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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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
We haveNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced[∵ i2 = - 1]
=  [(1  i)(1 + i)]n
=    [1   i2]n = [1 + 1]n = 2n
Hence, (1  i)nNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.2. Evaluate NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced, where n ∈ N .
Ans. 
We have NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
= (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,NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= – 1 + i.

Q.3. IfNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced=  x + iy, then find (x, y).
Ans.
We have NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= x + iy
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
 (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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= x + iy, then find the value of x + y.
Ans.
Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced [∵ i2 = - 1]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Comparing the real and imaginary parts, we get
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.5. If NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= a + ib, then find (a, b).
Ans. 
We have NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ ( - 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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans. 
Given that: a = cos θ + i sin θ
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence,NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.7. If (1 + i) z = (1 – i) z , then show that z = – iNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that: (1 + i)z = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence proved.

Q.8. If z = x + iy , then show that NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedwhere b ∈ R, represents a circle.
Ans.
Given that: z = x + iy
To prove: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ (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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced is 4, then show that the locus of the point representing z in the complex plane is a circle.
Ans.
Let   z = x + iy
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced 
So  NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Real part = 4
∴  NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 argNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced lies on a circle.
Ans.
Let z = x + iy
Given that:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ arg (z – 1) – arg (z + 1) =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ arg [x + iy – 1] – arg [x + iy + 1] =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ arg [(x – 1) + iy] – arg [(x + 1) + iy] = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ x2 + y2 – 1 = 2y
⇒ x2 + y2 – 2y – 1 = 0 which is a circle.
Hence, z lies on a circle.

Q.11. Solve the equation NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced = z + 1 + 2i.
Ans.
Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced = z + 1 + 2i
Let z = x + iy
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= (z + 1) + 2i
Squaring both sides
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of z = x + yi =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

LONG ANSWER TYPE QUESTIONS
Q.12. If NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced = z + 2 (1 + i), then find z.
Ans.
Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced = z + 2(1 + i)
Let z = x + iy
So, NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced = (x + iy) + 2(1 + i)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= x + iy + 2 + 2i
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= (x + 2) + (y + 2)i
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= (x + 2) + (y + 2)i
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ xy = (x – 1)(y + 3) ⇒ xy = xy + 3x – y – 3
⇒ 3x – y = 3 ⇒ 3x - 3 = y
⇒ 3(x – 1) = y ⇒ NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced⇒ x – 1 : y = 1 : 3
Hence, x – 1 : y = 1 : 3.

Q.14. Show thatNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedrepresents a circle. Find its centre and radius.
Ans.
Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + iy
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Here g = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the required equation of the circle is
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.15. IfNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedis a purely imaginary number (z ≠ – 1), then find the value of NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced.
Ans.
Given that NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedis purely imaginary number
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Since, the number is purely imaginary, then real part = 0
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ x2 + y2 – 1 = 0 ⇒ x2 + y2 = 1
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.16. z1 and z2 are two complex numbers such that NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced and arg (z1) + arg (z2) = π, then show that z1 =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced .
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]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced [∴ r1 = r2]…(ii)
From eqn. (i) and (ii) we get,
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence proved.

Q.17. If NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced then show that the real part of z2 is zero.
Ans.
 Let z1 = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ x2 + y2 = 1 ...(i)
Now NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
 Let the polar form of z1 = r1 (cos θ1 + i sin θ1)
    NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= r1 (cos θ1 – i sin θ1) = r1 [cos (– θ1) + i sin (– θ1)]
Similarly, z3 = r2 (cos θ2 + i sin θ2)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= r2 (cos θ2 – i sin θ2) = r2 [cos (– θ2) + i sin (– θ2)]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= arg (z1) – arg (z4) + arg (z2) – arg (z3)
= θ1  (- θ2) + (- θ1)  θ2
= θ1 + θ2  θ1  θ2 = 0
Hence, NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.19. IfNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced then
show that NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
We haveNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced...(i)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.21. Solve the system of equations Re (z2) = 0,NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that: Re(z2) = 0 andNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= 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 =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, z = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.22. Find the complex number satisfying the equationNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedin polar form.
Ans.
Given that:
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
So r = √2
Now arg(z) =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the polar is
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.24. If z and w are two complex numbers such thatNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced=1 and arg (z) – arg (w) =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedthen show thatNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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)]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= r1r2 = 1 (given)
Now arg (z) – arg (w) = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= 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)]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
HereNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & AdvancedHence proved.

Fill in the Blanks
Q.25. (i) For any two complex numbers z1, z2 and any real numbers a, b,NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(ii) The value ofNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(iii) The numberNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedis 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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(viii) IfNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedthen the greatest and least values ofNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedare ............... and ...............
(ix) IfNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedthen the locus of z is ............
(x) IfNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= 4 and arg (z) =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced, then z = ............
Ans.
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the filler isNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the filler is – 15.
(iii) NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(iv) i + i2 + i3 + ... upto 1000 terms
= i + i2 + i3 + ... + i1000 = 0
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the filler is 0.
(v) Multiplicative inverse of
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the filler = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(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 = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the filler isNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced.
(vii) arg ( z) + arg NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
If arg (z) = θ, then arg ( NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced ) = - θ
So θ + (– θ) = 0
Hence, the value of the filler is 0.
(viii) Given that:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
For the greatest value of
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the greatest value ofNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedis 6 and for the least value ofNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= 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:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + iy
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Which represents are equation of a circle.
Hence, the value of the filler is circle.
(x) Given that:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
From eqn. (i) and (ii)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of the filler isNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

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 ofNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(iv) The locus represented by NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced 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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced represents the region given by x > 3.
(vii) Let z1 and z2 be two complex numbers such that NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced , 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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
The value of NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced is minimum when x = 0, y = 0 i.e., 1.
Hence, it is ‘True’.
(iv) Let z = x + yi
Given that:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒(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)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ (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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Squaring both sides, we get
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Again squares on both sides, we get
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
∴ 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.
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
(a) Given that z = i + √3
Polar form of z = r [cos θ + i sin θ]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Since x > 0, y > 0
∴ Polar form of z =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Here argument (z) = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
So, NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Since x < 0 and y > 0
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, (b)  (iii).
(c) Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Which represents a circle on or outside having centre (0, – 4) and radius 3.
Hence, (e) ↔ (ii).
(f) NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Which is a circle having centre (– 4, 0) and r=NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced and is on or inside the circle.
Hence, (f) ↔ (vi).
(g) LetNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedwhich lies in third quadrant.
Hence, (g) ↔ (viii).
(h) Given that: z = 1 – i
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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 ofNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.29. If NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced is it necessary that z1 = z2?
Ans.
Let z1 = x1 + y1i and z2 = x2 + y2i
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= x + iy, what is the value of x2 + y2?
Ans.
Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= x + iy ...(i)
Taking conjugate on both sides
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= x – iy ...(ii)
Multiplying eqn. (i) and (ii) we have
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the value of x2 + y2 = NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.31. Find z ifNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= 4 and arg (z) =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= 4 ⇒ r = 4
So Polar form of z = r [ cos θ + i sin θ]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
= - 2 √3 + 2i
Hence, z = - 2 √3 + 2i .

Q.32. FindNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
= 1
Hence,NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Now Re(z) < 0 and image (z) > 0.
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the principal arg =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

Q.34. Where does z lie, ifNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given that:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(C) x = 0
(D) No value of x
Ans.
Let z = sin x + i cos 2x
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced = sin x – i cos 2x
But we are given that NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced = 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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ x = 2x ⇒ 2x - x = 0 ⇒ x = 0
Hence, the correct option is (c).

Q.36. The real value of α for which the expression NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedis purely real is :
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(C) n π
(D) None of these, where n ∈N
Ans.
Let NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Since, z is purely real, then
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
So, α = nπ, n ∈ N.

Q.37. If z = x + iy lies in the third quadrant, thenNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced 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.
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
When z lies in third quadrant thenNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedwill also be lie in third  quadrant  

NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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)NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedis equivalent to
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(C) z2 + 3
(D) None of these
Ans.
Given that: ( z + 3)NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
So ( z + 3)NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= (x + yi + 3)(x – yi + 3)
= [(x + 3) + yi][(x + 3) – yi]
= (x + 3)2 – y2i2 = (x + 3)2 + y2
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.39. IfNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedthen

(A) x = 2n+1
(B) x = 4n
(C) x = 2n
(D) x = 4n + 1,  where n ∈N
Ans.
Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ (i)x = (i)4n
⇒x = 4n, n ∈ N
Hence, the correct option is (b).

Q.40. A real value of x satisfies the equation NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= α − iβ(α, β ∈ R) if α2 + β2 =
(A) 1
(B) – 1
(C) 2
(D) – 2
Ans. 
Given that:NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced...(i)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced...(ii)
Multiplying eqn. (i) and (ii) we get
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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?
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(B) arg (z1z2) = arg (z1). arg (z2)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Let z1 = r1 (cos θ1 + i sin θ1)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= r1
and z2 = r2 (cos θ2 + i sin θ2)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= 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)]
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.42. The point represented by the complex number 2 – i is rotated about origin through an angle NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedin 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 ofNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedabout the origin in clockwise  direction.
Then the new position = z.e– (π/2)
= (2 – i) e– (π/2)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
= (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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Squaring both sides, we get a2 + b2 = c2 + d2
Hence, the correct option is (d).

Q.45. The complex number z which satisfies the conditionNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedlies on
(A) circle x2 + y2 = 1
(B) the x-axis
(C) the y-axis
(D) the line x + y = 1.
Ans.
Given that: NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Let z = x + yi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced= x2 + y2 ...(i)
Now z2 = x2 + y2i2 + 2xyi
z2 = x2 – y2 + 2xyi
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (b).

Q.47. NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedis possible if
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(C) arg (z1) = arg (z2)
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Ans.
Let z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2)
SinceNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
z1 + z2 = r1 cos θ1 + i r1 sin θ1 + r2 cos θ2 + i r2 sin θ2
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Squaring both sides, we get
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedis a real number is:
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(D) none of these.
Ans.
LetNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
If z is a real number, then
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
⇒ 3 cos θ = 0 ⇒ cos θ = 0
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
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) =NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advancedwhere z = 1 + 2i, thenNCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
(D) none of these.
Ans.
Given that: z = 1 + 2i
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
Now NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Complex Numbers & Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced

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