find C and D.
1.06
111
1
Conjugate Of A Complex Number
2
–1Re 0
1
–1 –1 –1 –1 0 0
1 1 1 1
– –1 – –1 0
1 1
1 Hence Proved.
z
z
z z z z
z z z z
zz z z zz z z
zz z
z
1.07
111
1
Rotation Theorem
–1arg
1 3
1–arg
–1– 3
z
z
z
z
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 7
Solution: 3 4
0
Let affix of C and D are respectively
Considering 90
z z
DAB AD AB
4
4
4
3
3
3
– (2 3 ) (3 4 ) – 2 3we get
2
– (2 3 ) (1 )
2 3 –1 1
– (3 4 ) ( 3 ) – (3 – 4 ) and –
2
3 4 – (1 )(– )
3 4 –1 5
z i i i ie
AD AB
z i i i
z i i zi
z i z i i ie
CB AB
z i i i
z i i z i
Illustration 15: Find the value of 192 194 .
Solution:
Illustration 16: 2If 1, , are cube roots of unity prove.
2 2
2 5 2 5
2 4 8
2 2 4 4 8 2
( ) (1– )(1 – ) 4
( ) (1– ) (1 – ) 32
( ) (1– )(1– )(1– )(1– ) 9
( ) (1– )(1– )(1– )..........to 2n factors 2 n
i
ii
iii
iv
Solution:
1.08
111
1
Cube Root Of Unity
192 194
21 – Ans.
2 2
2
( ) (1– )(1 – )
(–2 )(–2 )
4
i
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 8
Illustration 17: Find the roots of the equation 6 64 0z where real part is positive.
Solution:
Illustration 18: Find the value 6
1
2 2sin – cos
7 7k
k k.
Solution:
Illustration 19:
1.09
111
1
nth
Roots Of Unity
6
6 6 (2 1)
(2 1)6
5 7 3 11
6 6 62 2 2 2
11
6 6
–6
–64
.
2 , 2 , , , ,
roots with +ve real part are
2 Ans.
i n
i n
i i ii i i i
ii
i
z
z z e x z
z ze
z e e ze ze e ze ze
e e
e
6 6
1 1
6 6
0 0
6
0
6
0
2 2sin – cos
7 7
2 2 sin – cos 1
7 7
(Sum of imaginary part of seven seventh roots of unity)
– (Sum of real part of seven seventh roots of unity) 1
0 – 0
k k
k k
k
k
k k
k k
1 1
i Ans.
1.10 Logarithm Of A Complex Quantity
If cos cos cos 0 and also sin sin sin 0, then prove that
(i) cos2 +cos2 +cos2 = sin2 +sin2 +sin2 = 0
(ii) sin3 +sin3 +sin3 = 3sin( )
(iii) cos3 +cos3 +cos3 = 3cos( )
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 9
Solution:
1 2
3
1 2 3
1
Let cos sin , cos sin ,
cos sin .
(cos cos cos ) (sin sin sin )
0 .0 0 (1)
1( ) Also (
z i z i
z i
z z z i
i
iz
–1
1 3
1 2 3
2 2 2 2
1 2 3 1 2 3 1 2 2 3 3 1
cos sin ) cos – sin
1 1 cos – sin , – cos – sin
1 1 1 (cos cos cos ) – (sin sin sin ) (2)
0 – .0 0
Now ( ) – 2( )
i i
iz z
iz z z
i
z z z z z z z z z z z z
1 2 3
3 1 2
1 2 3
2 2 2
2
1 1 10 – 2
0 – 2 .0 0, sin (1) (2)
(cos sin ) (cos sin ) (cos sin ) 0
(cos 2 sin 2 ) cos 2 sin 2 2 sin 2 0 .0
Equation real and imaginary parts on bot
z z zz z z
z z z u g and
or i i i
or i i cos i i
3 3 3 3 3
1 2 3 1 2 1 2 1 2 3
3 3
3 1 2 3 3
1 2 3
3
h sides, cos 2 cos 2 cos 2 0 and sin 2 sin 2 sin 2 0
( ) ( ) – 3 ( )
(– ) – 3 (– ) , using (1)
3
(cos sin )
ii z z z z z z z z z z
z z z z z
z z z
i 3 3(cos sin ) (cos sin )
3(cos sin )(cos sin )(cos sin )
i i
i i i
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 10
cos3 sin 3 cos3 sin 3 cos3 sin 3
3{cos( ) sin( )}
Equation imaginary parts on both sides, sin 3 sin 3 sin 3 3sin( )
Alternative method
Let cos cos cos 0
sin sin sin
or i i i
i
C
S
– – –
– 2 – 2 – 2
2 2 2
0
0 (1)
– 0 (2)
From (1) ( ) ( ) ( ) ( )( ) ( )( ) ( )( )
i i i
i i i
i i i i i i i i i
i i i i i
C iS e e e
C iS e e e
e e e e e e e e e
e e e e e e –2 –
(2 ) 2 2
3 3 3
3 3 3 ( )
( )
0( 2)
Comparing the real and imaginary parts we
cos 2 cos 2 cos 2 – sin 2 sin 2 sin 2 0
Also from (1) ( ) ( ) ( ) 3
3
i i i
i i i
i i i i i i
i i i i
e e e
e e e from
e e e e e e
e e e e
Comparing the real and imaginary parts we obtain the results.
Illustration 20: If 1 2 and z z are two complex numbers and c > 0, then prove that
2 2 2–1
1 2 1 2 + (I C) (I C )z z z z .
Solution:
2 2 2–1
1 2 1 2
3 2 2 3–1
1 2 1 2 2 2 1 2
2 2 2 2–1
1 2 2 2 1 2 1 2 1 2 2 2
1 2 1 2
2
1 2
We have to prove that:
(1 ) (1 )
i.e. (1 ) (1 )
1 or – – 0
(using Re ( ) )
1– 0 wh
z z C z C z
z z z z z z C z C z
orz z z z c z C z c z z z z z zc
z z z z
or c z zc
ich is always true.
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 11
Illustration 21: 4 3 3
1 2 3 4 5If , [ / 6, / 3], 1, 2, 3, 4, 5 and cos cos cos cos cosi z z z z
32 3, then show that | | .
4z
Solution:
4 3 2
4 3 2
2 3 4 5
3 3 3 3 3 2 3 | | | | | | | |
2 2 2 2 2
3 | | | | | | | |
3 | | | | | | | | | | ..........
| | 3 3 – | | | |
1– | |
3 4 | | 3 | |
4
z z z z
z z z z
z z z z z
ze z z
z
z z
Illustration 21: Two different non parallel lines cut the circle |z| = r in point a, b, c, d respectively.
Prove that these lines meet in the point z given by –1 –1 –1 –1
–1 –1 –1 –1
– –z =
–
a b c d
a b c d.
Solution: Since point P, A, B are collinear
4 3 2
1 2 3 4 5
4 3 2
1 2 3 4 5
4 3 2
1 2 3 4 5
Given that
cos . cos . cos . cos . cos 2 3
or cos . cos . cos . cos . cos 2 3
2 3 cos . cos . cos . cos . cos
[ / 6, / 3]
1 3 cos
2 2i
z z z z
z z z z
z z z z
i
Progression / APEX INSTITUTE FOR IIT-JEE / AIEEE / PMT, 0120-4901457, +919990495952, +919910817866 Page 12
2
1
1 0 – – ( – ) – 0( )
1
Similarlym, since points P, C, D are collinear
– – – – – – – – – – ( )
( ) , , .
From equati
zz
a a z a b z a b ab ab i
b b
z a b c d z c d a b cd cd a b ab ab c d ii
k k kzz r k say a b c etc
a b c
–1 –1 –1 –1
–1 –1 –1 –1
on ( ) we get
– ( – ) – – ( – ) – ( – ) – – ( – )
– –
–
ii
k k k k ck kd ak bkz c d z a b a b c d
a b c d d c b a
a b c dz
a b c d