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Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The value of \(\sqrt{(-16)}\) is

(a) -4i
(b) 4i
(c) -2i
(d) 2i

2. The value of \(\sqrt{(-144)}\) is

(a) -12i
(b) 12i
(c) -10i
(d) 2i

3.  The value of \(\sqrt{-25}+3\sqrt{-4}+2\sqrt{-9}\) is

(a) -17i
(b) 15i
(c) 17i
(d) 8i

4. if z lies on |z| = 1, then 2/z lies on

(a) a circle
(b) an ellipse
(c) a straight line
(d) a parabola

5. If ω is an imaginary cube root of unity, then (1 + ω – ω²)⁷ equals

(a) 128 ω
(b) -128 ω
(c) 128 ω²
(d) -128 ω²

6. The value of i¹³⁵ is

(a) 1
(b) -1
(c) i
(d) -i

7. Let z1 and z2 be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z1 and z1 form an equilateral triangle. Then

(a) a² = b
(b) a²= 2b
(c) a² = 3b
(d) a² = 4b

8. The curve represented by Im(z²) = k, where k is a non-zero real number, is

(a) a pair of striaght line
(b) an ellipse
(c) a parabola
(d) a hyperbola 

9. The value of x and y if (3y – 2) + i(7 – 2x) = 0

(a) x = 7/2, y = 2/3
(b) x = 2/7, y = 2/3
(c) x = 7/2, y = 3/2
(d) x = 2/7, y = 3/2

10. Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary

(a) θ = nπ ± π/2 where n is an integer
(b) θ = nπ ± π/3 where n is an integer
(c) θ = nπ ± π/4 where n is an integer
(d) None of these

11. If arg (z) < 0, then arg (-z) – arg (z) =

(a) π
(b) -π
(c) -π/2
(d) π/2

12. The modulus of 3 + 4i is

(a) 5
(b) 25
(c) 15
(d) -5

13. The modulus of 5 + 3i is

(a) \(\sqrt{34}\)
(b) 25
(c) 15
(d) -5

14. The value of 1+ i² + i⁴ + i⁶ + i⁸ + ........+ i²⁰

(a) 1
(b) 2
(c) -1
(d) 3

15. Number of solutions of the equation z²+|z|² =0  is

(a) 1
(b) 2
(c) 3
(d) infinitely many

16. If ∣z−4∣<∣z−2∣, its solution is given by

(a) Re(z)>0
(b) Re(z)<0
(c) Re(z)>3
(d) Re(z)>2

17. If a + ib = c + id, then

(a) a² + c² = 0
(b) b² + c² = 0
(c) b² + d² = 0
(d) a² + b² = c² + d²

18.  If a complex number z lies in the interior or on the boundary of a circle of radius 3 units and centre (– 4, 0), the greatest value of |z +1| is

(a) 4
(b) 6
(c) 3
(d) 10

19. If 1 – i, is a root of the equation x2 + ax + b = 0, where a, b ∈ R, then the value of a – b is

(a) -4
(b) 0
(c) 2
(d) 1

20. The simplified value of (1 – i)³/(1 – i³) is

(a) 1
(b) -2
(c) -i
(d) 2i

Answer:

1. Answer: (b) 4i

Explanation: Given,

  \(=\sqrt{(16)}\times\sqrt{(-1)}\)

= 4i {since i = \(\sqrt{(-1)}\)}

2. Answer: (b) 12i

Explanation: Given,

  \(=\sqrt{(144)}\times\sqrt{(-1)}\)

= 12i {since i = \(\sqrt{(-1)}\)}

3.  Answer: (c) 17i

Explanation: \(\sqrt{-25}+3\sqrt{-4}+2\sqrt{-9}\)

\(=5\sqrt{-1}+6\sqrt{-1}+6\sqrt{-1}\) \(\therefore \sqrt{-1}=i\)

=5i+6i+6i

= 17i

4. Answer: (a) a circle

Explanation: Let \(w=\frac{2}{z}\)

 \(w=2\overset{-}{z}\) .....(Multiplying and dividing by \(\overset{-}{z}\) and ∣z∣² =1

Now locus of w represents a circle.

5. Answer: (d) -128 ω²

Explanation: ​​​​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 ω¹⁴

⇒ (1 + ω – ω²)⁷ = -128 ω¹² × ω²

⇒ (1 + ω – ω²)⁷ = -128 (ω³)⁴ ω²

⇒ (1 + ω – ω²)⁷ = -128 ω²

6. Answer: (d) -i

Explanation: 135 leaves remainder as 3 when it is divided by 4. 

∴ i¹³⁵ = i³ = -i

7. Answer: (c) a² = 3b

Explanation:  Given, z₁ and z₂ be two roots of the equation z₂ + az + b = 0

Now, z₁ + z₂ = -a and z₁ × z₂ = b

Since z₁ and z₂ and z₃ from an equilateral triangle.

⇒ z₁² + z₂² + z₃² = z₁ × z₂ + z₂ × z₃ + z₁ × z₃

⇒ z₁²+ z₂² = z₁ × z₂ {since z₃ = 0}

⇒ (z₁ + z₂)² – 2z₁ × z₂ = z₁ × z₂

⇒ (z₁ + z₂)² = 2z₁ × z₂ + z₁ × z₂

⇒ (z₁ + z₂)²  = 3z₁ × z₂

⇒ (-a)² = 3b

⇒ a² = 3b

8. Answer: (d) a hyperbola 

Explanation: 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.

9. Answer: (a) x = 7/2, y = 2/3

Explanation: Given, (3y – 2) + i(7 – 2x) = 0

3y – 2 = 0

⇒ y = 2/3 and 

7 – 2x = 0

⇒ x = 7/2

So, the value of x = 7/2 and y = 2/3

10. Answer: (a) x = 7/2, y = 2/3

Explanation: 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) …………. (i)

3 – 4sin² θ = 0

⇒ 4sin² θ = 3

⇒ sin² θ = 3/4

⇒ sin θ = \(\pm\frac{\sqrt3}{2}\)

⇒ θ = nπ ± π/3 where n is an integer

11. Answer: (a) π

Explanation: 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}

12. Answer: (a) 5

Explanation: \(\sqrt{3^2+4^2}\)

= 5

13. Answer: (a) 

Explanation: \(\sqrt{5^2+3^2}\)

= \(\sqrt{34}\)

14. Answer: (a) 1

Explanation: 1+ i² + i⁴ + i⁶ + i⁸ + i¹⁰ + i¹² + i¹⁴ + i¹⁶ + i¹⁸.+ i²⁰

= 1+(−1)+1+(−1)+1+(−1)+1+(−1)+1+(−1)+1

= 1

15. Answer: (d) infinitely many

Explanation: z²+|z|²=0 ,

z ≠ 0

⇒x²−y² +i²xy+x²+y² =0

⇒2x²+i²xy = 0

⇒2x(x+iy) = 0

⇒x=0 or x+iy=0 (not possible)

∴x=0 and z≠0

So y can have any real value.

Hence infinitely many solutions.

16. Answer: (c) Re(z)>3

Explanation: Given ∣z−4∣<∣z−2∣ all complex numbers lie farther from (2,0) than from (4,0) as

∣z−4∣<∣z−2∣

So they lie on the side of (z,0) which is closer to (4,0) than (2,0)

z = x+iy

⇒ x>3

∴Re(z)>3 

17. Answer: (d) a² + b² = c² + d²

Explanation: a + ib = c + id

⇒ |a + ib| = |c + id|

⇒ \(\sqrt{(a^2+b^2)}\) = \(\sqrt{(c^2+d^2)}\)

Squaring on both sides, we get;

a² + b² = c² + d²

18.  Answer: (b) 6

Explanation: The distance of the point representing z from the centre of the circle is |z – (-4 + i0)| = |z + 4|

|z + 4| ≤ 3

|z + 1| = |z + 4 – 3| ≤ |z + 4| + |-3| ≤ 3 + 3 ≤ 6

Hence, the greatest value of |z + 1| is 6.

19. Answer: (a) -4

Explanation: Given that 1 – i is the root of x² + ax + b = 0.

Thus, 1 + i is also the root of the given equation since non-real complex roots occur in conjugate pairs.

Sum of roots = −a/1 = (1 – i) + (1 + i)

⇒ a = – 2

Product of roots, b/1 = (1 – i)(1 + i)

b = 1 – i²

b = 1 + 1 {since i2 = -1}

⇒ b = 2

Now, a – b = -2 – 2 = -4

20. Answer: (b) -2

Explanation: (1 – i)³/(1 – i3)

= (1 – i)³/(1³ – i³)

= (1 – i)³/ [(1 – i)(1 + i + i²)]

= (1 – i)²/(1 + i – 1)

= (1 – i)²/i

= (1 + i² – 2i)/i

= (1 – 1 – 2i)/i

= -2i/i

= -2

Click here to practice MCQ Questions for Complex Numbers and Quadratic Equations 11

Given: A + iB = (a + ib)(c + id)(e + if)(g + ih) 

∴ A-iB = (a- ib)(c – id)(e – if)(g – ih) 

But A²  + B²  = (A + iB)(A – iB) 

= (a²  + b²)(c² + d²)(c² + f²)(g² + h²)

Let Z = 2i = r(cosθ + isinθ)

Now, separating real and complex part, we get

0 = rcosθ ……….eq.1

2 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

4 = r²

Since r is always a positive no., therefore,

r = 2,

Hence its modulus is 2.

Now, dividing eq.2 by eq.1, we get,

\(\frac{rsin\theta}{rcos\theta}=\frac{2}{0}\)

Tanθ = ∞

Since cosθ = 0, sinθ = 1 and tanθ = ∞.

Therefore the θ lies in first quadrant.

tanθ = ∞, therefore θ = π/2

Representing the complex no. in its polar form will be

Z = 2{cos(π/2)+i sin(π/2)}

Given |(z - 5i)/(z + 5i)| = 1.

⇒ |z – 5i| = |z + 5i|

⇒ |x + iy – 5i| = |x + iy + 5i|

⇒ |x + i (y – 5) |² = |x + i (y + 5) |²

⇒ x² + (y – 5)² = x² + (y + 5)²

⇒ 20y = 0

⇒ y = 0

∴ z lies on the x – axis.

Let, z = -2i

Let 0 = r cosθ and -2 = r sinθ

By squaring and adding, we get

(0)² + (-2)² = (r cosθ)² + (r sinθ)²

⇒ 0+4 = r² (cos²θ + sin²θ)

⇒ 4 = r²

⇒ r = 2

∴ cosθ = 0 and sinθ = -1

Since, θ lies in fourth quadrant, we have

θ = -π/2

Since, θ ∈ (-π ,π ] it is principal argument.

A. |z + 3|³

Given (z + 3) (z̅ + 3) = (x + iy + 3) (x –iy + 3)

= (x + 3)² – (iy)²

= (x + 3)² + y²

= |x + 3 + iy|²

= |z + 3|²

ω is the complex cube root of unity.

ω³ = 1 and 1 + ω + ω² = 0

Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(1 – ω – ω²)³ + (1 – ω + ω²)³

= [1 – (ω + ω²)]³ + [(1 + ω ) – ω²]³

= [1 – (-1)²] + (-ω – ω)³

= 2³ + (-2ω)³

= 8 – 8ω³

= 8 – 8(1)

= 0

ω is the complex cube root of unity.

ω³ = 1 and 1 + ω + ω² = 0

Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

(1 + ω²)³

= (-ω)³

= -ω³

= -1

ω is the complex cube root of unity.

ω³ = 1 and 1 + ω + ω² = 0

Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

ω² + ω³ + ω⁴ = ω² (1 + ω + ω² )

= ω² (0)

= 0

ω is the complex cube root of unity.

ω³ = 1 and 1 + ω + ω² = 0

Also, 1 + ω² = -ω, 1 + ω = -ω² and ω + ω² = -1

L.H.S. = (2 – ω)(2 – ω²)

= 4 – 2ω² – 2ω + ω³

= 4 – 2(ω² + ω) + 1

= 4 – 2(-1) + 1

= 4 + 2 + 1

= 7

= R.H.S.

Given i + i² + i³ + …. Upto 1000 terms

= (i + i² + i³ + i⁴) + (i⁵ + i⁶ + i⁷ + i⁸) + … 250 brackets

Since iⁿ + iⁿ⁺¹ + i^{n + 2} + i^{n + 3} = 0 where n ∈ N

= 0 + 0 + 0 … + 0

= 0

As we know that, z = a+ib

z = (1+i\(\sqrt{3}\))²

= 1² + \((\sqrt{3i})^2\) + 2 x 1 x \(\sqrt{3i}\) 

= 1+i²+2i√3 

= 1-3+2i√3 

= -2+2i√3 

a = -2 b = 2√3

tanα = \(|\frac{b}{a}|\) 

= \(|\frac{2\sqrt{3}}{-2}|\)

= |√3|

= α = \(\frac{\pi}{3}\)  or 60°

α<0,b>1 

∴z lies in second quadrant 

arg(z) = θ

=π - a

= π = - \(\frac{π}{3}\)

= \(\frac{2π}{3}\)

Given, 

6i − i √−49 = 6i – i(7i) 

= 6i –7i² 

= 6i –7 

∴ Additive inverse = -6 – 7

A complex number z is said to be: 

•  Purely real, if Im(z) = 0, 
•  Purely imaginary, if Re(z) = 0

D. |z₁ + z₂| ≥ ||z₁| - |z₂|

Let z₁ = |z₁| (cos θ₁ + i sin θ₁) and z₂ = |z₂| (cos θ₂ + i sin θ₂)

Now, z₁z₂ = |z₁| |z₂| (cos θ₁ + i sin θ₁) (cos θ₂ + i sin θ₂)

= |z₁| |z₂| [cos θ₁ cos θ₂ + i sin θ₁ cos θ₂ + i cos θ₁ sin θ₂ + i² sin θ₁ sin θ₂]

= |z₁| |z₂| [cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

⇒ |z₁ z₂| = |z₁| |z₂|

And arg (z₁ z₂) = θ₁ + θ₂ = arg (z₁) + arg (z₂)

⇒ |z₁ + z₂| = |z₁| + |z₂| is true only when z₁, z₂ and O are collinear.

Also, |z₁ + z₂| ≥ ||z₁| - |z₂|

-2 = -2 + 0i

Let -2 + 0i = r(cos θ + i sin θ)

⇒ r cos θ = -2 ….(i)

r sin θ = 0 ………(ii)

⇒ tan θ = 0/-2  = 0

Hence, principal argument of -z is π.

Answer is (B)

statement |z₁ – z₂| ≤ |z₁| + |z₂| is true.

Given: (x – iy) (3 + 5i) is the conjugate of (-6 – 24i)

We know that,

Conjugate of – 6 – 24i = - 6 + 24i

∴ According to the given condition,

(x – iy) (3 + 5i) = -6 + 24i

⇒ x(3 + 5i) – iy(3 + 5i) = -6 + 24i

⇒ 3x + 5ix – 3iy – 5i2y = -6 + 24i

⇒ 3x + i(5x – 3y) – 5(-1)y = -6 + 24i [∵ i² = -1]

⇒ 3x + i(5x – 3y) + 5y = - 6 + 24i

⇒ (3x + 5y) + i(5x – 3y) = - 6 + 24i

Comparing the real parts, we get

3x + 5y = - 6 …(i)

Comparing the imaginary parts, we get

5x – 3y = 24 …(ii)

Solving eq. (i) and (ii) to find the value of x and y

Multiply eq. (i) by 5 and eq. (ii) by 3, we get

15x + 25y = -30 …(iii)

15x – 9y = 72 …(iv)

Subtracting eq. (iii) from (iv), we get

15x – 9y – 15x – 25y = 72 – (-30)

⇒ - 34y = 72 + 30

⇒ - 34y = 102

⇒ y = -3

Putting the value of y = -3 in eq. (i), we get

3x + 5(-3) = - 6

⇒ 3x – 15 = - 6

⇒ 3x = - 6 + 15

⇒ 3x = 9

⇒ x = 3

Hence, the value of x = 3 and y = - 3

Given:

x² + 5 = 0

⇒x² = -5

⇒ x = ±√(-5)

⇒ x = ±√5 i

Ans: x = ±√5 i

This equation is a quadratic equation.

Solution of a general quadratic equation ax² + bx + c = 0 is given by:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Given:

⇒x² + 2 = 0

⇒x² = -2

⇒x = ± √(-2)

But we know that √(-1) = i

⇒ x = ±√2 i

Ans: x = ±√2 i

Given: z = 2 – 3i

To Prove: z² – 4z + 13 = 0

Taking LHS, z² – 4z + 13

Putting the value of z = 2 – 3i, we get

(2 – 3i)² – 4(2 – 3i) + 13

= 4 + 9i² – 12i – 8 + 12i + 13

= 9(-1) + 9 = - 9 + 9 = 0

= RHS

Hence, z² – 4z + 13 = 0 …(i)

Now, we have to deduce 4z³ – 3z² + 169

Now, we will expand 4z³ – 3z² + 169 in this way so that we can use the above equation i.e. z² – 4z + 13

= 4z³ – 16z² + 13z² +52z – 52z + 169

Re – arrange the terms,

= 4z³ – 16z² + 52z + 13z² – 52z + 169

= 4z(z² – 4z + 13) + 13(z² – 4z + 13)

= 4z(0) + 13(0) [from eq. (i)]

= 0 = RHS

Hence Proved

We have, (1 – i) x + (1 + i) y = 1 – 3i

⇒ x - ix+y+iy = 1- 3i

⇒ (x+y)+i(-x+y) = 1 -3i

On equating the real and imaginary coefficients we get,

⇒ x+y = 1 (i) and –x+y = -3 (ii)

From (i) we get

x = 1 - y

Substituting the value of x in (ii), we get

-(1 - y)+y = -3

⇒ -1+y+y = -3

⇒ 2y = -3+1

⇒ y = -1

⇒ x = 1 - y = 1- (-1) = 2

Hence, x = 2 and y = -1

(1 – i) x + (1 + i) y = 1 – 3i

⇒ x – ix + y + iy = 1 – 3i

⇒ (x + y) – i(x – y) = 1 – 3i

Comparing the real parts, we get

x + y = 1 …(i)

Comparing the imaginary parts, we get

x – y = -3 …(ii)

Solving eq. (i) and (ii) to find the value of x and y

Adding eq. (i) and (ii), we get

x + y + x – y = 1 + (-3)

⇒ 2x = 1 – 3

⇒ 2x = -2

⇒ x = -1

Putting the value of x = -1 in eq. (i), we get

(-1) + y = 1

⇒ y = 1 + 1

⇒ y = 2

[i¹⁸ + (1/i)²⁵]³ = [i¹⁶. i² + (-i)²⁵]³

= [-1 - i]³ = - 1 - 3i + 3 + i = 2 - 2i

Given: z = (-5 – 2i)

Here, we have to find the conjugate of (-5 – 2i)

So, the conjugate of (- 5 – 2i) is (-5 + 2i)

Given complex number is Z=1+i 

We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ) 

Where, 

|Z|=modulus of complex number= \(\sqrt{x^2+y^2}\)

θ =arg(z)=argument of complex number= tan^-1 \(\Big(\frac{|y|}{|x|}\Big)\)

Now for the given problem,

⇒ |z| = \(\sqrt{(1)^2+(1)^2}\) 

⇒ |z| = \(\sqrt{1+1}\) 

⇒ |z| = \(\sqrt{2}\)

⇒ θ = tan^-1\(\Big(\frac{1}{1}\Big)\) 

Since x>0,y>0 complex number lies in 1^st quadrant and the value of θ will be as follows 0°≤θ≤90°.

⇒ θ = tan^-1(1)

⇒ θ = \(\frac{\pi}{4}\)

⇒  z = \(\sqrt{2}\Big(cos\Big(\frac{\pi}{4}\Big) + isin\Big(\frac{\pi}{4}\Big)\Big)\) 

∴ The Polar form of Z=1+i is z =\(\sqrt{2}\Big(cos\Big(\frac{\pi}{4}\Big) + isin\Big(\frac{\pi}{4}\Big)\Big)\) 

Let, (a + ib)² = 5 + 12i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)² + 2abi = 5 + 12i

Since i² = -1

a² - b² + 2abi = 5 + 12i

Now, separating real and complex parts, we get

⇒ a² - b² = 5…………..eq.1

⇒ 2ab = 12……..eq.2

⇒ a = 6/b

Now, using the value of a in eq.1, we get

⇒ (6/b)² – b² = 5

⇒ 36 – b⁴ = 5b²

⇒ b⁴ + 5b² - 36 = 0

Simplify and get the value of b², we get,

b² = - 9 or b² = 4

As b is real no. so, b² = 4

b = 2 or b = - 2

Therefore, a = 3 or a = - 3

Hence the square root of the complex no. is 3 + 2i and - 3 -2i.

Let, (a + ib)² = - 2 + 2√3 i

Now using, (a + b)² = a² + b² + 2ab

⇒ a² + (bi)² + 2abi = - 2 + 2√3 i

Since i² = -1

⇒ a² - b² + 2abi = - 2 + 2√3 i

Now, separating real and complex parts, we get

⇒ a² - b² = - 2…………..eq.1

⇒ 2ab = 2√3 ……..eq.2

⇒ a = √3/b

Now, using the value of a in eq.1, we get

⇒ (√3/b)² – b² = -2

⇒3 – b⁴ = -2b²

⇒ b⁴ - 2b² - 3= 0

Simplify and get the value of b², we get,

⇒ b² = -1 or b² = 3

As b is real no. so, b² = 3

b = √3 or b = -√3

Therefore, a = 1 or a = -1

Hence the square root of the complex no. is 1 + √3 i and -1 - √3 i.

Given: (1 + i) y² + (6 + i) = (2 + i)x

Consider, (1 + i) y² + (6 + i) = (2 + i)x

⇒ y² + iy² + 6 + i = 2x + ix

⇒ (y² + 6) + i(y² + 1) = 2x + ix

Comparing the real parts, we get

y² + 6 = 2x

⇒ 2x – y² – 6 = 0 …(i)

Comparing the imaginary parts, we get

y² + 1 = x

⇒ x – y² – 1 = 0 …(ii)

Subtracting the eq. (ii) from (i), we get

2x – y² – 6 – (x – y² – 1) = 0

⇒ 2x – y² – 6 – x + y² + 1 = 0

⇒ x – 5 = 0

⇒ x = 5

Putting the value of x = 5 in eq. (i), we get

2(5) – y² – 6 = 0

⇒ 10 – y² – 6 = 0

⇒ -y² + 4 = 0

⇒ - y² = -4

⇒ y² = 4

⇒ y = √4

⇒ y = ± 2

Hence, the value of x = 5 and y = ± 2

L.H.S = (1 + i¹⁰ + i²⁰ + i³⁰)

= (i + i^{4 x 2 + 2} + i^{4 x 5} + i^{4 x 7 + 2})

Since ⇒ i⁴ⁿ = 1

⇒ i⁴ⁿ⁺¹ = i

⇒ i⁴ⁿ⁺² = -1

⇒ i⁴ⁿ⁺³ = -1

= 1 + i² + 1 + i²

= 1 + -1 +1 + -1

= 0, which is a real no.

Hence, (1 + i¹⁰ + i²⁰ + i³⁰) is a real number.

Let. z = a + ib be a complex number. Then, the modulus of z, denoted by |z| to be the non-negative real number √(a² + b²)

Let z = a + ib be a complex number. Then, the conjugate of complex number z, denoted   as , is the complex number a – ib, 

i.e., i= a – ib

(i) root α = 5 and β = – 2

Then, sum of roots = α + β = 5 – 2

⇒ α + β = 3

and product of roots = αβ = 5 × (-2)

⇒ αβ = -10

Hence, required equation whose roots are 5 and -2.

x² – (sum of roots) x + product of roots = 0

⇒ x² – 3x + (-10) = 0

⇒ x² – 3x – 10 = 0

Hence, required equation is x² – 3x – 10 = 0 whose roots are 5 and -2.

(ii) Roots α = 1 + 2i and β = 1 – 2i

Then, sum of roots α + β = 1 + 2i + 1 – 2i = 2

and product of roots αβ = (1 + 2i)(1 – 2i) = 1 – 4i² = 1 + 4 = 5

Hence, required equation whose roots are 1 + 2i and 1 – 2i,

x² – (sum of roots) x + Product of roots = 0

⇒ x² – 2x + 5 = 0.

Remark: If one root is 1 + 2i, then second not will be 1 – 2i.

(i) -216 = (-6) (1)^1/3 = -6, -6ω, -6ω²

Hence, cube root of -216 = -6, -6ω, -6ω²

(ii) -512 = (-8) (1)^1/3 = -8, - 8ω, -8ω²

Hence, cube root of -512 = -8, -8ω, – 8ω²

False

It is a false statement because 2 = 2 + i(0), which is of the form α + ib, and the number of the form a + ib, where a and b are real numbers is called a complex number.

False

The given statement is false because the order relation “greater than” and “less than” are not defined on the set of complex numbers.

We know that 

i⁴ⁿ⁺¹ = i 

i⁴ⁿ⁺² = i² = -1 

i⁴ⁿ⁺³ = i³ = -i

i⁴ⁿ⁺⁴ = i⁴ = 1 

i⁵ + i⁶ + i⁷ + i⁸ + i⁹ = i + (-1) + (-i) + 1 + i 

= i

\(\frac{(i^5 + i^6 + i^7 + i^8 + i^9)}{(1 + i)}\) = \(\frac{1}{1+i}\)    

=\(\frac{1}{1+i}\times\frac{(1-i)}{(1-i)}\) 

= \(\frac{1}{2}(1+i)\)

(i) z = 5 – 2i, w = -1 + 3i 

z + w = (5 – 2i) + (-1 + 3i) 

= (5 – 1) + (-2i + 3i) 

= 4 + i

(ii) z – iw = (5 – 2i) – i (-1 + 3i) 

= 5 – 2i + i + 3 

= (5 + 3) + (-2i + i) 

= 8 – i 

(iii) 2z + 3w = 2(5 – 2i) + 3 (-1 +3i) 

= 10 – 4i – 3 + 9i 

= 7 + 5 i 

(iv) zw = (5 – 2i) (-1 + 3i) 

= -5 + 15i + 2i – 6i 

= -5 + 17i + 6 

= 1 + 17i 

(v) z² + 2zw + w² = (z + w)² [from (i)] 

= (4 + i)² 

= 16 – 1 + 8i 

= 15 + 8i 

(vi) (z + w)² 

= 15 + 8z [from (v)] 

Let √18i = a + bi, where a, b ∈ R.

Squaring on both sides, we get

18i = a² + b² i² + 2abi

0 + 18i = a² – b² + 2abi …..[∵ i²= -1]

Equating real and imaginary parts, we get

a² – b² = 0 and 2ab = 18

a² – b² = 0 and b = \(\frac9a\) 

a² - (\(\frac9a\))² = 0

a² - \(\frac{81}{a^2}\) = 0

 a⁴ – 81 = 0

(a² – 9) (a² + 9) = 0

a² = 9 or a² = -9

But a ∈ R

∴ a² ≠ -9

∴ a² = 9

∴ a = ± 3

When a = 3, b = 9/3 = 3

When a = -3, b = 9/-3 = -3

\(\therefore\) √18i = ±(3 + 3i) = ±3(1 + i)

Correct option   (c)  54

Explanation:

∴ (ω + 1/ω)² + (ω² + 1/ω²) + (ω² + 1/ω³)² .....+ (ω²⁷ + 1/ω²⁷)²

= (–1)² + (–1)² + (2)² + .................+(2)²

= 18 x 1 + 9 x 4 = 54

Let z₁= a + ib and z₂ = c + id be any two complex numbers. Then addition of z₁ and z₂ defined as z₁ + z₂ = (a + c) + i(b + d) which is again a complex number. 

Properties of addition of complex numbers: 

(i) Closure property: The sum of two complex numbers is always a complex number, i.e., addition is closed in C set of complex numbers.

(ii) Commutative property: For any two complex numbers z₁ and z₂ we have z₁ + z₂ = z₂ + z₁ 

(iii) Associative properly: For any complex numbers z₁, z₂ and Z₃, we have z₁ + (z₂ + z₃) = (z₁ + z₂) + z₃

(iv) Existence of additive identity: For any complex number z, we have z + 0 = 0 +-z = z 

Thus, 0 is the additive identity for complex numbers. 

(v) Existence of additive inverse: Every complex number z-a + ib has -z = (-a) + i(-b) as its additive inverse, as z + (-z) = (-z) + z = 0.

(i) We have, (z₁ + z₂)²  = (z, + z₂)(z₁ + z₂) = (z₁ + z₂)z₁ + (z₁ + z₂)z₂ (by distributive law)

= z²₁ + z₂z₁ +z₁z₂ + z₂ (by distributive law)

= z²₁ + 2z₁z₂ + z²₂

∵ z₁z₂ = z₂z₁

Let z₁ and z₂ be any two complex numbers. The different z₁ – z₂ is defined as follows: 

z₁ – z₂ =z₁ + (-z₂) 

Let z₁ = a + ib and z₂ = c + id be any two complex numbers. Then, the product z₁z₂ is defined as follows: 

z₁z₂ = (ac – bd) + i(ad + bc) 

Properties of product of complex numbers: 

• Closure property: The product of two complex numbers is always a complex number. 
• Commutative law: For any two complex numbers z₁ and z₂ we have z₁ z₂ = z₂z₁ 
• Associative law: For any three complex numbers z₁ z₂ and z₃ we have (z₁z₂)z₃ = z₁(z₂z₃). 
• Existence of multiplicative identity: For every complex number z, we have z . 1 = 1. z = z. Thus, 1 is the multiplicative identity. 
• Existence of multiplicative inverse: Let z = a + ib. Then

z^-1 = 1/z = 1/(a + ib) = 1/(a + ib) . (a - ib)/(a - ib) = (a - ib)/(a² + b²)

Clearly

z. 1/z = 1/z . z = 1

Thus, every z = a + ib has its multiplicative inverse, given

z^-1 = 1/z = a/(a² + b²) - a/(a² + b²)i (z ≠ 0)

Given:

⇒ a +ib =  \(\frac{2+3i}{4+5i}\)

Multiplying and dividing with 4-5i

⇒ a +ib =  \(\frac{2+3i}{4+5i}\) x  \(\frac{4-5i}{4-5i}\)

⇒ a +ib =  \(\frac{2(4-5i)+3i(4-5i)}{(4)^2-(5i)^2}\)

⇒ a +ib = \(\frac{8-10i+12i-15i^2}{16-25i^2}\) 

We know that i²=-1

⇒ a +ib = \(\frac{8+2i-15(-1)}{16-25(-1)}\) 

⇒ a +ib =  \(\frac{23+2i}{41}\) 

∴ The values of a, b are \(\frac{23}{41}\) , \(\frac{2}{41}\) .

Given: 

⇒ a+ ib = \(\frac{(2+i)^3}{2+3i}\)

⇒ a+ ib = \(\frac{2^3+i^3+3(2)^2(i)+3(i)^2(2)}{2+3i}\) 

⇒ a+ ib = \(\frac{8+(i^2.i)+3(4)(i)+6i^2}{2+3i}\)

We know that i²=-1 

⇒ a+ ib = \(\frac{8+(-1)i+12i+6(-1)}{2+3i}\)

⇒ a+ ib =\(\frac{2+11i}{2+3i}\)

Multiplying and dividing with 2-3i

⇒ a+ ib = \(\frac{2+11i}{2+3i}\) x \(\frac{2-3i}{2-3i}\) 

⇒ a+ ib = \(\frac{2(2-3i)+11i(2-3i)}{(2)^2-(3i)^2}\) 

⇒ a+ ib = \(\frac{4-6i+22i-33i^2}{4-9i^2}\)

We know that i²=-1 

⇒ a+ ib = \(\frac{4+16i-33(-1)}{4-9(-1)}\)

⇒ a+ ib = \(\frac{37+16i}{13}\)

∴ The values of a, b are \(\frac{37}{13}\) , \(\frac{16}{13}\).

Given: 

⇒ a + ib = \(\frac{(1-i)^3}{1-i^3}\) 

⇒ a + ib = \(\frac{1^3-3(i)^2(i)+3(1)(i)^2-i^3}{1-i^2.i}\) 

We know that i² = -1 

⇒ a + ib = \(\frac{1-3i+3(-1)-i^2.i}{1-(-1)i}\) 

⇒ a + ib = \(\frac{-2-3i-(-1)i}{1+i}\) 

⇒ a + ib = \(\frac{-2-4i}{1+i}\) 

Multiplying and diving with 1-i 

⇒ a + ib = \(\frac{-2-4i}{1+i} \times \ \frac{1-i}{1-i}\) 

⇒ a + ib = \(\frac{-2(1-i)-4i(1-i)}{1^2-i^2}\) 

We know that i² = -1 

⇒  a+ib = \(\frac{-2+2i-4i+4i^2}{1-(-1)}\)

⇒ a + ib = \(\frac{-2-2i+4(-1)}{2}\)  

⇒ a + ib = \(\frac{-6-2i}{2}\)

⇒ a + ib = -3-i

∴ The values of a, b are -3, -1.

Given: 

⇒ a + ib = \(\frac{1-i}{1+i}\)

Multiplying and dividing by 1-i

⇒ a + ib = \(\frac{1-i}{1+i}\)  x  \(\frac{1-i}{1-i}\)

⇒ a + ib =  \(\frac{1^2+i^2-2(1)(i)}{1^2-(i)^2}\)

We know that i² = -1

⇒ a + ib = \(\frac{1+(-1)-2i}{1-(-1)}\) 

⇒ a + ib = \(\frac{-2i}{2}\) 

⇒ a + ib = -i

⇒ a + ib = 0 - i

∴ The values of a, b is 0, -1.

Given: 

⇒ a + ib = \(\frac{3+2i}{-2+i}\)

Multiplying and dividing with -2-i 

⇒ a + ib = \(\frac{3+2i}{-2+i}\) x \(\frac{-2-i}{-2-i}\)

⇒  a + ib = \(\frac{3(-2-i)+2i(-2-i)}{(-2)^2-(i)^2}\) 

We know that i²=-1

⇒ a + ib = \(\frac{-6-3i-4i-2i^2}{4-i^2}\) 

⇒ a + ib = \(\frac{-6-7i-2(-1)}{4-(-1)}\) 

⇒ a + ib = \(\frac{-4-7i}{5}\) 

∴ The values of a, b are \(-\frac{4}{5}\), \(-\frac{7}{5}\) .

⇒ a+ib = (1+i)(1+2i) 

⇒ a+ib = 1(1+2i)+i(1+2i) 

⇒ a+ib = 1+2i+i+2i² 

We know that i²=-1 

⇒ a+ib = 1+3i+2(-1) 

⇒ a+ib = 1+3i-2 

⇒ a+ib=-1+3i 

∴ The values of a, b are -1, 3.