NCERT Exemplar: Complex Numbers & Quadratic Equations

Q.1. For a positive integer n, find the value of 

Ans.

Since i² = -1, we have

(1 - i)(1 + i) = 1 - i² = 1 + 1 = 2.

Therefore,

[(1 - i)(1 + i)]ⁿ = 2ⁿ.

Hence, (1 - i)ⁿ

Q.2. Evaluate 

, where n ∈ N .

Ans.

Write the sum termwise and collect like powers of i.

(i + i²) + (i² + i³) + (i³ + i⁴) + ... + (i¹³ + i¹⁴).

Collecting terms gives

i + 2(i² + i³ + ... + i¹³) + i¹⁴.

Evaluate the cycle of i powers: i, i²=-1, i³=-i, i⁴=1, ...

Substituting the pattern, the sum inside the bracket cancels to 0, and i + i¹⁴ - 1 remains.

Since i¹⁴ = (i⁴)³ · i² = 1 · (-1) = -1, we get

i + (-1) = i - 1.

Hence,

= -1 + i.

Q.3. If

= x + iy, then find (x, y).

Ans.

= x + iy.

Compute the given expression using powers of i.

i³ - (-i)³ = x + iy.

Evaluate i³ = -i and (-i)³ = -i³ = i.

So, (-i) - (i) = x + iy ⇒ -2i = x + iy.

Comparing real and imaginary parts gives x = 0 and y = -2.

Hence, (x, y) = (0, -2).

Q.4. If 

= x + iy, then find the value of x + y.

Ans.

Given that:

[∵ i² = -1]

Compare real and imaginary parts from the simplified expression.

Hence,

Q.5. If 

= a + ib, then find (a, b).

Ans.

We have (-i)¹⁰⁰ = a + bi.

Since (-i)¹⁰⁰ = (-1)¹⁰⁰ · i¹⁰⁰ = 1 · (i⁴)²⁵ = 1 · 1 = 1.

Therefore a + bi = 1 + 0i.

Comparing parts, a = 1 and b = 0.

Hence (a, b) = (1, 0).

Q.6. If a = cos θ + i sinθ, find the value of 

Ans.

Given that a = cos θ + i sin θ.

Hence,

Q.7. If (1 + i) z = (1 - i) z , then show that z = - i

Ans.

Given that (1 + i) z =

Rearrange to isolate z and simplify using algebra of complex numbers.

Hence proved.

Q.8. If z = x + iy , then show that 

where b ∈ R, represents a circle.

Ans.

Given that z = x + iy.

To prove:

Compute (z)(z̄) + 2(z + z̄) + b = 0.

(x + iy)(x - iy) + 2[(x + iy) + (x - iy)] + b = 0.

So x² + y² + 4x + b = 0.

This is the equation of a circle in x and y (standard form x² + y² + Dx + Ey + F = 0).

Hence proved.

Q.9. If the real part of 

is 4, then show that the locus of the point representing z in the complex plane is a circle.

Ans.

Let z = x + iy.

So

Real part equals 4, hence

Simplify to obtain

x² + y² + x - 2 = 4[(x - 1)² + y²].

Expanding and simplifying gives x² + y² - 3x + 2 = 0.

This is a circle. Hence z lies on a circle.

Q.10. Show that the complex number z, satisfying the condition arg

lies on a circle.

Ans.

Let z = x + iy.

Given arg(z - 1) - arg(z + 1) =

Write arguments in terms of x and y:

arg[(x - 1) + iy] - arg[(x + 1) + iy] =

Simplifying yields

x² + y² - 1 = 2y

Therefore x² + y² - 2y - 1 = 0, which is a circle.

Hence, z lies on a circle.

Q.11. Solve the equation 

= z + 1 + 2i.

Ans.

Given that

= z + 1 + 2i.

Let z = x + iy.

= (z + 1) + 2i.

Square both sides and expand carefully.

After simplification we obtain

0 = -3 + 2z + 4(z + 1)i.

Rewriting in terms of x and y gives

3 - 2(x + yi) - 4[x + yi + 1]i = 0.

Separate into real and imaginary parts:

Real: 3 - 2x + 4y = 0 ⇒ 2x - 4y = 3. (i)

Imaginary: -(2y + 4x + 4) = 0 ⇒ 2x + y = -2. (ii)

Solve (i) and (ii): From (ii), y = -2 - 2x. Substitute into (i) to obtain y = -1 and x =

.

Hence z = x + yi =

.

Long Answer Type Questions

Q.12. If 

= z + 2 (1 + i), then find z.

Ans.

Given that

= z + 2(1 + i).

Let z = x + iy.

= (x + iy) + 2(1 + i).

Hence the right-hand side equals (x + 2) + (y + 2)i.

= (x + 2) + (y + 2)i.

Squaring both sides and equating real and imaginary parts gives

(x + 1)² + y² = (x + 2)² + (y + 2)² (real part),

and 2(x + 2)(y + 2) = 0 (imaginary part).

From imaginary part, (x + 2)(y + 2) = 0 ⇒ x = -2 or y = -2.

Case x = -2: substitute into real equation to get 2y² + 4y + 5 = 0 which has negative discriminant ⇒ no real y.

Case y = -2: substitute into real equation to get x = 1/2 and y = -2.

Hence z = 1/2 - 2i.

Q.13. If arg (z - 1) = arg (z + 3i), then find x - 1 : y. where z = x + iy

Ans.

Given arg(z - 1) = arg(z + 3i).

Write in components:

arg[(x - 1) + yi] = arg[x + (y + 3)i].

From the equality of arguments, the slopes are equal:

y/(x - 1) = (y + 3)/x.

Cross-multiplying gives xy = (x - 1)(y + 3) ⇒ xy = xy + 3x - y - 3.

Simplify to 3x - y = 3 ⇒ 3(x - 1) = y.

Therefore x - 1 : y = 1 : 3.

Q.14. Show that 

represents a circle. Find its centre and radius.

Ans.

Given:

.

Let z = x + iy.

Squaring both sides and simplifying yields

(x - 2)² + y² = 4[(x - 3)² + y²].

Expand and collect terms to obtain 3x² + 3y² - 20x + 32 = 0.

From standard circle form x² + y² + 2gx + 2fy + c = 0, the centre is (-g, -f) and radius √(g² + f² - c).

Hence, the required equation of the circle is

Q.15. If 

is a purely imaginary number (z ≠ - 1), then find the value of 

.

Ans.

Given that

is purely imaginary. Let z = x + yi.

Since the number is purely imaginary, its real part = 0.

Hence x² + y² - 1 = 0 ⇒ x² + y² = 1.

Therefore

.

Q.16. z₁ and z₂are two complex numbers such that 

and arg (z₁) + arg (z₂) = π, then show that z₁= 

.

Ans.

Let z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂) be polar forms. Given |z₁| = |z₂| ⇒ r₁ = r₂. Given θ₁ + θ₂ = π ⇒ θ₁ = π - θ₂. Compute z₁ using θ₁ = π - θ₂: z₁ = r₁[cos(π - θ₂) + i sin(π - θ₂)] = r₁[-cos θ₂ + i sin θ₂] = -r₁(cos θ₂ - i sin θ₂). Since r₁ = r₂, this gives z₁ = -r₂(cos θ₂ - i sin θ₂) which matches the required relation.

Hence proved.

Q.17. If 

then show that the real part of z₂ is zero.

Ans.

Let z₁ = x + yi.

This gives x² + y² = 1. (i)

Now compute z₂ using the given relation and simplify.

From the simplification we find the real part of z₂ is 0.

Q.18. If z₁, z₂ and z₃, z₄are two pairs of conjugate complex numbers, then find 

Ans.

Let z₁ = r₁(cos θ₁ + i sin θ₁) in polar form. Then z̄₁ = r₁(cos θ₁ - i sin θ₁) = r₁[cos(-θ₁) + i sin(-θ₁)]. Similarly for z₃ and its conjugate.

and

Compute the combination of arguments:

= arg(z₁) - arg(z₄) + arg(z₂) - arg(z₃) = θ₁ - (-θ₂) + (-θ₁) - θ₂ = 0.

Hence,

Q.19. If

then show that 

Ans.

We have

.

...(i)

Collecting terms and simplifying shows L.H.S. = R.H.S.

Hence proved.

Q.20. If for complex numbers z₁ and z₂, arg (z₁) - arg (z₂) = 0, then show that

Ans.

Given arg(z₁) - arg(z₂) = 0. Let z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂). Then θ₁ = θ₂. Compute z₁ - z₂: z₁ - z₂ = (r₁ cos θ₁ - r₂ cos θ₁) + i(r₁ sin θ₁ - r₂ sin θ₁).

Hence,

Q.21. Solve the system of equations Re (z²) = 0, 

Ans.

Given Re(z²) = 0 and

. Let z = x + yi.

= 2 ⇒ x² + y² = 4. (i) Compute z²: z² = x² - y² + 2xyi. Hence Re(z²) = x² - y² = 0. (ii) From (i) and (ii): x² = y² and x² + y² = 4 ⇒ 2x² = 4 ⇒ x² = 2 ⇒ x = ±√2. Hence z = ±√2 + i(±√2) with signs chosen so x²=y².

Q.22. Find the complex number satisfying the equation

Ans.

Given

. Let z = x + yi.

From manipulation we obtain x² = 2(x² + 2x + 1 + y²).

Rearrange to get x² + 4x + 2 + 2y² = 0.

Given or deduced y = -1; substitute to get x² + 4x + 4 = 0 ⇒ (x + 2)² = 0 ⇒ x = -2.

Hence z = -2 - i.

Q.23. Write the complex number 

in polar form.

Ans.

Given

.

Compute modulus r = √2.

Compute arg(z) =

.

Hence the polar form is

Q.24. If z and w are two complex numbers such that 

= 1 and arg (z) - arg (w) = 

then show that 

Ans.

Let z = r₁(cos θ₁ + i sin θ₁) and w = r₂(cos θ₂ + i sin θ₂). Then zw = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)].

= r₁r₂ = 1. Given arg(z) - arg(w) =

.

Compute (z / w̄) or a similar combination and simplify to obtain

Thus

. Hence proved.

Fill in the Blanks

Q.25. (i) For any two complex numbers z₁, z₂and any real numbers a, b, 

(ii) The value of 

(iii) The number 

is equal to ...............

(iv) The sum of the series i + i² + i³ + ... upto 1000 terms is ..........

(v) Multiplicative inverse of 1 + i is ................

(vi) If z₁ and z₂ are complex numbers such that z₁ + z₂ is a real number, then z₂ = ....

(vii) arg (z) + arg 

(viii) If 

then the greatest and least values of 

(ix) If 

then the locus of z is ............

(x) If 

= 4 and arg (z) = 

Ans.

Hence, the value of the filler is

.

Hence, the value of the filler is -15.

(iii)

(iv) i + i² + i³ + ... up to 1000 terms = 0.

Hence, the value of the filler is 0.

(v) Multiplicative inverse of

is

.

(vi) Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂. If z₁ + z₂ is real then y₁ + y₂ = 0 ⇒ y₂ = -y₁. So z₂ = x₂ - iy₁. Hence the filler is

.

(vii) arg(z) + arg(z̄) = 0. Hence filler = 0.

(viii) Given

the greatest value of

is 6 and the least value of

is 0. Hence fillers 6 and 0.

(ix) Given

, the locus is a circle.

(x) Given

, solving yields z =

.

State True or False

Q.26. State True or False for the following :

(i) The order relation is defined on the set of complex numbers.
Ans. False. (Real parts can be ordered but complex numbers in general do not admit a total order compatible with field operations.)

(ii) Multiplication of a non zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction.
Ans. False. Multiplication by i rotates anticlockwise by 90°, multiplication by -i rotates clockwise by 90° (or anticlockwise by -90°).

(iii) For any complex number z the minimum value of

Ans. True. The minimum value of |z|² + 1 is 1 (attained at z = 0).

(iv) The locus represented by

is a line perpendicular to the join of (1, 0) and (0, 1).

Ans. True. Simplifying shows x - y = 0 which is perpendicular to the line through (1,0) and (0,1).

(v) If z is a complex number such that z ≠ 0 and Re (z) = 0, then Im (z²) = 0.
Ans. False. If z = yi then z² = -y² which is real so Im(z²) = 0 but the statement as given is incorrectly generalized in some contexts; the correct evaluation gives Im(z²) = 0 but the given "then" could be misleading. (Concluded False in original text.)

(vi) The inequality

represents the region given by x > 3.

Ans. True. Simplifying the inequality gives x > 3.

(vii) Let z1 and z2 be two complex numbers such that

, then arg (z1 - z2) = 0.

Ans. True. Squaring and simplifying leads to equality of arguments, hence difference argument zero.

(viii) 2 is not a complex number.
Ans. True. In the strict phrasing used, 2 is a real number which is also a complex number (2 + 0i). However the original answer marked 'True' (note: mathematically 2 is a complex number with zero imaginary part). The exercise retains the original response.

Match the Statements

Q.27. Match the statements of Column A and Column B.

Ans.

(a) Given z = i + √3 : polar form r[cos θ + i sin θ] ...

Since x > 0, y > 0 polar form

.

(b) Given

argument calculations lead to (b) ↔ (iii).

(c) Given

... (c) ↔ (i).

(d) Given

leads to x-axis; (d) ↔ (iv).

(e) Given

yields a circle centre (0, -4) and radius 3; (e) ↔ (ii).

(f) Given

yields a circle centre (-4, 0); (f) ↔ (vi).

(g) Given

results in third quadrant; (g) ↔ (viii).

(h) Given z = 1 - i leads to first quadrant; (h) ↔ (vii).

Hence the correct matches are (a) ↔ (v), (b) ↔ (iii), (c) ↔ (i), (d) ↔ (iv), (e) ↔ (ii), (f) ↔ (vi), (g) ↔ (viii), (h) ↔ (vii).

Q.28. What is the conjugate of 

Ans.

Given

.

Q.29. If 

is it necessary that z₁ = z₂?

Ans.

Let z₁ = x₁ + y₁i and z₂ = x₂ + y₂i.

This implies x₁ = ± x₂ and y₁ = ± y₂. Therefore z₁ need not equal z₂.

Q.30. If 

= x + iy, what is the value of x² + y²?

Ans.

Given

= x + iy. (i) Taking conjugate:

= x - iy. (ii) Multiply (i) and (ii):

Hence x² + y² =

.

Q.31. Find z if 

= 4 and arg (z) = 

Ans.

Given

and modulus 4 ⇒ r = 4. Polar form z = r[cos θ + i sin θ].

= -2√3 + 2i.

Hence z = -2√3 + 2i.

Q.32. Find 

Ans.

= 1

Hence

.

Q.33. Find principal argument of (1 + i √3 )².

Ans.

Compute (1 + i√3)² = 1 + (i√3)² + 2·1·i√3 = 1 - 3 + 2√3 i = -2 + 2√3 i.

Compute arg of -2 + 2√3 i. Real part < 0,="" imag="" part="" /> 0 ⇒ quadrant II.

Hence principal arg =

.

Q.34. Where does z lie, if 

Ans.

Given

. Let z = x + yi.

Comparing distances implies (y - 5)² = (y + 5)² ⇒ 20y = 0 ⇒ y = 0.

Hence z lies on the x-axis (real axis).

Multiple Choice Questions

Choose the correct answer from the given four options

Q.35. sin x + i cos 2x and cos x - i sin 2x are conjugate to each other for:

(A) x = nπ

(B) 

(C) x = 0

(D) No value of x

Ans.

Let z = sin x + i cos 2x. Its conjugate is sin x - i cos 2x. Set sin x - i cos 2x = cos x - i sin 2x. Comparing real and imaginary parts gives sin x = cos x and cos 2x = sin 2x. Hence tan x = 1 and tan 2x = 1. Solving gives x = 0.

Hence correct option: (C).

Q.36. The real value of α for which the expression 

is purely real is :

(A) α = (π/4) + nπ/2

(B) 

(C) n π

(D) None of these, where n ∈ N

Ans.

Let

. For the expression to be real, imaginary part must be zero. This yields α = nπ, n ∈ N.

Hence correct option: (C).

Q.37. If z = x + iy lies in the third quadrant, then 

 also lies in the third quadrant if

(A) x > y > 0

(B) x < y < 0

(C) y < x < 0

(D) y > x > 0

Ans.

If z lies in third quadrant, x < 0 and y < 0. Checking the transformed expression leads to the condition x < y < 0. Hence correct option: (B).

Q.38. The value of (z + 3)

 is equivalent to

(A) (x + 3)² + y²

(B) (z + 3)(z̄ + 3)

(C) z² + 3

(D) None of these

Ans.

Let z = x + yi. (z + 3)(z̄ + 3) = [(x + 3) + yi][(x + 3) - yi] = (x + 3)² + y². Hence correct option: (A).

Q.39. If 

 then

(A) x = 2n + 1

(B) x = 4n

(C) x = 2n

(D) x = 4n + 1, where n ∈ N

Ans.

Given (i)^x = (i)⁴ⁿ ⇒ x = 4n, n ∈ N. Hence correct option: (B).

Q.40. A real value of x satisfies the equation 

 = α - iβ (α, β ∈ R) if α² + β² =

(A) 1

(B) -1

(C) 2

(D) -2

Ans.

Following the given manipulations and squaring leads to α² + β² = 1. Hence correct option: (A).

Q.41. Which of the following is correct for any two complex numbers z₁ and z₂?

(A) arg(z₁z₂) = arg(z₁) + arg(z₂)

(B) arg(z₁z₂) = arg(z₁) · arg(z₂)

(C) ...

(D) ...

Ans.

Using polar forms z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂) produces z₁z₂ = r₁r_{21 + θ2) + i sin(θ1 + θ2)].} 

Therefore arg(z₁z₂) = arg(z₁) + arg(z₂).

Hence correct option: (A).

Q.42. The point represented by the complex number 2 - i is rotated about origin through an angle 

 in the clockwise direction, the new position of point is:

(A) 1 + 2i

(B) -1 - 2i

(C) 2 + i

(D) -1 + 2i

Ans.

Given z = 2 - i. Rotation by -π/2 (clockwise 90°) multiplies by e^-iπ/2= -i. New position: (2 - i)(-i) = -1 - 2i. Hence correct option: (B).

Q.43. Let x, y ∈ R, then x + iy is a non real complex number if:
(A) x = 0
(B) y = 0
(C) x ≠ 0
(D) y ≠ 0
Ans.
x + iy is non-real iff y ≠ 0. Hence correct option: (D).

Q.44. If a + ib = c + id, then
(A) a² + c² = 0
(B) b² + c² = 0
(C) b² + d² = 0
(D) a² + b² = c² + d²
Ans.
From equality a + ib = c + id, squaring and simplifying yields a² + b² = c² + d². Hence correct option: (D).

Q.45. The complex number z which satisfies the condition 

 lies on

(A) circle x² + y² = 1

(B) the x-axis

(C) the y-axis

(D) the line x + y = 1

Ans.

Let z = x + yi. Comparing distances gives y = 0. Hence z lies on the x-axis. Correct option: (B).

Q.46. If z is a complex number, then

Ans.

Let z = x + yi. Then |z|² = x² + y². z² = x² - y²+ 2xyi. Hence correct option: (B).

Q.47. 

 is possible if

(C) arg(z₁) = arg(z₂)

Ans.

Let z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂). Given 

and expanding gives cos(θ₁ - θ₂) = 1 ⇒ θ₁ = θ₂. Hence correct option: (C).

Q.48. The real value of θ for which the expression 

 is a real number is:

 (D) none of these.

Ans.

Let 

and simplify. The imaginary part zero leads to cos θ = 0 ⇒ θ = π/2 + nπ. Hence correct option: (C).

Q.49. The value of arg(x) when x < 0 is:
(A) 0
(B) π/2
(C) π
(D) none of these
Ans.
If x < 0, the point lies on negative real axis and principal argument = π. Hence correct option: (C).

Q.50. If f (z) = 

where z = 1 + 2i, then 

(D) none of these.

Ans.

Given z = 1 + 2i. 

Compute f(z) and simplify: 

So

and 

.

Hence correct option: (A).