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Class 11 Maths
Complex Numbers And Quadratic Equations
Quiz 1
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16
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19
20
if z lies on |z| = 1, then 2/z lies on
0%
a circle
0%
an ellipse
0%
a straight line
0%
a parabola
Explanation
a circle
Hint:
Let w = 2/z
Now, |w| = |2/z|
=> |w| = 2/|z|
=> |w| = 2
This shows that w lies on a circle with center at the origin and radius 2 units.
The value of √(-16) is
0%
-4i
0%
4i
0%
-2i
0%
2i
Explanation
4i
Hint:
Given, √(-16) = √(16) × √(-1)
= 4i {since i = √(-1) }
The value of √(-144) is
0%
12i
0%
-12i
0%
±12i
0%
None of these
Explanation
12i
Hint:
Given, √(-144) = √{(-1) × 144}
= √(-1) × √(144)
= i × 12 {Since √(-1) = i}
= 12i
So, √(-144) = 12i
The value of √(-25) + 3√(-4) + 2√(-9) is
0%
13i
0%
-13i
0%
17i
0%
-17i
Explanation
17i
Hint:
Given, √(-25) + 3√(-4) + 2√(-9)
= √{(-1) × (25)} + 3√{(-1) × 4} + 2√{(-1) × 9}
= √(-1) × √(25) + 3{√(-1) × √4} + 2{√(-1) × √9}
= 5i + 3×2i + 2×3i {since √(-1) = i}
= 5i + 6i + 6i
= 17i
So, √(-25) + 3√(-4) + 2√(-9) = 17i
If ω is an imaginary cube root of unity, then (1 + ω – ω²)
7
equals
0%
128 ω
0%
-128 ω
0%
128 ω²
0%
-128 ω²
Explanation
-128 ω²
Hint:
Given ω is an imaginary cube root of unity.
So 1 + ω + ω² = 0 and ω³ = 1
Now, (1 + ω – ω²)
7
= (-ω² – ω²)
7
⇒ (1 + ω – ω²)
7
= (-2ω2)7
⇒ (1 + ω – ω²)
7
= -128 ω
14
⇒ (1 + ω – ω²)
7
= -128 ω
12
× ω³
⇒ (1 + ω – ω²)
7
= -128 (ω³)
4
ω³
⇒ (1 + ω – ω²)
7
= -128 ω²
The least value of n for which {(1 + i)/(1 – i)}
n
is real, is
0%
1
0%
2
0%
3
0%
4
Explanation
2
Hint:
Given, {(1 + i)/(1 – i)}
n
= [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]
n
= [{(1 + i)²}/{(1 – i²)}]
n
= [(1 + i² + 2i)/{1 – (-1)}]
n
= [(1 – 1 + 2i)/{1 + 1}]
n
= [2i/2]
n
= i
n
Now, in is real when n = 2 {since i2 = -1 }
So, the least value of n is 2
Let z be a complex number such that |z| = 4 and arg(z) = 5π/6, then z =
0%
-2√3 + 2i
0%
2√3 + 2i
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2√3 – 2i
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-√3 + i
Explanation
-2√3 + 2i
Hint:
Let z = r(cos θ + i × sin θ)
Then r = 4 and θ = 5π/6
So, z = 4(cos 5π/6 + i × sin 5π/6)
⇒ z = 4(-√3/2 + i/2)
⇒ z = -2√3 + 2i
The value of i
-999
is
0%
1
0%
-1
0%
i
0%
-i
Explanation
i
Hint:
Given, i
-999
= 1/i
999
= 1/(i
996
× i³)
= 1/{(i
4
)
249
× i³}
= 1/{1
249
× i³} {since i
4
= 1}
= 1/i³
= i
4
/i³ {since i
4
= 1}
= i
So, i
-999
= i
Let z
1
and z
2
be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z
1
and z
1
form an equilateral triangle. Then
0%
a² = b
0%
a² = 2b
0%
a² = 3b
0%
a² = 4b
Explanation
a² = 3b
Hint:
Given, z
1
and z
2
be two roots of the equation z² + az + b = 0
Now, z
1
+ z
2
= -a and z
1
× z
2
= b
Since z1 and z2 and z3 from an equilateral triangle.
⇒ z
1
² + z
2
² + z
3
² = z
1
× z
2
+ z
2
× z
3
+ z
1
× z
3
⇒ z
1
²+ z
2
² = z
1
× z
2
{since z
3
= 0}
⇒ (z
1
+ z
2
)² – 2z
1
× z
2
= z
1
× z
2
⇒ (z
1
+ z
2
)² = 2z
1
× z
2
+ z
1
× z
2
⇒ (z
1
+ z
2
)² = 3z
1
× z
2
⇒ (-a)² = 3b
⇒ a² = 3b
The complex numbers sin x + i cos 2x are conjugate to each other for
0%
x = nπ
0%
x = 0
0%
x = (n + 1/2) π
0%
no value of x
Explanation
no value of x
Hint:
Given complex number = sin x + i cos 2x
Conjugate of this number = sin x – i cos 2x
Now, sin x + i cos 2x = sin x – i cos 2x
⇒ sin x = cos x and sin 2x = cos 2x {comparing real and imaginary part}
⇒ tan x = 1 and tan 2x = 1
Now both of them are not possible for the same value of x.
So, there exist no value of x
The curve represented by Im(z²) = k, where k is a non-zero real number, is
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a pair of striaght line
0%
an ellipse
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a parabola
0%
a hyperbola
Explanation
a hyperbola
Hint:
Let z = x + iy
Now, z² = (x + iy)²
⇒ z² = x² – y² + 2xy
Given, Im(z²) = k
⇒ 2xy = k
⇒ xy = k/2 which is a hyperbola.
The value of x and y if (3y – 2) + i(7 – 2x) = 0
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x = 7/2, y = 2/3
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x = 2/7, y = 2/3
0%
x = 7/2, y = 3/2
0%
x = 2/7, y = 3/2
Explanation
x = 7/2, y = 2/3
Hint:
Given, (3y – 2) + i(7 – 2x) = 0
Compare real and imaginary part, we get
3y – 2 = 0
⇒ y = 2/3
and 7 – 2x = 0
⇒ x = 7/2
So, the value of x = 7/2 and y = 2/3
Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary
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θ = nπ ± π/2 where n is an integer
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θ = nπ ± π/3 where n is an integer
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θ = nπ ± π/4 where n is an integer
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None of these
Explanation
θ = nπ ± π/3 where n is an integer
Hint:
Given,
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 + 2i × sin θ)×(1 – 2i × sin θ)}/(1 – 4i² × sin² θ)
(3 + 2i × sin θ)/(1 – 2i × sin θ) = {(3 – 4sin² θ) + 8i × sin θ}/(1 + 4sin² θ) …………. 1
Now, equation 1 is imaginary if
3 – 4sin² θ = 0
⇒ 4sin² θ = 3
⇒ sin² θ = 3/4
⇒ sin θ = ±√3/2
⇒ θ = nπ ± π/3 where n is an integer
If {(1 + i)/(1 – i)}
n
= 1 then the least value of n is
0%
1
0%
2
0%
3
0%
4
Explanation
4
Hint:
Given, {(1 + i)/(1 – i)}n = 1
⇒ [{(1 + i) × (1 + i)}/{(1 – i) × (1 + i)}]
n
= 1
⇒ [{(1 + i)²}/{(1 – i²)}]
n
= 1
⇒ [(1 + i² + 2i)/{1 – (-1)}]
n
= 1
⇒ [(1 – 1 + 2i)/{1 + 1}]
n
= 1
⇒ [2i/2]
n
= 1
⇒ i
n
= 1
Now, in is 1 when n = 4
So, the least value of n is 4
If arg (z) < 0, then arg (-z) – arg (z) =
0%
π
0%
-π
0%
-π/2
0%
π/2
Explanation
π
Hint:
Given, arg (z) < 0
Now, arg (-z) – arg (z) = arg(-z/z)
⇒ arg (-z) – arg (z) = arg(-1)
⇒ arg (-z) – arg (z) = π {since sin π + i cos π = -1, So arg(-1) = π}
if x + 1/x = 1 find the value of x
2000
+ 1/x
2000
is
0%
0
0%
1
0%
-1
0%
None of these
Explanation
-1
Hint:
Given x + 1/x = 1
⇒ (x² + 1) = x
⇒ x² – x + 1 = 0
⇒ x = {-(-1) ± √(1² – 4 × 1 × 1)}/(2 × 1)
⇒ x = {1 ± √(1 – 4)}/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(-1)×√3}/2
⇒ x = {1 ± i√3}/2 {since i = √(-1)}
⇒ x = -w, -w²
Now, put x = -w, we get
x
2000
+ 1/x
2000
= (-w)
2000
+ 1/(-w)
2000
= w
2000
+ 1/w
2000
= w2000 + 1/w2000
= {(w³)
666
× w²} + 1/{(w³)
666
× w²}
= w² + 1/w² {since w³ = 1}
= w² + w³ /w²
= w² + w
= -1 {since 1 + w + w² = 0}
So, x
2000
+ 1/x
2000
= -1
If the cube roots of unity are 1, ω, ω², then the roots of the equation (x – 1)³ + 8 = 0 are
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-1, -1 + 2ω, – 1 – 2ω²
0%
– 1, -1, – 1
0%
– 1, 1 – 2ω, 1 – 2ω²
0%
– 1, 1 + 2ω, 1 + 2ω²
Explanation
– 1, 1 – 2ω, 1 – 2ω²
Hint:
Note that since 1, ω, and ω² are the cube roots of unity (the three cube roots of 1), they are the three solutions to x³ = 1 (note: ω and ω² are the two complex solutions to this)
If we let u = x – 1, then the equation becomes
u³ + 8 = (u + 2)(u² – 2u + 4) = 0.
So, the solutions occur when u = -2 (giving -2 = x – 1 ⇒ x = -1), or when:
u² – 2u + 4 = 0,
which has roots, by the Quadratic Formula, to be u = 1 ± i√3
So, x – 1 = 1 ± i√3
⇒ x = 2 ± i√3
Now, x³ = 1 when x³ – 1 = (x – 1)(x² + x + 1) = 0, giving x = 1 and
x² + x + 1 = 0
⇒ x = (-1 ± i√3)/2
If we let ω = (-1 – i√3)/2 and ω₂ = (-1 + i√3)/2
then 1 – 2ω and 1 – 2ω² yield the two complex solutions to (x – 1)³ + 8 = 0
So, the roots of (x – 1)³ + 8 are -1, 1 – 2ω, and 1 – 2ω²
(1 – w + w²)×(1 – w² + w
4
)×(1 – w
4
+ w
8
) × …………… to 2n factors is equal to
0%
2
n
0%
2
2n
0%
2
3n
0%
2
4n
Explanation
2
2n
Hint:
Given, (1 – w + w²)×(1 – w² + w
4
)×(1 – w
4
+ w
8
) × …………… to 2n factors
= (1 – w + w²)×(1 – w² + w )×(1 – w + w²) × …………… to 2n factors
{Since w
4
= w, w
8
= w²}
= (-2w) × (-2w²) × (-2w) × (-2w²)× …………… to 2n factors
= (2² w³)×(2² w³)×(2² w³) …………… to 2n factors
= (2²)
n
{since w³ = 1}
= 2
2n
The modulus of 5 + 4i is
0%
41
0%
-41
0%
√41
0%
-√41
Explanation
√41
Hint:
Let Z = 5 + 4i
Now modulus of Z is calculated as
|Z| = √(5² + 4²)
⇒ |Z| = √(25 + 16)
⇒ |Z| = √41
So, the modulus of 5 + 4i is √41
The complex number $ \frac {2 + 6 \sqrt 3 i}{5 + \sqrt 3 i} $ in polar form will be
0%
$2(cos \frac {\pi}{4} + sin \frac {\pi}{4})$
0%
$2(cos \frac {2\pi}{3} + sin \frac {2\pi}{3})$
0%
$2(cos \frac {\pi}{3} + sin \frac {\pi}{3})$
0%
$2(cos \frac {\pi}{6} + sin \frac {\pi}{6})$
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