NCERT Exemplar: Complex Numbers & Quadratic Equations

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)+ y2]
⇒ x2 + y2 + x – 2 = 4[x2 + 1 – 2x + y2]
⇒ x2 + y2 + x – 2 = 4x2 + 4 – 8x + 4y2
⇒ x2 – 4x+ 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 =

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 + y= 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   θ2) + i sin      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]
= (rcos 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 θsin 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 z= ....
(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 z= x1 + iy1 and z2
= x2 + iy2 z+ 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

Hence, (g) ↔ (viii).
(h) Given that: z = 1 – i

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 zand 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
θ1θ2 = 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).

The document NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

## FAQs on NCERT Exemplar: Complex Numbers & Quadratic Equations - Mathematics (Maths) for JEE Main & Advanced

 1. What are complex numbers and how are they represented?
Ans. Complex numbers are numbers that consist of a real part and an imaginary part. They are represented in the form a + bi, where a is the real part and bi is the imaginary part with i being the imaginary unit.
 2. How do you perform operations with complex numbers?
Ans. To perform operations with complex numbers, we can add, subtract, multiply, and divide them. Addition and subtraction is done by adding or subtracting the real and imaginary parts separately. Multiplication is done by using the distributive property and combining like terms. Division is done by multiplying both the numerator and denominator by the conjugate of the denominator.
 3. What is the geometric interpretation of complex numbers?
Ans. Complex numbers can be represented as points on a complex plane, where the real part is represented on the x-axis and the imaginary part is represented on the y-axis. The magnitude of a complex number represents its distance from the origin, while the argument represents the angle it makes with the positive x-axis.
 4. How are quadratic equations solved using complex numbers?
Ans. Quadratic equations with complex solutions can be solved using the quadratic formula. The discriminant of the quadratic equation is calculated, and if it is negative, the equation has complex solutions. The complex solutions can be found by using the formula x = (-b ± √(b^2 - 4ac))/(2a), where a, b, and c are the coefficients of the quadratic equation.
 5. How are complex numbers used in real-life applications?
Ans. Complex numbers have numerous applications in fields such as engineering, physics, and computer science. They are used in electrical engineering for analyzing AC circuits, in signal processing for analyzing and manipulating signals, in quantum mechanics for describing wave functions, and in computer graphics for representing and manipulating 3D objects.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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