NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

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Q.1. For a positive integer n, find the value of NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
We haveNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev[∵ i2 = - 1]
=  [(1  i)(1 + i)]n
=    [1   i2]n = [1 + 1]n = 2n
Hence, (1  i)nNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

Q.2. Evaluate NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev, where n ∈ N .
Ans. 
We have NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
= (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 Notes | EduRev= – 1 + i.

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

Q.5. If NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= a + ib, then find (a, b).
Ans. 
We have NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ ( - 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 Notes | EduRev
Ans. 
Given that: a = cos θ + i sin θ
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence,NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

Q.7. If (1 + i) z = (1 – i) z , then show that z = – iNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
Given that: (1 + i)z = NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence proved.

Q.8. If z = x + iy , then show that NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevwhere b ∈ R, represents a circle.
Ans.
Given that: z = x + iy
To prove: NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ (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 Notes | EduRev 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 Notes | EduRev 
So  NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Real part = 4
∴  NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ 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 Notes | EduRev lies on a circle.
Ans.
Let z = x + iy
Given that:NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ arg (z – 1) – arg (z + 1) =NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ arg [x + iy – 1] – arg [x + iy + 1] =NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ arg [(x – 1) + iy] – arg [(x + 1) + iy] = NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ 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 Notes | EduRev = z + 1 + 2i.
Ans.
Given that: NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev = z + 1 + 2i
Let z = x + iy
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= (z + 1) + 2i
Squaring both sides
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ 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 Notes | EduRev
Hence, the value of z = x + yi =NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

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

Q.14. Show thatNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevrepresents a circle. Find its centre and radius.
Ans.
Given that: NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Let z = x + iy
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
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 Notes | EduRev
Here g = NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the required equation of the circle is
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

Q.15. IfNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevis a purely imaginary number (z ≠ – 1), then find the value of NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev.
Ans.
Given that NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevis purely imaginary number
Let z = x + yi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Since, the number is purely imaginary, then real part = 0
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ x2 + y2 – 1 = 0 ⇒ x2 + y2 = 1
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

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

Q.17. If NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev then show that the real part of z2 is zero.
Ans.
 Let z1 = x + yi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ x2 + y2 = 1 ...(i)
Now NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
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 Notes | EduRev
Ans.
 Let the polar form of z1 = r1 (cos θ1 + i sin θ1)
    NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= 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 Notes | EduRev= r2 (cos θ2 – i sin θ2) = r2 [cos (– θ2) + i sin (– θ2)]
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= arg (z1) – arg (z4) + arg (z2) – arg (z3)
= θ1  (- θ2) + (- θ1)  θ2
= θ1 + θ2  θ1  θ2 = 0
Hence, NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

Q.19. IfNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev then
show that NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
We haveNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev...(i)
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
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 Notes | EduRev
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 Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

Q.21. Solve the system of equations Re (z2) = 0,NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
Given that: Re(z2) = 0 andNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Let z = x + yi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= 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 Notes | EduRev
Hence, z = NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

Q.22. Find the complex number satisfying the equationNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
Given that: NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Let z = x + yi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ 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 Notes | EduRevin polar form.
Ans.
Given that:
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
So r = √2
Now arg(z) =NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the polar is
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

Q.24. If z and w are two complex numbers such thatNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev=1 and arg (z) – arg (w) =NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevthen show thatNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
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 Notes | EduRev= r1r2 = 1 (given)
Now arg (z) – arg (w) = NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= 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 Notes | EduRev
HereNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevHence 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 Notes | EduRev
(ii) The value ofNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
(iii) The numberNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevis 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 Notes | EduRev
(viii) IfNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevthen the greatest and least values ofNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevare ............... and ...............
(ix) IfNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevthen the locus of z is ............
(x) IfNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= 4 and arg (z) =NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev, then z = ............
Ans.
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the value of the filler isNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the value of the filler is – 15.
(iii) NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
(iv) i + i2 + i3 + ... upto 1000 terms
= i + i2 + i3 + ... + i1000 = 0
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the value of the filler is 0.
(v) Multiplicative inverse of
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the value of the filler = NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
(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 Notes | EduRev
Hence, the value of the filler isNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev.
(vii) arg ( z) + arg NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
If arg (z) = θ, then arg ( NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev ) = - θ
So θ + (– θ) = 0
Hence, the value of the filler is 0.
(viii) Given that:NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
For the greatest value of
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the greatest value ofNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevis 6 and for the least value ofNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= 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 Notes | EduRev
Let z = x + iy
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Which represents are equation of a circle.
Hence, the value of the filler is circle.
(x) Given that:NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Let z = x + yi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
From eqn. (i) and (ii)
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the value of the filler isNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

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

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

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

Q.32. FindNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
= 1
Hence,NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

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 Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Now Re(z) < 0 and image (z) > 0.
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the principal arg =NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev

Q.34. Where does z lie, ifNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
Given that:NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Let z = x + yi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ 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 Notes | EduRev
(C) x = 0
(D) No value of x
Ans.
Let z = sin x + i cos 2x
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev = sin x – i cos 2x
But we are given that NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev = 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 Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ 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 Notes | EduRevis purely real is :
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
(C) n π
(D) None of these, where n ∈N
Ans.
Let NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Since, z is purely real, then
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
So, α = nπ, n ∈ N.

Q.37. If z = x + iy lies in the third quadrant, thenNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev 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 Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
When z lies in third quadrant thenNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevwill also be lie in third  quadrant  

NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ 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 Notes | EduRevis equivalent to
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
(C) z2 + 3
(D) None of these
Ans.
Given that: ( z + 3)NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Let z = x + yi
So ( z + 3)NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= (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 Notes | EduRev
Hence, the correct option is (a).

Q.39. IfNCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRevthen

(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 Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ (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 Notes | EduRev= 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 Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev...(i)
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev...(ii)
Multiplying eqn. (i) and (ii) we get
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
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 Notes | EduRev
(B) arg (z1z2) = arg (z1). arg (z2)
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
Let z1 = r1 (cos θ1 + i sin θ1)
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= r1
and z2 = r2 (cos θ2 + i sin θ2)
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= 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 Notes | EduRev
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 Notes | EduRevin 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 Notes | EduRevabout the origin in clockwise  direction.
Then the new position = z.e– (π/2)
= (2 – i) e– (π/2)
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
= (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 Notes | EduRev
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 Notes | EduRevlies 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 Notes | EduRev
Let z = x + yi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
⇒ 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 Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Ans.
Let z = x + yi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev= x2 + y2 ...(i)
Now z2 = x2 + y2i2 + 2xyi
z2 = x2 – y2 + 2xyi
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the correct option is (b).

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

So  NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
NCERT Exemplar - Complex Numbers & Quadratic Equations Notes | EduRev
Hence, the correct option is (a).

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