Chapter-5
COMPLEX NUMBERS AND QUADRATIC EQUATIONS
1.
Complex numbers and quadratic equations
Number of teaching hours
Total marks
:
:
8
10
Blue - Print
1 Mark
1
2 Marks
3
3 Marks
1
Total
10
2
10
2
10
OR
-
2
OR
1
1
1 Mark Questions
1.
Express the following in the form a + i b
a)
2.
3.
4.
5.
√
√
b)
i-35
Find the real and imaginary part of √3
2−
√
a)
4 - 3i
b)
-1 + √3
Write the complex conjugate of (1 + i )2
Find the real numbers x and y if ( x - iy ) ( 1 + i ) is the conjugate
of -3 -2i
Write the multiplicative inverse of
+ 1
6.
Evaluate : 7.
Find the sqare roots of
x2
+3=0
8.
Solve :
9.
Find the modulus of
10.
Find the amplitude of 1 + i
1
2 Marks Questions
1.
Express the following in the form a + ib
(b)
(a)
(5 - 3i)3
(c)
(e)
(g)
2.
√ √
√ √ √√
1
3 + 3
(f)
(h)
b)
(b)
+
+
(2 + 5i)2
If 'n' is any integer find in + in+1 + in+2 + in+3
5.
If Z1 = 2 - i, Z2 = -2+i. Find imaginary part of
6.
Find the values of x and y if
a)
( x + 2y) + i(2x - 3y) = 5 - 4i
b)
(1-i) x + (1+i)y = 1-3i
c)
(x-iy) (3+5i) is the conjugate of -6 -24i.
!
c)
4.
!
+
(1 + 2i) (2 - 3i)
Find the multiplicative inverse of
(a)
(d)
Find the conjugate of
a)
3.
(1 + i) (1+2i)
3 (7 + i7) + i (7 + i7)
Prove that x2 + y2 = 1.
7.
If x + iy =
8.
9.
If Z1 = 2 - i, Z2 = 1 + i. Find "
10.
Find real and imaginary parts of :
11.
Find the least +ve integer m such that
12.
If x - iy =
If 1 + 4√3 = (a + ib)2 prove that a2 - b2 = 1 and ab = 2√3
'
!
()
"
. Prove that
√
√
(x2
2
+
%
#
y2)2
Also find its modulus.
=
$
! ( )
=1
13.
14.
Solve : * + *
+ 1 = 0
√2
Find the modulus and amplitudes of
(a)
1.
1+ i
(b)
-1 + i
(d)
3 Marks Questions
Express complex numbers in Polar form :
(a)
1-i
(b)
-1 + i
©
(e)
√3 + (d)
√
Find the conjugate of :
3.
Find 'θ' so that
4.
a)
Purely real
b)
Convert complex number
(a)
/- .
√
(b)
6
√
(f)
,- .
2.
5.
-√3 − (c)
is
Purely imaginary
01, 2
3452
into polar form.
7
− 8 = −2 9 + 8 If a + ib = (x + iy)1/3 where a, b, x & y are real.
Prove that
#
. $
6.
If Z1 = 1 - i, & Z2 = -2 + 4i. Find imaginary part of
7.
If α and β are different complex numbers with | β |=1 then
Prove that
8.
9.
If p + iq =
Solve :
"
;<
=;
<
<
<
(a)
(b)
©
(d)
"=1
show that p2 + q2 =
x2
+ 3x + 9 = 0
3x2 - 4x + 20/3 = 0
27x2 - 10x + 1 = 0
ix2 - x + 12i = 0
3
< < SOLUTIONS
1 Mark Questions
1.
>=
a)
√
√
=
b)
@A
×
=
√
√
@ . @
=
=
√
. @
=
=
×
√
=
2.
Z = √3
2 −
3.
a)
Z = 4 - 3i, Multiplicative inverse of Z =
b)
Z = -1 + √3 Multiplicative inverse of Z =
√
√
; CD E = √3
2 , GHI. E = −
Z = (1 + i)2 = 1 - 1 + 2i = 2i
>J = −2
(x - iy) (1 + i) = -3 + 2i
(x + y) + i (x - y) = -3 + 2i
x + y = -3
x-y=2
∴ x = −1
2 , 7 = −5
2
4.
5.
i18 + # $
6.
'
7.
x = + √3
8.
x2 + 3 = 0
9.
Z=
10.
Z = 1 + i = r ( Cos θ + i sin θ)
r = √2, PQR S = 1
X
√2
N
=> x2 = -3
∴ |z| =
||
||
=
√
= 1
√5
TU S = 1
W
√2V
∴ Amp θ = Y
4
4
= 1 + √2
= = 0 + 1. J
=
||
√
= . + − = −1 − . = −1 − = ±
TWO MARKS
1.
a)
Z
b)
3 (7 + i7) + i (7 + i7) = 21 + 2i i + 7i - 7 = 14 + 28 i
c)
(1+i) (1+2i) = 1 + 2i + i-2 = -1 + 3i
d)
e)
f)
= (5 - 3i)3 = 53 - 3.52 (-3i) + 3.5 (-3i)2 - (3i)3
= 125 - 225i - 135 + 27 i
= -10 - 198 i
=
√√√√
1
3 + 3 =
h)
=
+
=
+
=
√ √
=
g)
=
√√
√
=
√ =
=
√
− 26
=
= 1
2 + 1
2
+
+
− 3
2 + 4
5 + + 1 + =
+ − 27 − 9
=
+ 3 . 3 + 3. −9 − 27
=
= 9
25 −
2 −3
2 + 1 = 3
2 + 4
5 + 1 + −5
^
−
^
5
=0−
√
2)a)
b)
c)
3.a)
Z = 1 + 2 2 − 3 = 2 − 3 + 4 + 6 = 8 + EJ = 8 − E=
EJ =
×
1 − 7
10
EJ = -21 - 20i
=
4.
5.
6.a)
z=
×
=
=
^
E=
=
multiplicative inverse of
z=
in + in+1 + in+2 + in+3
=
=
=
=
= 2 − −2 − =
Z1 . >
= -4 - 2i + 2i - 1 = -5
∴
=
= −1
5
.a
x + 2y = 5 2x - 3y = -4
x+y= 1
- x + y = -3
∴ y = -1,
=
×
=
in (i + i + i2 + i3 )
in ( 1 + i - 1 - i ) = in - 0
0
2x + 4y = 10
2x - 3y = -4
---------------7y = 14
y = 2, x = 1
b)
z = (2 + 5i)2 = 4 ≠ 20i - 25 = -21 + 20 i
multiplicative inverse of z =
b)
=
2y = -2
x=2
6
=
^
c)
3x + 5xi - 3iy + 5y = -6 + 24i
(3x + 5y) + i(5x - 3y) = -6 + 24i
3x + 5y = -6
5x - 3y = 24
7.
=>
x + iy =
a + ib
a - ib
∴ x - iy =
a - ib
a + ib
∴ (x + iy) (x - iy) =
x2 + y2 = 1
8.
!
!
×
!
!
1 + 4 √3 = 9 + 8 = 9 − 8 + 298 equating real and imaginary
parts. a2 - b2 = 1, ab = 2√3
9.
"
10.
z=
=
"= "
×
=
=
Re z = 1, Img. z = 1,
11.
9x + 15y = -18
25x - 15y = 120
-------------------34x
= 102
x = 102 = 3
34
y = -3
%
#
$
"= "
"=
√
=
√
= 2√2
= 1 + & |z| = √1 + 1 = √2
%
= 1 => b
c
i2m = 1 = i4
2m = 4 => m = 2
7
%
= 1 => #
$
=1
12.
x – iy = '
!
()
!
()
9 − 8
e − f
9 − 8
|* − 7 − 2 *7| = g
g
e − f
∴ (x – iy)2 =
* − 7 − 2 *7 =
|9 − 8|
|P − f |
√9 + 8
h* + 7 − 2* 7 + 4* 7 =
√P + f √9 + 8
h* + 7 + 2* 7 =
√P + f √9 + 8
h* + 7 =
√P + f Ti. QU 8Qjℎ RfDR.
9 + 8
* + 7 = e + f
h* − 7 + 4* 7 =
13.
* + *
+ 1 = 0
√2
√2* + * + √2 = 0
*=
=
± h√.√
√
± √
*=
√
± √
√
14.a) Z = 1 + i = r (Cos θ + isin θ)
l = √1 + 1 = √2
PQR S = *⁄l = 1
√2
7
TU S = l = 1
√2
Amp (θ) = Y
4
8
b)
Z = -1 +i=r (Cos θ + isin θ)
Modulus = √1 + 1 = √2
PQR S = *⁄l = −1
√2
7
TU S = l = 1
√2
Amp (θ) = Y − Y
4 = 3Y
4
c)
Z = -√3 − = l eQRS + sin S
Modulus = √3 + 1 = 2
PQR S = − √3
2
TU S = −1
2
Amp (θ) = −5Y
6
d)
>=
=
l = √0 + 1 = 1
PQR S = 0
TU S = 1
qHr S = Y
2
=
=
9
3 Marks Questions
1.
a)
Let Z = 1 – i = r ( Cos θ + I sin θ)
l = √1 + 1 = √2
1
PQR S =
√2 X
S = − Y
4
−1
W
√2V
1 − = √2 sPQR −Y
4 + sin−Y
4t
N
TU S =
Let Z = -1 + i = r ( Cos θ + I sin θ)
l = √1 + 1 = √2
PQR S = −1
X
√2
b)
N
1
W
√2 V
TU S =
S = Y − Y
4 = 3Y
4
−1 + = √2 PQR 3Y
4 + TU 3Y
4
Let Z = √3 + c)
l = √3 + 1 = 2
√3
X
2u
PQR S =
S = Y
6
1
TU S =
2 W
u
V
√3 + = 2 PQR Y
6 + TU Y
6
N
d)
Let Z =
√
l = v1
4 + 3
4 = 1
PQR S =
1
X
2 u
S = − Y
3
3
TU S = − √ 2W
u
V
√
= PQR −Y
3 + sin−Y
3
N
10
e)
Let Z =
√
l = v1
4 + 3
4 = 1
N
− 1
2 X
u
PQR S =
S = Y − Y
3 = 2 Y
3
+ √3
2W
u
V
√
= PQR 2Y
3 + sin 2Y
3
TU S =
f)
Let Z =
=
×
=
= -1 + i
l = √1 + 1 = √2
N
PQR S =
TU S =
−1
√2X
1
W
√2 V
S = Y − Y
4 = 3Y
4
> = √2 PQR 3Y
4 + sin 3Y
4
2.
>=
=
=
>J =
=
=
=
^
11
>=
3.
/- .
×
/- .
/- .
/- . /- . /- .
/- .
=
=
a)
/- .
/- . /- .
/- .
=0
/- .
‘Z’ is purely real
/- .
=>
Sin θ = 0
∴ θ = n π : n ∈ I.
b)
Z is purely imaginary
/- .
/- .
=>
=0
3 – 4 Sin2 θ = 0
Sin2 θ = 3 .
4
= TU ± Y
3
∴ θ = n π + (-1)n . Y
3
Sin θ = +
4. a) Z =
√
√
×
√
√
=
√
=
√
= 4 (-1 + i √3) = 4 + i4 √3 = r (Cos θ + I sin 0)
l = √46 + 48 = √64 = 8
N
PQR S =
TU S =
−4
= −1
X
8
2u
+
√
+ √3
2W
u
V
S = Y − Y
3 = 2Y
3
Z = 8 PQR 2Y
3 + sin 2Y
3
12
>=
b)
=
=
2
01, 3452
=
√√
√
= #
$+ #
√
=
√
√ √
√
$
×
√
√
= l PQR S + sin S
√
√
$ + #
$ =2
l = '#
N
√3 − 1 X
2√2 u
PQR S =
√3 + 1 W
u
2√2 V
TU S =
5.
S = 5Y
12
(a + ib) = (x+iy)1/3
(a+ib)3 = x + iy
a3 + 3a2. ib – 3ab2 – ib3 = x+iy
(a3 – 3ab2) + i (3a2b – b3) = x + iy
∴ x = a3 – 3ab2
y = 3a2b – b3
x = a2 - 3b2
y = 3a2 – b2
a
b
x - y = a2 – 3b2 – 3a2 + b2 = - 2 (a2 + b2)
a
6.
. a
J
=
=
b
=
×
=
= 4 + 2
Imaginary part of
#
=
.
J
$=2
|z|2 = z.EJ
13
7.
"
; < "
=;
<
=
=
=
|<
= ;|
= <
;< ;
=
=
<
= ; <;
xDUeD "
;<
"
=;
<
y → ri =
+
=
= |<|
|;| <
= ;<;
= <;
= |<w |;|
<;
=1
<
<
<
< .
< <
(p+iq) (p-iq) =
q2
;< ;<
<
= ; <
= ;
<
∴ r − i =
p2
=
=
= |<|
= ; <;
<
= |<|
= ; <;
<
=1
8.
|;<|
<
<
< 9. a) x2 + 3x + 9 = 0
x=
=
±√
± √
=
± √
14
b)
3x2 – 4x + 20 = 0
3
9x2 – 12x + 20 = 0
*=
±√^
=
c)
±
±√
=
^ ±√
27x2 – 10x + 1 =0
*=
^ ± √^^^
× =
d)
=
±√
×
ix2 – x + 12i=0
x=
*=
± √
*=
=
±
x = -4i x = 3i.
15
=
±