1.
Express the following in the standard form a + ib.
- Solution:
= = =
= = =
= 1 + 3i
2.
Express the following in the standard form a + ib.
- Solution:
= =
= =
= = - i
3.
Express the following in the standard form a + ib.
(-3 + i) (4 – 2i)
- Solution:
(-3 + i) (4 – 2i) = -12 + 4i + 6i – 2i2
= -12 + 10i + 2
= -10 + 10i
4.
Express the following in the standard form a + ib.
- Solution:
=
= since i2 = -1 and i3 = -i
= = 1
5.
Find the real and imaginary parts of the following complex numbers.
- Solution:
= = = =
R.P. = and I.P. = -
6.
Find the real and imaginary parts of the following complex numbers.
- Solution:
=
= = = -
RP = IP =
7.
Find the real and imaginary parts of the following complex numbers.
(2 + i) ( 3 – 2i)
- Solution:
(2 + i) ( 3 – 2i) = 6 + 3i – 4i – 2i2
= 6 – i + 2 = 8 – i
RP = 8, IP = -1
8.
Find the least positive integer n such that
- Solution:
(i)n = 1
This is possible for n = 4, 8, 12, ………………….
The least positive integer n for this to be true = 4.
9.
Find the real values of x and y for which the following equations are satisfied.
(1 – i)x + (1 + i)y = 1 – 3i
- Solution:
(1 – i)x + (1 + i)y = 1 – 3i
x – ix + y + iy = 1 – 3i
Equating real and imaginary parts
x + y = 1
-x + y = -3
Adding 2y = -2
y = -1 and so x = 2
10.
Find the real values of x and y for which the following equations are satisfied.
+ = i
- Solution:
+ = i
= i
(3 + 2i – i2)x – 6i + 2i2 +(6 – 7i – 3i2)y + 3i + i2 = 10i
(4 + 2i)x – 6i – 2 + (9 – 7i)y + 3i – 1 = 10i
[x(4) + 9y] + i(2x – 7y) = 3 + 13i
Equating real and imaginary parts
4x + 9y = 3 4x + 9y = 3
2(2x – 7y = 13) 4x - 14y = 26
23y = -23
y = -1
Put y = -1 in 4x + 9y = 3 then x = 3
x = 3, y = -1
- x = -3i, y = -i
x = -1, y = 3
We take the square root of –8 – 6i as 1 – 3i and –1 + 3i.
11.
Find the real values of x and y for which the following equations are satisfied.
= y(2 + i)
- Solution:
= y(2 + i)
Equating real and imaginary parts
= 2y
= 4y2
and x + 4 = y
x2 + 3x + 8 = 4(x + 4)2
x2 + 3x + 8 = 4(x2 + 8x + 16)
x2 + 3x + 8 = 4x2 + 32x + 64
3x2 + 29x + 56 = 0
3x2 + 8x + 21x+ 56 = 0
x(3x + 8) + 7(3x + 8) = 0
(3x + 8) (x + 7) = 0
x = - or –7
When x = -
y = 4 - = =
When x = -7
y = 4 – 7 = -3
x = -, y = and x = -7, y = -3
12.
For what values of x and y, the numbers –3 + ix2 y and x2 + y + 4i are complex conjugate of each other?
- Solution:
Given conjugate of x2 + y + 4i is x2 + y + 4i.
But, conjugate of –3 + ix2y is –3 –ix2y.
Hence, x2 + y = -3
x2 + y = -3
-x2y = 4
x2y = -4
Put x2 = t
t + y = - 3 and ty = -4 or y = -
t - = -3
t2 – 4 = -3t
(or) t2 + 3t – 4 = 0
(t + 4) (t – 1) = 0
t = -4 or t = 1
y = 1 or y = -4
Substituting y in x2 = t
x2 = -4 or + 1 and y = 1 or –4
x = 2i or + 1 and y = 1 or –4
x | 2i | 1 |
y | 1 | -4 |
13.
If (1 + i) (1 + 2i) (1 + 3i)….(1 + ni) = x + iy. Show that 2.5.10….(1 + n2) = x2 + y2.
- Solution:
Given (1 + i) (1 + 2i) (1 + 3i) ……. (1 + ni) = (x + iy)
Taking the modulus on each side
….. =
Squaring both sides
(2) (5) (10) ……(1 + n2) = (x2 + y2)
14.
Find the square root of (-8, -6i).
- Solution:
Let = x + iy
-8 – 6i = (x + iy)2 = x2 – y2 + 2ixy
Equating real part and imaginary parts
x2 – y2 = -8 and 2xy = -6 or xy = -3 y = -
Hence x2- = -8
Put x2 = t, t-= -8
t2 – 9 = -8t
t2 + 8t – 9 = 0
(t + 9) (t – 1) = 0
t = 9 or t = 1
Substituting x2 = t in x2 = -9
x = 3i
x2 = 1
x = 1
When x = 3i, y = i
x = -3i, y = -i
x = -1, y = 3
We take the square root of –8 – 6i as 1 – 3i and –1 + 3i.
15.
If z2 = (0, 1), find z.
- Solution:
z2 = (0, 1)
= i
Let z =
Let = x + iy
Squaring both sides i = x2 – y2 + 2i xy
x2 – y2 = 0 or x = y and 2xy = 1
2y2 = 1
y =
Hence z = or
16.
Prove that the triangle formed by the points representing the complex numbers (10 + 8i), (-2 + 4i) and (-11 + 31i) on the Argand plane is right angled.
- Solution:
Let A, B, C represents the complex numbers 10 + 8i, -2 + 4i and (-11 + 31i) on the Argand diagram respectively.
AB =
= =
BC =
= = =
CA =
= = =
CA2 = AB2 + BC2
ABC is right angled.
17.
Prove that the points representing the complex numbers (7 + 5i), (5 + 2i), (4 + 7i) and (2 + 4i) form parallelogram. (Plot the points and use midpoint formula).
- Solution:
Let A, B, C, D represent the complex numbers (7 + 5i), (5 + 2i), (4 + 7i) and (2 + 4i) in the Argand diagram respectively.
(7 + 5i) (7, 5)
(5 + 2i) (5, 2)
(4 + 7i) (4, 7)
(2 + 4i) (2, 4)
The midpoint of AD is
=
The mid point of BC is
=
ABCD is a parallelogram.
18.
Express the following complex numbers in polar form.2 + 2
- Solution:
2 + 2i = r(cos + i sin )
r cos = 2
r = = 4
r sin = 2
Hence cos = and sin = ; =
2 + 2i = 4
19.
Express the following complex numbers in polar form.–1 + i
- Solution:
–1 + i = r(cos + i sin )
-1 = r cos
= r sin
r = == 2
-1 = 2 cos
cos = -
= 2 sin
sin =
= - = 2
Polar form of –1 + i is 2
20.
Express the following complex numbers in polar form.–1 – i
- Solution:
Let –1 – i = r(cos - i sin )
-1 = r cos
r = =
-1 = r sin
Hence -1 = cos
cos = -
-1 = sin
sin = -
= - =
Polar form of –1 – i is
21.
Express the following complex numbers in polar form. 1 – i
- Solution:
22.
If arg(z - 1) = and arg (z +1) = 2, then prove that = 1.
- Solution:
arg(x + iy - 1) =
tan-1 =
= tan =
x – 1 = y ……………………… (1)
arg ( z + 1) =
arg(x + iy + 1) =
tan-1 =
= tan = tan
= -
y = - ……………….. (2)
-= y
Multiply (1) and (2)
- = y(y
- = y2
(x2 – 1) = y2
x2 + y2 = 1
z2 = 1
= 1
23.
P represents the variable complex number z. Find the locus of P, if Im = -2,
- Solution:
Im = -2
Let z = x + iy
Im = -2
Im
Im = -2
Im = -2
= -2
2y – 2y2 – 2x2 – x = -2[1 – 2y + y2 + x2]
2y – 2y2 – 2x2 – x = -2+ 4y - 2y2 - 2x2]
-x - 2y = -2 x + 2y = 2
24.
P represents the variable complex number z. Find the locus of P, if =
- Solution:
Let x = x + iy
=
Squaring both sides
x2 + y2 – 10y + 25 = x2 + y2 + 10y + 25
-20y = 0
y = 0
25.
P represents the variable complex number z. Find the locus of P, if Re = 1
- Solution:
Let z = x + iy
Re= 1
Re= 1
Re= 1
Re= 1
= 1
x2 + x + y2 + y = x2 + y2 + 2y + 1
x – y = 1
26.
P represents the variable complex number z. Find the locus of P, if = 2
- Solution:
Let z = x + iy
Re= 1
Re= 1
Re= 1
Re= 1
= 1
x2 + x + y2 + y = x2 + y2 + 2y + 1
x – y = 1
27.
P represents the variable complex number z. Find the locus of P, if arg = -
- Solution:
arg =
Let z = x + iy
arg =
arg =
arg =
arg =
tan-1 =
= tan
=
(x – 1) (x +3) + y2 = 0
x2 + y2 + 2x - 3 = 0
28.
Solve the equation x4 – 8x3 + 24x2 – 32x +20 = 0, if 3 + i is a root.
- Solution:
If 3 + i is a root then 3 – i is also a root.
Sum of the roots = 6
Product of the roots = 9 + 1 = 10
The corresponding factor is x2 –6x + 10
Hence x4 – 8x3 + 24x2 – 32x +20 = (x2 – 6x + 10) (x2 + px + 2)
Equating the coefficient of x
-32 = 10p – 12
p = -2
x2 – 2x + 2 = 0
x = = = 1 i
The roots are 3 i and 1 i.
29.
Solve the equation x4 – 4x3 + 11x2 – 14x + 10 = 0, if one root is 1 + 2i.
- Solution:
If 1+ 2i is a root then 1 – 2i is also a root.
Sum of the roots = 2
Product of the roots = (1 + 2i) (1 – 2i) = 1 + 4 = 5
The corresponding factor is x2 – 2x + 5
Hence x4 – 4x3 + 11x2 –14x + 10 = (x2 – 2x + 5) (x2 + px + 2)
Equating the coefficient of x
-14 = -4 + 5p
p = -2
x2 – 2x + 2 = 0
x = = = 1 i
The roots are 1 2i and 1 i.
30.
Solve: 6x4 – 25x3 + 32x3 + 3x – 10 = 0 given that one of the roots is 2 – i.
- Solution:
Since 2 – i is a root 2 + i is also a root.
Sum of the roots = 4
Product of the roots = 4 + 1 = 5
The corresponding factor is x2 – 4x + 5
Hence 6x4 – 25x3 + 32x2 + 3x – 10 = (x2 – 4x + 5) (6x2 + px – 2)
Equating the coefficient of x.
3 = 5p + 8
p = -1
6x2 – x – 2 = 0
x = = = or -
The roots are 2 i, - ,
31.
Simplify:
- Solution:
== = =
= cos (-107 ) – i sin(-107 )
32.
Simplify:
- Solution:
=
= = = since =
=
= cos (3 + 4 ) + isin(3 + 4 )
33.
If cos + cos + cos = 0 = sin + sin + sin , prove that cos 3 + cos 3 + cos 3 = 3 cos( + + )
- Solution:
Given cos + cos + cos = 0
sin + sin + sin = 0
cos + cos + cos +i (sin + sin + cos ) = 0
(cos + i sin ) + (cos + i sin ) + (cos + i sin ) = 0
Let a = cos + i sin ; b = cos + i sin ; c = cos + i sin
Then, a + b + c = 0
We know that if a + b + c = 0, then a3 + b3 + c3 = 3 abc
(cos + i sin )3 + (cos + i sin )3 + (cos + i sin )3
= 3(cos + i sin ) (cos + i sin ) (cos + i sin )
cos 3 + i sin 3 + cos 3 + i sin 3 + cos 3 + i sin 3
= 3[cos( + + ) + i sin ( + + )]
Equating real and imaginary parts, we have
cos 3 + cos 3 + cos 3 = 3 cos( + + )
and sin 3 + sin 3 + sin 3 = 3 sin ( + + )
34.
If cos + cos + cos = 0 = sin + sin + sin , prove that sin 3 + sin 3 + sin 3 = 3 sin( + + )
- Solution:
Given cos + cos + cos = 0
sin + sin + sin = 0
cos + cos + cos +i (sin + sin + cos ) = 0
(cos + i sin ) + (cos + i sin ) + (cos + i sin ) = 0
Let a = cos + i sin ; b = cos + i sin ; c = cos + i sin
Then, a + b + c = 0
We know that if a + b + c = 0, then a3 + b3 + c3 = 3 abc
(cos + i sin )3 + (cos + i sin )3 + (cos + i sin )3
= 3(cos + i sin ) (cos + i sin ) (cos + i sin )
cos 3 + i sin 3 + cos 3 + i sin 3 + cos 3 + i sin 3
= 3[cos( + + ) + i sin ( + + )]
Equating real and imaginary parts, we have
cos 3 + cos 3 + cos 3 = 3 cos( + + )
and sin 3 + sin 3 + sin 3 = 3 sin ( + + )
35.
If cos + cos + cos = 0 = sin + sin + sin , prove that cos 2 + cos 2 + cos 2 = 0
- Solution:
a = cos + i sin
= cos - i sin
b = cos + i sin
= cos - i sin
c = cos + i sin
= cos - i sin
Hence
++ = (cos + cos + cos ) – i(sin + sin + sin )
= 0
And so = 0 ab + bc + ca = 0
Consider (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
0 = a2 + b2 + c2 + 2(ab + bc + ca)
0 = a2 + b2 + c2 + 0
a2 + b2 + c2 = 0
(cos + isin )2 +(cos + i sin )2 + (cos + i sin )2 = 0
cos 2 + isin 2 + cos 2 + i sin 2 + cos 2 + i sin 2 = 0
Equating real and imaginary parts, we get
cos 2 + cos 2 + cos 2 = 0
sin 2 + 2 sin 2 + sin 2 = 0
36.
If cos + cos + cos = 0 = sin + sin + sin , prove that sin 2 + sin 2 + sin 2 = 0
- Solution:
a = cos + i sin
= cos - i sin
b = cos + i sin
= cos - i sin
c = cos + i sin
= cos - i sin
Hence
++ = (cos + cos + cos ) – i(sin + sin + sin )
= 0
And so = 0 ab + bc + ca = 0
Consider (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
0 = a2 + b2 + c2 + 2(ab + bc + ca)
0 = a2 + b2 + c2 + 0
a2 + b2 + c2 = 0
(cos + isin )2 +(cos + i sin )2 + (cos + i sin )2 = 0
cos 2 + isin 2 + cos 2 + i sin 2 + cos 2 + i sin 2 = 0
Equating real and imaginary parts, we get
cos 2 + cos 2 + cos 2 = 0
sin 2 + 2 sin 2 + sin 2 = 0
37.
If cos + cos + cos = 0 = sin + sin + sin , prove that cos2
+ cos2
+ cos2
= sin2
+ sin2
+ sin2
=
- Solution:
cos2
+ cos2
+ cos2
=
=
But cos 2 + cos2 + cos2 = 0
Hence cos2
+ cos2
+ cos2
=
and in the same way
sin2
+ sin2
+ sin2
= 1 - cos2
+ 1- cos2
+ 1 - cos2
= 3 – (cos2 + cos2 + cos2
)
= 3 - =
38.
Prove that (1 + i)n + (1 - i)n =
- Solution:
Let 1 + i = r(cos + i sin )
r = =
cos = ; sin =
=
1 + i =
Replace 'i' by –i
1 – i =
Now, (1 + i)n + (1 – i)n
= +
= 2
= 2 = 2.2n/2 cos = 2 cos
39.
Prove that (1 + + (1 - = 2n+1cos deduce the value when n = 9
- Solution:
First express 1 + iand 1 - iin the modulus and amplitude form.
Let 1 + i= r(cos + i sin ). Equating real and imaginary parts, we get
r cos = 1
r sin = r = = 2
sin =
=
1 + = 2
Let 1 - i = r (cos + i sin )
r cos = 1
r sin = -
r = = 2
cos =
sin =
= -
1 -i =
L.H.S. = (1 -i)n + (1 - i)n
= 2n
= 2n
= 2n. 2 cos
= 2n+1 cos
= R.H.S
When n = 9, we get
= 29+1 cos
= 210 cos 3 = 210(-1) = -1024
40.
Prove that (1 + cos + i sin )n + (1 + cos - isin )n = 2n+1 cosn cos
- Solution:
L.H.S. = (1 + cos + isin )n + (1 + cos - isin
=
= 2ncosn
= 2ncosn
= 2ncosn2
= 2n+1cosn = R.H.S.
41.
Prove that (1 + i)4n and (1 + i)4n + 2 are real and imaginary respectively.
- Solution:
(1 + i)4n =
= [cos n + i sin n ] = 22n cos n Since sin n = 0
= purely real
(1 + i)4n+2 =
=
=
= 22n+1 cos n = purely imaginary
42.
If and are the roots of the equation x2 – 2px +(p2 + q2) = 0 and tan = show that = qn-1 .
- Solution:
x2 – 2px + (p2 + q2) = 0
Let and be the roots of the equation
Then + = 2p
= p2 + q2
Also =
= p iq
= p +iq and
= p - iq
And - = 2 iq ………….. (1)
Consider, =
= ………(2)
Let y + p = r cos
q = r sin
tan =
(y + p + iq)n = (r cos + r i sin )n
= rn[cos + i sin ]n
= rn[cosn + i sin n ]
Similarly, [(y + p)- iq]n = rn[cosn - i sin n ]
Hence substitute in (2),
= …….(3)
Now, r2 = r2 cos2
+ r2 sin2
= (y + p)2+ q2
= [(y + p)2 + q2]n/2 = qn
= qn [cot2
+1]n/2
= qn (cosec2
)n/2 =
Substitute in (3)
= .
= qn-1
43.
If x + = 2 cos prove that xn + = 2cos
- Solution:
x + = 2 cos
x2 – 2x cos + 1 = 0
x =
= cos + i sin
i.e., x = cos + i sin
= cos - i sin
and so x + = 2 cos which is given.
Therefore xn = (cos + i sin )n
xn = cosn + i sinn
= cos n - i sin n
xn + = 2cos n
and xn - = 2isin n
44.
If x + = 2 cos prove that xn - - 2i sin n
- Solution:
x + = 2 cos
x2 – 2x cos + 1 = 0
x =
= cos + i sin
i.e., x = cos + i sin
= cos - i sin
and so x + = 2 cos which is given.
Therefore xn = (cos + i sin )n
xn = cosn + i sinn
= cos n - i sin n
xn + = 2cos n
and xn - = 2isin n
45.
If x + = 2 cos and y + = 2cos , show that = 2 cos(m - n )
- Solution:
x + = 2 cos
x = cos + i sin
Similarly y + = 2cos
y = cos + i sin
Now, = = =
= = cos(m - n ) + i sin(m - n )
Hence = cos(n - m ) + i sin(n - m ) = + cos(m - n ) – i sin (m - n )
Therefore = 2cos(m - n )
and = 2isin(m - n )
46.
If x + = 2 cos and y + = 2cos , show that = 2isin(m - n )
- Solution:
x + = 2 cos
x = cos + i sin
Similarly y + = 2cos
y = cos + i sin
Now, = = =
= = cos(m - n ) + i sin(m - n )
Hence = cos(n - m ) + i sin(n - m ) = + cos(m - n ) – i sin (m - n )
Therefore = 2cos(m - n )
and = 2isin(m - n )
47.
If x = cos + i sin ; y = cos - i sin prove that
xm yn + = 2 cos(m + n )
- Solution:
Let x = cos + i sin
Then, = cos - i sin
y = cos + i sin
= cos - i sin
x + = 2 cos
y + = 2 cos
x = cos + i sin
y = cos + i sin
xm = (cos + i sin )m = cos m + i sin m
yn = (cos + i sin )n = cos n + i sin n
=
= cos(m - n ) + i sin(m - n )
= cos(m - n ) – i sin(m - n )
Adding
= 2 cos(m - n )
One of the values: 2 cos(m - n )
48.
If a = cos2 + i sin 2 , b = cos 2 + i sin 2 and c = cos 2 + i sin 2 prove that
= 2 cos( + + )
- Solution:
abc = (cos 2 i + sin 2 ) (cos 2 + i sin 2 ) (cos 2 + i sin 2 )
= cos 2( + + ) + i sin 2( + + )
= [cos 2( + + ) + i sin 2( + + )]1/2
= cos ( + + ) + i sin ( + + )
= [cos ( + + ) + i sin( + + )]-1
= cos( + + ) – i sin( + + )
+ = 2 cos( + + )
49.
If a = cos2 + i sin 2 , b = cos 2 + i sin 2 and c = cos 2 + i sin 2 prove that
= 2 cos 2( + - )
- Solution:
=
=
= cos 2( + - ) + i sin 2( + - )
= = cos 2( + - ) – i sin 2( + - )
= 2 cos 2( + - )
i.e. = 2 cos 2( + - )
50.
Find all the values of the following: i1/3
- Solution:
i1/3 = since i =
=
= cos k = 0 , 1 ,2
The values are cos + i sin.cos+ i sin, cos
51.
Find all the values of the following:(8i)1/3
- Solution:
(8i)1/3 = (8)1/3 (i)1/3
= 2 (i)1/3
The different values are 2 and
52.
Find all the values of the following:(-.
- Solution:
(-
Let -- i = r(cos + i sin )
- = r cos
r = = 2
-1 = r sin
cos = - = - += -
sin = -
(- - i ) = 2
(- - i )2/3 = 22/3
= (
= (
= 21/3
= 21/3
The values are 21/3
21/3
and 21/3
53.
If x = a + b, y = a + b
2, z = a
2 + b show that xyz = a3 + b3
- Solution:
x = a + b
y = a + b
2
z = a
2 + b
where is the complex cube root of unity.
That is, 1 + +
2 = 0 and
3 = 1
xyz = (a + b) (a + b
2) (a
2 + b )
= (a2
+ ab + ab
2 + b2 2 ) (a
2 + b )
= a3 3 + a2b
3+ a2b
4 + ab2 4+ a2b
2 + ab2 2 + ab2 3 + b3 3
= a3 + a2b(1 + +
2) + ab2( +
2 + 1) + b3
= a3 + b3
54.
If x = a + b, y = a + b
2, z = a
2 + b show that x3 + y3 + z3 = 3(a3+b3) where is the complex cube root of unity.
- Solution:
x3 + y3 + z3 = 3(a3 + b3)
Consider x + y + z = a + b + a + b
2 + a
2 + b
= a(1 + +
2) + b(1 + +
2)
= 0 + 0 = 0
If (x + y + z) = 0 then x3 + y3 + z3 = 3 xyz
Hence x3 + y3 + z3 = 3(a2 + b2) from previous problem.
55.
Prove that if
3 = 1, then (a + b + c)(a + b + c
2)(a + b
2 + c ) = a3 + b3 + c3 – 3abc
- Solution:
LHS = (a + b + c)(a + b + c
2)(a + b
2 + c )
= (a2 + ab + ac + ab + b2
+ bc + ac
2 + bc
2 + c2 2) (a + b
2 + c )
= [a2 + ab(1 + ) + ac(1 +
2) + b2
+ bc( +
2) + c2 2][a + b
2 + c ]
= (a2 - ab
2 - ac + b2
- bc + c2 2](a + b
2 + c )
= a3 - a2b
2 – a2c + ab2
- abc + ac2 2 + a2b
2 – ab2 4 - abc
3 + b3 3- b2 c
2 + bc2 4 + a2 c - abc
3 – ac2 2 + b2c
2 - bc2
+ c3 3
= a3 + b3 + c3 – 3abc = R.H.S
56.
Prove that if
3 = 1, then = -1
- Solution:
= and =
2
LHS =
5 + (
2)5
=
5 +
10
=
2.
3 +
3.
3.
3.
=
2+ = -1
= R.H.S. (since 1 + +
2 = 0)
57.
Prove that if
3 = 1, then
- Solution:
LHS = -+
=
=
=
= = 0 = R.H.S
58.
Solve: x4 + 4 = 0
- Solution:
x4 + 4 = 0
x4 = - 4 = 4(-1) = 4(cos + isin )
x = 41/4[ cos + i sin ]1/4
= (22)1/4 [cos (2k + ) + i sin(2k + )]1/4
= , k = 0,1,2,3
The values are
59.
Solve: x4 – x3 + x2 – x + 1 = 0
- Solution:
x4 – x3 + x2 - x + 1 = 0
= 0
Hence x5 = -1 = cos + i sin
x = [cos(2k + ) + i sin(2k + )]1/5
= cos , k = 0, 1, 3, 4
The values are cos
cos
cos
cos
[because x = -1 = cos , does not satisfy …………….(1)
60.
Solve:x9 + x5 – x4 – 1 = 0
- Solution:
x9 + x5 – x4 – 1 = 0
x5 (x4+ 1) – (x4 + 1) = 0
(x4+ 1) (x5- 1) = 0
x4 + 1 = 0
x = (-1)1/4
= (cos + isin )1/4
= [cos(2k + ) + isin(2k + )]1/4
= cos
k = 0,1,2,3
x5- 1 = 0
x = (1)1/5 = (cos 0 – i sin 0)1/5
= (cos 2k + i sin 2k )1/5
= cos where k = 0,1,2,3,4
The values are cos ,,, ,
, ,
61.
Find all the values of and hence prove that the product of the values is l.
- Solution:
Let -i = r(cos + i sin )= r cos
r = = 1
- = r sin
Hence cos =
sin = -
= -
-i = cos
=
=
=
= [cos(- ) + i sin ( )]1/4
= [cos(2k - ) + i sin(2k - )]1/4
= cos
where k = 0,1,2,3
The values are cos
cos
cos
cos
62.
Write the following as complex numbers
(i) (ii) 3-.
- Solution:
(i) = . = i
(ii) 3-= 3- = 3- = 3-i.
63.
Write the real and imaginary parts of the following numbers:
(i) 4-i (ii).
- Solution:
Let z =4-i ; Re(z) = 4, Im(z) = -
Let z = ; Re(z) = 0, Im(z) = .
64.
Find the complex conjugate of (i) 2+i, (ii) -4-i9 (iii)
- Solution:
By definition , the complex conjugate is obtained by reversing the sign of the imaginary part of the complex number. Hence the required conjugate are (i) 2-i, (ii) -4+i9 and (iii) ( the conjugate of any real number is itself).
65.
Express the following in the standard form of a + ib
(i) (3+2i) + (-7-i)
(ii) (8-6i)-(2i-7)
(iii) (2-3i)(4+2i)
(iv) .
- Solution:
(3+2i)+(-7-i)=3+2i-7-i=-4+i
(8-6i)-(2i-7)=8-6i-2i+7=15-8i
(2-3i)(4+2i)=8+4i-12i-6i =14-8i
Note : i
i
i
(i)4n= 1
(i)4n-1 = -i
(i).
66.
Find the real and imaginary parts of the complex number
Z=.
- Solution:
z=
=
=
=
=
=
Re(z) = - and Im(z) =-.
67.
If z1 = 2+i, z = 3-2i and z= and find the conjugate of zz.
- Solution:
Conjugate of zz is
i.e.,
= (2-i) (3+2i)
= (2-i) (3+2i)
= 6+4i-3i-2i=6+4i-3i+2
= 8+i.
68.
If z1 = 2+i, z = 3-2i and z= and find the conjugate of (z).
- Solution:
= ()=
=
=.
69.
Find the modulus of the following complex numbers:
(i) -2+4i (ii) 2-3i (iii)-3-2i (iv) 4+3i.
- Solution:
(i)
(ii)
(iii)
(iv) .
70.
Find the modulus or the absolute value of .
- Solution:
=
71.
Find the modulus and argument of the following complex numbers.
- .
- Solution:
Let -)
Equating the real and imaginary parts separately
rcos
r
sin,r
r
r=
cos, sin quadrant
modulus r =2, arguments
Hence -.
72.
Find the modulus and argument of the following complex numbers 1+i.
- Solution:
Let 1+i
Equating the real and imaginary parts separately
rcos rsin
=1 , r
r
cos
sin, in the 1 quadrant
modulus r =2, argument =
Hence 1 +i.
73.
Find the modulus and argument of the following complex numbers -1-i.
- Solution:
Let -1-i
Equating the real and imaginary parts separately
rcos=-1
r=1
rsin
r
r(cos)=4
r=2
cos= -,sin=- in the 3rd quadrant
modulus r=2, argument =
Hence -1-i.
74.
If (a+ib1)(a + ib)…..(a+ ib) = A + iB
Prove That (i) (a+b)(a+b)……(a+b)=A+B
(ii) tan+tan+……..tan=k+tan,kZ.
- Solution:
Given (a+ib)(a+ib)……(a+ib)=A+iB
squaring on both sides
(a+b)(a
also
arg[(a+ib)(a+ib) ……..(a+ib)] =arg(A+iB)
arg (a+ib) + arg(a+ib) ………+arg(a+ib)=arg(A+iB) ……..(1)
Now arg (a+ib)=tan
Hence (1) becomes
tantan+……….+tan=tan
By taking the general value.
tan.
75.
P represents the variable complex number z, find the locus of P if
Re.
- Solution:
Let z = x + iy Then
Given that Re
x+y
x - y = 1 which is a straight line.
The locus of P is a straight line.
76.
P represents the variable complex number z, find the locus of P if arg.
- Solution:
arg
arg(z-1)-arg(z+1) =
arg(x+iy-1)-arg(x+iy+1)=
arg[(x-1)+iy]-arg[(x+1)+iy] =
tan
tan
is the required locus.
77.
Graphically prove that .
- Solution:
By triangle inequality in
……..(1)
By triangle inequality in
from (1)
.
78.
Prove that the complex numbers 3+3i, -3-3i,-3I are the vertices of an equilateral triangle in the complex plane.
- Solution:
Let A,B and C represent the complex numbers (3 + 3i),(-3-3i) and
(-3in the argand diagram.
AB =
=
BC =
=
CA =
=
AB=BC=CA
is an equilateral triangle.
79.
Prove that the points representing the complex numbers 2i, 1+ i ,4 + 4i and 3 + 5i on the Argand plane are the vertices of a rectangle.
- Solution:
Let A, B, C and D represents the complex number 2i,(1+i),(4+4i) and (3+5i) in the argand diagram respectively.
Fig3.14
AB =
= =
BC =
=
=
CD =
=
DA =
AB=CD and BC=DA
AC =
=
=
AB
AC
Hence AB
As pairs of opposites sides are equal and =90 ABCD is a rectangle.
80.
Show that the points representing the complex numbers 7 + 9i, -3 + 7i, 3 + 3i form a right angled triangle on the argand diagram.
- Solution:
Let A, B and C represents the complex numbers
7+9i, -3+7i and 3+3i in the Argand diagram respectively.
AB =
=
=
BC=
=
=
CA =
=
=
AB
=90
Hence is a right angled isosceles triangle.
81.
Find the square root of (-7 + 24i).
- Solution:
Let
On squaring,
-7+24i = (x)+2ixy
Equating the real and imaginary parts
x and 2xy=24
x
=
Solving, x and x
We get x and y=16
x = 3 and y = 4
Since xy is positive, x and y have the same sign.
(x = 3, y = 4) or (x = -3, y = -4)
= (3 + 4i) or (-3 - 4i).
82.
Solve the equation x - 4x, if one of its roots is 2+i.
- Solution:
Since 2 + i is a root, 2 - i is also a root.Sum of the roots = 4
Product of the roots (2+i)(2-i) =4+3=7
The corresponding factor is x
x
Equating x term, we get 8 = 7p-20
Other factor is (x
x
Thus the roots are 2 and -2i.
83.
Simplify:.
- Solution:
The given expression =
=
=e
=e
Alternative method:
The given expression =
= (cos
=(cos
=cos 31.
84.
Simplify: .
- Solution:
= cos
= cos
= cos
= cos
=sin9
Alternative method:
= -i(cos4
= -i[cos9
= sin9
Result : .
85.
If n is a positive integer, prove that
.
- Solution:
Let z=sin
=
= .
86.
If n is a positive integers, prove that .
- Solution:
Let
Equating real and imaginary parts separately , we have
r cos and r sin =1
r=
cos
Hence (
(
= 2 ………..(1)
To determine we replace I in the above result by –I we get
( …………….(2)
Adding (1) and (2) we have
(
= 2.
87.
If and are the roots of x
and cot show that .
- Solution:
The roots of the equation x are 1 i.
Let and
Then (y+)
= (cot)
=
=
(y+ =
(y+)-(y+)=
further
.
88.
Solve the equation
x.
- Solution:
x
= (x(x=0
= x
x = (1)
x = (1)
= (cos 2k
os
(-1)
={cos (2k+1)}
=cos
Thus we have 9 roots.
89.
Solve the equation x.
- Solution:
x
x
x =(-1)
= (cos)
i.e., = [cos(2k
=
(ii) x
= (cos
= [cos(2k
= cos (2k +1).
90.
Find all the values of ().
- Solution:
Let
rcos
r =
cos
(
=2
= 2
= 2
= 2 where k =0, 1, 2.