NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced PDF Download

Q.1. For a positive integer n, find the value of
Ans.
We have
[∵ i² = - 1]
=  [(1 – i)(1 + i)]ⁿ
=    [1 –  i²]ⁿ = [1 + 1]ⁿ = 2ⁿ
Hence, (1 – i)ⁿ

Q.2. Evaluate , where n ∈ N .
Ans. 
We have
= (i + i²) + (i² + i³) + (i³ + i⁴) + (i⁴ + i⁵) + (i⁵ + i⁶) + (i⁶ + i⁷) + (i⁷ + i⁸) + (i⁸ + i⁹) + (i⁹ + i¹⁰) + (i¹⁰ + i¹¹) + (i¹¹ + i¹²) + (i¹² + i¹³) + (i¹³ + i¹⁴)
= i + 2(i² + i³ + i⁴ + i⁵ + i⁶ + i⁷ + i⁸ + i⁹ + i¹⁰ + i¹¹ + i¹² + i¹³) + i¹⁴
= i + 2[– 1 – i + 1 + i – 1 – i + 1 + i – 1 – i + 1 + i] + (– 1)
= i + 2(0) – 1 ⇒ 1 + i
Hence,= – 1 + i.

Q.3. If=  x + iy, then find (x, y).
Ans.
We have
= x + iy

⇒
⇒ (i)3 – (- i)3 = x + iy ⇒ i2.i + i2.i = x + iy
⇒ - i – i = x + iy ⇒ 0 – 2i = x + iy
Comparing the real and imaginary parts, we get
x = 0, y = - 2. Hence, (x , y) = (0 , - 2).

Q.4. If = x + iy, then find the value of x + y.
Ans.
Given that:
⇒
⇒
⇒  [∵ i2 = - 1]
⇒
Comparing the real and imaginary parts, we get

Hence,

Q.5. If = a + ib, then find (a, b).
Ans. 
We have

⇒ ( - i)¹⁰⁰ = a + bi ⇒ i¹⁰⁰ = a + bi
⇒ (i⁴)²⁵ = a + bi ⇒ (1)²⁵ = a + bi ⇒ 1 = a + bi
⇒ 1 + 0i = a + bi
Comparing the real and imaginary parts, we have
a = 1, 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 =
⇒

⇒
∴
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:
⇒ (x + iy) (x – iy) + 2(x + iy + x – iy) + b = 0
⇒x² + y² – 2(x + x) + b = 0
⇒x² + y² – 4x + b = 0 Which represents a circle.
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 = 4
∴  
⇒ x² + y² + x – 2 = 4[(x – 1)² + y²]
⇒ x² + y² + x – 2 = 4[x² + 1 – 2x + y²]
⇒ x² + y² + x – 2 = 4x² + 4 – 8x + 4y²
⇒ x² – 4x² + y² – 4y² + x + 8x – 2 – 4 = 0
⇒ – 3x² – 3y² + 9x – 6 = 0
⇒ x² + y² – 3x + 2 = 0
Which represents 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 that:
⇒ arg (z – 1) – arg (z + 1) =

⇒ arg [x + iy – 1] – arg [x + iy + 1] =
⇒ arg [(x – 1) + iy] – arg [(x + 1) + iy] =
⇒

⇒
⇒
⇒
⇒ x² + y² – 1 = 2y
⇒ 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
Squaring both sides

⇒
⇒ 0 = – 3 + 2z + 4(z + 1)i
⇒ 3 – 2z – 4(z + 1)i = 0
⇒ 3 – 2(x + yi) – 4[x + yi + 1]i = 0
⇒ 3 – 2x – 2yi – 4xi – 4yi² – 4i = 0
⇒ 3 – 2x + 4y – 2yi – 4i – 4xi = 0
⇒ (3 – 2x + 4y) – i(2y + 4x + 4) = 0
⇒ 3 – 2x + 4y = 0 ⇒ 2x - 4y = 3 ...(i)
and 4x + 2y + 4 = 0 ⇒ 2x + y = - 2 ...(ii)
Solving eqn. (i) and (ii), we get
y = – 1 and x =
Hence, the value of 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
So,  = (x + iy) + 2(1 + i)
⇒ = x + iy + 2 + 2i
⇒ = (x + 2) + (y + 2)i
= (x + 2) + (y + 2)i

Squaring both sides, we get
(x + 1)² + y² = (x + 2)² + (y + 2)².i² + 2(x + 2)(y + 2)i
⇒ x² + 1 + 2x + y² = x² + 4 + 4x – y² – 4y – 4 + 2(x + 2)(y + 2)i
Comparing the real and imaginary parts, we get
x² + 1 + 2x + y² = x² + 4x – y² – 4y and 2(x + 2)(y + 2) = 0
⇒ 2y² – 2x + 4y + 1 = 0 ...(i)
and (x + 2)(y  + 2) = 0 ...(ii)
x + 2 = 0 or y + 2 = 0
∴ x = – 2 or y = – 2
Now put x = – 2 in eqn. (i)
2y² – 2 × (- 2) + 4y + 1 = 0
⇒ 2y² + 4 + 4y + 1 = 0
⇒ 2y² + 4y + 5 = 0
b² – 4ac = (4)² – 4 × 2 × 5
= 16 – 40 = - 24 < 0 no real roots.
Put y = – 2 in eqn. (i)
2(– 2)² – 2x + 4(– 2) + 1 = 0
8 - 2x - 8 + 1 = 0 ⇒ x = 1/2 and y = -2
Hence, z = x + iy = (1/2 - 2i)

Q.13. If arg (z – 1) = arg (z + 3i), then find x – 1 : y. where z = x + iy
Ans.
Given that: arg (z – 1) = arg (z + 3i)
⇒ arg [x + yi – 1] = arg [x + yi + 3i]
⇒ arg [(x – 1) + yi] = arg [x + (y + 3)i]
⇒
⇒
⇒ xy = (x – 1)(y + 3) ⇒ xy = xy + 3x – y – 3
⇒ 3x – y = 3 ⇒ 3x - 3 = y
⇒ 3(x – 1) = y ⇒ ⇒ x – 1 : y = 1 : 3
Hence, x – 1 : y = 1 : 3.

Q.14. Show thatrepresents a circle. Find its centre and radius.
Ans.
Given that:
Let z = x + iy
∴
 


Squaring both sides, we get
(x – 2)² + y² = 4[(x – 3)² + y²]
⇒ x² + 4 – 4x + y² = 4[x² + 9 – 6x + y²]
⇒ x² + y² – 4x + 4 = 4x² + 4y² – 24x + 36
⇒ 3x² + 3y² – 20x + 32 = 0

Here g =

Hence, the required equation of the circle is


Q.15. Ifis a purely imaginary number (z ≠ – 1), then find the value of .
Ans.
Given that is purely imaginary number
Let z = x + yi


Since, the number is purely imaginary, then real part = 0
∴
⇒ x² + y² – 1 = 0 ⇒ x² + y² = 1
⇒

Q.16. z₁ and z₂ are 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) are polar form of two complex number z₁ and z₂.
Given that: |z₁| =|z₂| ⇒ r₁ = r₂ ….(i)
and arg (z₁) + arg (z₂) = π
⇒ θ₁ + θ₂ = π
⇒ θ₁ = π – θ₂
Now z₁ = r₁ [cos (π – θ₂) + i sin (π –    θ₂)]
⇒ z₁ = r₁  [- cos θ₂ + i sin θ₂]
⇒ z₁ = - r₁ (cos θ₂ – i sin θ₂) ….(i)
z₂ = r₂ [cos θ₂ + i sin θ₂]
 [∴ r₁ = r₂]…(ii)
From eqn. (i) and (ii) we get,

Hence proved.

Q.17. If  then show that the real part of z₂ is zero.
Ans.
 Let z₁ = x + yi
 
⇒ x² + y² = 1 ...(i)
Now

Hence, the real part of z₂ is 0.

Q.18. If z₁, z₂ and z₃, z₄ are two pairs of conjugate complex numbers, then find
Ans.
 Let the polar form of z₁ = r₁ (cos θ₁ + i sin θ₁)
∴    = r₁ (cos θ₁ – i sin θ₁) = r₁ [cos (– θ₁) + i sin (– θ₁)]
Similarly, z₃ = r₂ (cos θ₂ + i sin θ₂)
∴ = r₂ (cos θ₂ – i sin θ₂) = r₂ [cos (– θ₂) + i sin (– θ₂)]
= arg (z₁) – arg (z₄) + arg (z₂) – arg (z₃)
= θ₁ – (- θ₂) + (- θ₁) – θ₂
= θ₁ + θ₂ – θ₁ – θ₂ = 0
Hence,

Q.19. If then
show that
Ans.
We have
⇒ ...(i)
⇒
⇒



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 that for z₁ and z₂, arg (z₁) – arg (z₂) = 0
Let us represent z₁ and z₂ in polar form
z₁ = r₁ (cos θ₁ + i sin θ₁) and z₂ = r₂ (cos θ₂ + i sin θ₂)
arg (z₁) = θ₁ and arg (z₂) = θ₂
Since arg (z₁) – arg (z₂) = 0
⇒ θ₁ – θ₂ = 0 ⇒ θ₁ = θ₂
Now z₁ – z₂ = r₁ (cos θ₁ + i sin θ₁) – r₂ (cos θ₂ + i sin θ₂)
= r₁ cos θ₁ + i r₁ sin θ₁ – r₂ cos θ₁ – i r₂ sin θ₁
[∴ θ₁ = θ₂]
= (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 that: Re(z²) = 0 and
Let z = x + yi
∴
⇒ = 2 ⇒ x2 + y2 = 4 ...(i)
Since, z = x + yi
z² = x² + y2i² + 2xyi
⇒ z² = x² – y² + 2xyi
∴ Re(z²) = x² – y²
⇒ x² – y² = 0 ...(ii)
From eqn. (i) and (ii), we get x
x² + y² = 4 ⇒ 2x² = 4 ⇒ x² = 2 ⇒ x =
Hence, z =

Q.22. Find the complex number satisfying the equation
Ans.
Given that:
Let z = x + yi
∴


⇒ x² = 2(x² + 2x + 1 + y²)
⇒ x² = 2x² + 4x + 2 + 2y²
⇒ x² + 4x + 2 + 2y² = 0
⇒ x² + 4x + 2 + 2(– 1)² = 0    [∴ y = – 1]
⇒ x² + 4x + 4 = 0
⇒ (x + 2)² = 0
⇒ x + 2 = 0 ⇒ x = – 2
Hence, z = x + yi = – 2 – i.
Q.23. Write the complex number in polar form.
Ans.
Given that:


So r = √2
Now arg(z) =

Hence, the polar 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)
zw = r₁r₂ [(cos θ₁ + i sin θ₁)] [(cos θ₂ + i sin θ₂)]
= r₁r₂ = 1 (given)
Now arg (z) – arg (w) =

= r₁ (cos θ₁ – i sin θ₁) r₂ (cos θ₂ + i sin θ₂)
= r₁ r₂ [cos θ₁ cos θ₂ + i cos θ₁ sin θ₂ – i sin θ₁ cos θ₂ – i₂ sin θ₁ sin θ₂]
= r₁ r₂ [(cos θ₁ cos θ₂ + sin θ₁ sin θ₂) + i(cos θ₁ sin θ₂ – sin θ₁ cos θ₂)]
= r₁ r₂ [cos (θ₂ – θ₁) + i sin (θ₂ – θ₁)]

HereHence proved.

Fill in the Blanks
Q.25. (i) For any two complex numbers z₁, z₂ and any real numbers a, b,
(ii) The value of
(iii) The numberis 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) Ifthen the greatest and least values ofare ............... and ...............
(ix) Ifthen the locus of z is ............
(x) If= 4 and arg (z) =, then z = ............
Ans.


Hence, the value of the filler is
Hence, the value of the filler is – 15.
(iii)

(iv) i + i² + i³ + ... upto 1000 terms
= i + i² + i³ + ... + i¹⁰⁰⁰ = 0

Hence, the value of the filler is 0.
(v) Multiplicative inverse of

Hence, the value of the filler =
(vi) Let z₁ = x₁ + iy₁ and z₂
= x₂ + iy₂ z₁ + z₂
= (x₁ + iy₁) + (x₂ + iy₂) z₁ + z₂
= (x₁ + x₂) + (y₁ + y₂)i
If z₁ + z₂ is real then
y₁ + y₂ = 0⇒ y₁ = – y₂ 
∴ z₂ = x₂ – iy₁
z₂ = x₁ – iy₁ (when x₁ = x₂)
So z₂ =
Hence, the value of the filler is.
(vii) arg ( z) + arg
If arg (z) = θ, then arg (  ) = - θ
So θ + (– θ) = 0
Hence, the value of the filler is 0.
(viii) Given that:
For the greatest value of

Hence, the greatest value ofis 6 and for the least value of= 0.
[∴ The least value of the modulus of complex number is 0] Hence, the value of the filler are 6 and 0.
(ix) Given that:
Let z = x + iy

Which represents are equation of a circle.
Hence, the value of the filler is circle.
(x) Given that:
Let z = x + yi

From eqn. (i) and (ii)

Hence, the value of the filler is

Q.26. State True or False for the following :
(i) The order relation is defined on the set of complex numbers.
(ii) Multiplication of a non zero complex number by – i rotates the point about origin through a right angle in the anti-clockwise direction.
(iii) For any complex number z the minimum value of
(iv) The locus represented by  is a line perpendicular to the join of (1, 0) and (0, 1).
(v) If z is a complex number such that z ≠ 0 and Re (z) = 0, then Im (z²) = 0.
(vi) The inequality  represents the region given by x > 3.
(vii) Let z1 and z2 be two complex numbers such that  , then arg (z1 – z2) = 0.
(viii) 2 is not a complex number.
Ans.
(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex numbers can be compared.
So it is ‘False’.
(ii) Let z = x + yi
z.i = (x + yi) i = xi – y which rotates at angle of 180°
So, it is ‘False’.
(iii) Let z = x + yi

The value of  is minimum when x = 0, y = 0 i.e., 1.
Hence, it is ‘True’.
(iv) Let z = x + yi
Given that:

⇒(x – 1)² + y² = x² + (1 – y)²
⇒ x² – 2x + 1 + y² = x² + 1 + y² – 2y
⇒ – 2x + 2y = 0
⇒ x – y = 0 which is a straight line.
Slope = 1
Now equation of a line through the point (1, 0) and (0, 1)

⇒ y = – x + 1 whose slope = – 1.
Now the multiplication of the slopes of two lines = – 1 x 1 = - 1,
so they are perpendicular.
Hence, it is ‘True’.
(v) Let z = x + yi, z ≠ 0 and Re(z) = 0
Since real part is 0 ⇒ x = 0
∴ z = 0 + yi = yi
∴ lm (z²) = y²i² = - y² which is real.
Hence, it is ‘False’.
(vi) Given that:
Let z = x + yi

⇒ (x – 4)² + y² < (x – 2)² + y² 
⇒ (x – 4)² < (x – 2)²
⇒ x² + 16 – 8x < x² + 4 – 4x 
⇒ – 8x + 4x < – 16 + 4
⇒ – 4x < – 12 ⇒ x > 3
Hence, it is ‘True’.
(vii) Let z₁ = x₁ + y₁i and z₂ = x₂ + y₂i

Squaring both sides, we get

Again squares on both sides, we get

∴ arg (z₁) = arg (z₂)
⇒ arg (z₁) – arg (z₂) = 0
Hence, it is ‘True’.
(viii) Since 2 has no imaginary part.
So, 2 is not a complex number.
Hence, it is ‘True’.

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

Ans.
(a) Given that z = i + √3
Polar form of z = r [cos θ + i sin θ]

Since x > 0, y > 0
∴ Polar form of z =
Hence, (a) ⇒ 8y = 0 Þ y = 0. Which is the equation of x-axis and it is perpendicular to the line segment joining (0, – 2) and (0, 2).
Hence, (d) ↔ (iv).
(b) Given that
Here argument (z) =
So,
Since x < 0 and y > 0

Hence, (b) ↔ (iii).
(c) Given that:
Let z = x + yi

⇒ 8y = 0 ⇒ y = 0. Which is the equation of x-axis and it is perpendicular to the line segment joining (0, – 2) and (0, 2).
Hence, (d) ↔ (iv).
(e) Given that:
Let z = x + yi

Which represents a circle on or outside having centre (0, – 4) and radius 3.
Hence, (e) ↔ (ii).
(f)
Let z = x + yi

Which is a circle having centre (– 4, 0) and r= and is on or inside the circle.
Hence, (f) ↔ (vi).
(g) Let

∴which lies in third quadrant.
Hence, (g) ↔ (viii).
(h) Given that: z = 1 – i

Which lies in first quadrant.
Hence, (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 that




Q.29. If  is it necessary that z₁ = z₂?
Ans.
Let z₁ = x₁ + y₁i and z₂ = x₂ + y₂i

⇒ x₁ = ± x₂ and y₁ = ± y₂
So z₁ = x₁ + y₁i and z₂ = ± x₂ ± y₂i
⇒ z₁ ≠ z₂
Hence, it is not necessary that z₁ = z₂.

Q.30. If = x + iy, what is the value of x² + y²?
Ans.
Given that: = x + iy ...(i)
Taking conjugate on both sides
⇒ = x – iy ...(ii)
Multiplying eqn. (i) and (ii) we have

Hence, the value of x² + y² =

Q.31. Find z if= 4 and arg (z) =
Ans.
Given that:
= 4 ⇒ r = 4
So Polar form of 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.
Given that: (1 + i √3 )² = 1 + i² . 3 + 2 √3i
= 1 - 3 + 2 √3i = - 2 + 2 √3i

⇒
⇒
Now Re(z) < 0 and image (z) > 0.
∴
Hence, the principal arg =

Q.34. Where does z lie, if
Ans.
Given that:
Let z = x + yi
∴
⇒
⇒ x² + (y – 5)² = x² + (y + 5)²
⇒(y – 5)² = (y + 5)²
⇒ y² + 25 – 10y = y² + 25 + 10y
⇒ 20y = 0⇒ y = 0
Hence, z lies on x-axis i.e., real axis.

Choose the correct answer from the given four options
Q.35. sinx + i cos 2x and cos x – i sin 2x are conjugate to each other for:
(A) x = nπ

(C) x = 0
(D) No value of x
Ans.
Let z = sin x + i cos 2x
 = sin x – i cos 2x
But we are given that  = cos x – i sin 2x
∴ sin x – i cos 2x = cos x – i sin 2x
Comparing the real and imaginary parts, we get
sin x = cos x and cos 2x = sin 2x
⇒ tan x = 1 and tan 2x = 1
⇒
∴
⇒ x = 2x ⇒ 2x - x = 0 ⇒ x = 0
Hence, the correct option is (c).

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


(C) n π
(D) None of these, where n ∈N
Ans.
Let

Since, z is purely real, then

So, α = nπ, n ∈ N.

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.
Given that: z = x + iy
If z lies in third quadrant.
So x < 0 and y < 0.


When z lies in third quadrant thenwill also be lie in third  quadrant  

∴
⇒ 2 – y² < 0 and 2xy > 0
⇒ x² < y² and xy > 0
So x < y < 0.
Hence, the correct option is (b)

Q.38. The value of (z + 3)is equivalent to


(C) z² + 3
(D) None of these
Ans.
Given that: ( z + 3)
Let z = x + yi
So ( z + 3)= (x + yi + 3)(x – yi + 3)
= [(x + 3) + yi][(x + 3) – yi]
= (x + 3)² – y²i² = (x + 3)² + y²

Hence, the correct option is (a).

Q.39. Ifthen
(A) x = 2n+1
(B) x = 4n
(C) x = 2n
(D) x = 4n + 1,  where n ∈N
Ans.
Given that:

⇒ (i)^x = (i)⁴ⁿ
⇒x = 4n, n ∈ N
Hence, the correct option is (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. 
Given that:
⇒
⇒
⇒
⇒ ...(i)
⇒ ...(ii)
Multiplying eqn. (i) and (ii) we get

⇒
⇒
⇒
So, α² + β² = 1
Hence, the correct option is (a).

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

(B) arg (z₁z₂) = arg (z₁). arg (z₂)


Ans.
Let z₁ = r₁ (cos θ₁ + i sin θ₁)
∴ = r₁
and z₂ = r₂ (cos θ₂ + i sin θ₂)
∴ = r₂
z₁z₂ = r₁ (cos θ₁ + i sin θ₁) . r₂ (cos θ₂ + i sin θ₂)
= r₁r₂ (cos θ₁ + i sin θ₁) . (cos θ₂ + i sin θ₂)
= r₁r₂ (cos θ₁ cos θ₂ + i sin θ₂ cos θ₁ + i sin θ₁ cos θ₂ + i₂ sin θ₁ sin θ₂)
= r₁r₂ [(cos θ₁ cos θ₂ – sin θ₁ sin θ₂) + i(sin θ₁ cos θ₂ + cos θ₁ sin θ₂)]
= r₁r₂ [cos (θ₁ + θ₂) + i sin (θ₁ + θ₂)]

Hence, the correct option is (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 + 2 i
Ans.
Given that: z = 2 – i
If z rotated through an angle ofabout the origin in clockwise  direction.
Then the new position = z.e^{– (π/2)}
= (2 – i) e^{– (π/2)}

= (2 – i) (0 – i) = – 1 – 2i
Hence, the correct option is (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 + yi is a non-real complex number if y ≠ 0. If x, y ∈ R.
Hence, the correct option is (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.
Given that: a + ib = c + id

Squaring both sides, we get a² + b² = c² + d²
Hence, the correct option is (d).

Q.45. The complex number z which satisfies the conditionlies on
(A) circle x² + y² = 1
(B) the x-axis
(C) the y-axis
(D) the line x + y = 1.
Ans.
Given that:
Let z = x + yi

⇒ x² + (y + 1)² = x² + (y – 1)²
⇒ (y + 1)² = (y – 1)² 
⇒ y² + 2y + 1 = y² – 2y + 1
⇒ 2y = – 2y
⇒ 4y = 0 ⇒ y = 0 ⇒ x-axis.
Hence, the correct option is (b).

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




Ans.
Let z = x + yi

= x² + y² ...(i)
Now z² = x² + y²i² + 2xyi
z² = x² – y² + 2xyi


Hence, the correct option is (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)
Since
z₁ + z₂ = r₁ cos θ₁ + i r₁ sin θ₁ + r₂ cos θ₂ + i r₂ sin θ₂


Squaring both sides, we get

⇒ 2r₁r₂ – 2r₁r₂ cos (θ₁ – θ₂) = 0
⇒ 1 – cos (θ₁ – θ₂) = 0 ⇒ cos (θ₁ – θ₂) = 1
⇒ θ₁ – θ₂ = 0 ⇒ θ₁ = θ₂
So, arg (z₁) = arg (z₂)
Hence, the correct option is (c).

Q.48. The real value of θ for which the expression is a real number is:



(D) none of these.
Ans.
Let


If z is a real number, then

⇒ 3 cos θ = 0 ⇒ cos θ = 0
∴
Hence, the correct option is (c).

Q.49. The value of arg (x) when x < 0 is:
(A) 0
(B) π/2
(C) π
(D) none of these
Ans.
Let z = – x + 0i and x < 0
∴
Since, the point (–x, 0) lies on the negative side of the real axis
(∴ x < 0).
∴ Principal argument (z) = π
Hence, the correct option is (c).

Q.50. If f (z) =where z = 1 + 2i, then



(D) none of these.
Ans.
Given that: z = 1 + 2i

Now

So  

Hence, the correct option is (a).