Q.1. For a positive integer n, find the value of
Ans.
We have
[∵ i^{2} = - 1]
= [(1 – i)(1 + i)]^{n}
= [1 – i^{2}]^{n} = [1 + 1]^{n} = 2^{n}
Hence, (1 – i)^{n}
Q.2. Evaluate , where n ∈ N .
Ans.
We have
= (i + i^{2}) + (i^{2} + i^{3}) + (i^{3} + i^{4}) + (i^{4} + i^{5}) + (i^{5} + i^{6}) + (i^{6} + i^{7}) + (i^{7} + i^{8}) + (i^{8} + i^{9}) + (i^{9} + i^{10}) + (i^{10} + i^{11}) + (i^{11} + i^{12}) + (i^{12} + i^{13}) + (i^{13} + i^{14})
= i + 2(i^{2} + i^{3} + i^{4} + i^{5} + i^{6} + i^{7} + i^{8} + i^{9} + i^{10} + i^{11} + i^{12} + i^{13}) + i^{14}
= i + 2[– 1 – i + 1 + i – 1 – i + 1 + i – 1 – i + 1 + i] + (– 1)
= i + 2(0) – 1 ⇒ 1 + i
Hence,= – 1 + i.
Q.3. If= x + iy, then find (x, y).
Ans.
We have
= x + iy
⇒
⇒ (i)3 – (- i)3 = x + iy ⇒ i2.i + i2.i = x + iy
⇒ - i – i = x + iy ⇒ 0 – 2i = x + iy
Comparing the real and imaginary parts, we get
x = 0, y = - 2. Hence, (x , y) = (0 , - 2).
Q.4. If = x + iy, then find the value of x + y.
Ans.
Given that:
⇒
⇒
⇒ [∵ i2 = - 1]
⇒
Comparing the real and imaginary parts, we get
Hence,
Q.5. If = a + ib, then find (a, b).
Ans.
We have
⇒ ( - i)^{100} = a + bi ⇒ i^{100} = a + bi
⇒ (i^{4})^{25} = a + bi ⇒ (1)^{25} = a + bi ⇒ 1 = a + bi
⇒ 1 + 0i = a + bi
Comparing the real and imaginary parts, we have
a = 1, b = 0
Hence (a, b) = (1, 0)
Q.6. If a = cos θ + i sinθ, find the value of
Ans.
Given that: a = cos θ + i sin θ
∴
Hence,
Q.7. If (1 + i) z = (1 – i) z , then show that z = – i
Ans.
Given that: (1 + i)z =
⇒
⇒
∴
Hence proved.
Q.8. If z = x + iy , then show that where b ∈ R, represents a circle.
Ans.
Given that: z = x + iy
To prove:
⇒ (x + iy) (x – iy) + 2(x + iy + x – iy) + b = 0
⇒x^{2} + y^{2} – 2(x + x) + b = 0
⇒x^{2} + y^{2} – 4x + b = 0 Which represents a circle.
Hence proved.
Q.9. If the real part of is 4, then show that the locus of the point representing z in the complex plane is a circle.
Ans.
Let z = x + iy
∴
So
Real part = 4
∴
⇒ x^{2} + y^{2} + x – 2 = 4[(x – 1)^{2 }+ y^{2}]
⇒ x^{2} + y^{2} + x – 2 = 4[x^{2} + 1 – 2x + y^{2}]
⇒ x^{2} + y^{2} + x – 2 = 4x^{2} + 4 – 8x + 4y^{2}
⇒ x^{2} – 4x^{2 }+ y^{2} – 4y^{2} + x + 8x – 2 – 4 = 0
⇒ – 3x^{2} – 3y^{2} + 9x – 6 = 0
⇒ x^{2} + y^{2} – 3x + 2 = 0
Which represents a circle. Hence, z lies on a circle.
Q.10. Show that the complex number z, satisfying the condition arg lies on a circle.
Ans.
Let z = x + iy
Given that:
⇒ arg (z – 1) – arg (z + 1) =
⇒ arg [x + iy – 1] – arg [x + iy + 1] =
⇒ arg [(x – 1) + iy] – arg [(x + 1) + iy] =
⇒
⇒
⇒
⇒
⇒ x^{2} + y^{2} – 1 = 2y
⇒ x^{2} + y^{2} – 2y – 1 = 0 which is a circle.
Hence, z lies on a circle.
Q.11. Solve the equation = z + 1 + 2i.
Ans.
Given that: = z + 1 + 2i
Let z = x + iy
= (z + 1) + 2i
Squaring both sides
⇒
⇒ 0 = – 3 + 2z + 4(z + 1)i
⇒ 3 – 2z – 4(z + 1)i = 0
⇒ 3 – 2(x + yi) – 4[x + yi + 1]i = 0
⇒ 3 – 2x – 2yi – 4xi – 4yi^{2} – 4i = 0
⇒ 3 – 2x + 4y – 2yi – 4i – 4xi = 0
⇒ (3 – 2x + 4y) – i(2y + 4x + 4) = 0
⇒ 3 – 2x + 4y = 0 ⇒ 2x - 4y = 3 ...(i)
and 4x + 2y + 4 = 0 ⇒ 2x + y = - 2 ...(ii)
Solving eqn. (i) and (ii), we get
y = – 1 and x =
Hence, the value of z = x + yi =
LONG ANSWER TYPE QUESTIONS
Q.12. If = z + 2 (1 + i), then find z.
Ans.
Given that: = z + 2(1 + i)
Let z = x + iy
So, = (x + iy) + 2(1 + i)
⇒ = x + iy + 2 + 2i
⇒ = (x + 2) + (y + 2)i
= (x + 2) + (y + 2)i
Squaring both sides, we get
(x + 1)^{2} + y^{2} = (x + 2)^{2} + (y + 2)^{2}.i^{2} + 2(x + 2)(y + 2)i
⇒ x^{2} + 1 + 2x + y^{2} = x^{2} + 4 + 4x – y^{2} – 4y – 4 + 2(x + 2)(y + 2)i
Comparing the real and imaginary parts, we get
x^{2} + 1 + 2x + y^{2} = x^{2} + 4x – y^{2} – 4y and 2(x + 2)(y + 2) = 0
⇒ 2y^{2} – 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)
2y^{2} – 2 × (- 2) + 4y + 1 = 0
⇒ 2y^{2} + 4 + 4y + 1 = 0
⇒ 2y^{2} + 4y + 5 = 0
b^{2} – 4ac = (4)^{2} – 4 × 2 × 5
= 16 – 40 = - 24 < 0 no real roots.
Put y = – 2 in eqn. (i)
2(– 2)^{2} – 2x + 4(– 2) + 1 = 0
8 - 2x - 8 + 1 = 0 ⇒ x = 1/2 and y = -2
Hence, z = x + iy = (1/2 - 2i)
Q.13. If arg (z – 1) = arg (z + 3i), then find x – 1 : y. where z = x + iy
Ans.
Given that: arg (z – 1) = arg (z + 3i)
⇒ arg [x + yi – 1] = arg [x + yi + 3i]
⇒ arg [(x – 1) + yi] = arg [x + (y + 3)i]
⇒
⇒
⇒ xy = (x – 1)(y + 3) ⇒ xy = xy + 3x – y – 3
⇒ 3x – y = 3 ⇒ 3x - 3 = y
⇒ 3(x – 1) = y ⇒ ⇒ x – 1 : y = 1 : 3
Hence, x – 1 : y = 1 : 3.
Q.14. Show thatrepresents a circle. Find its centre and radius.
Ans.
Given that:
Let z = x + iy
∴
Squaring both sides, we get
(x – 2)^{2} + y^{2 }= 4[(x – 3)^{2} + y^{2}]
⇒ x^{2} + 4 – 4x + y^{2} = 4[x^{2} + 9 – 6x + y^{2}]
⇒ x^{2} + y^{2} – 4x + 4 = 4x^{2} + 4y^{2} – 24x + 36
⇒ 3x^{2} + 3y^{2} – 20x + 32 = 0
Here g =
Hence, the required equation of the circle is
Q.15. Ifis a purely imaginary number (z ≠ – 1), then find the value of .
Ans.
Given that is purely imaginary number
Let z = x + yi
Since, the number is purely imaginary, then real part = 0
∴
⇒ x^{2} + y^{2} – 1 = 0 ⇒ x^{2} + y^{2} = 1
⇒
Q.16. z_{1} and z_{2} are two complex numbers such that and arg (z_{1}) + arg (z_{2}) = π, then show that z_{1} = .
Ans.
Let z_{1} = r_{1} (cos θ_{1} + i sin θ_{1}) and z_{2} = r_{2} (cos θ_{2} + i sin θ_{2}) are polar form of two complex number z_{1} and z_{2}.
Given that: |z_{1}| =|z_{2}| ⇒ r_{1} = r_{2} ….(i)
and arg (z_{1}) + arg (z_{2}) = π
⇒ θ_{1} + θ_{2} = π
⇒ θ_{1} = π – θ_{2}
Now z_{1} = r_{1} [cos (π – θ_{2}) + i sin (π – θ_{2})]
⇒ z_{1} = r_{1} [- cos θ_{2} + i sin θ_{2}]
⇒ z_{1} = - r_{1} (cos θ_{2} – i sin θ_{2}) ….(i)
z_{2} = r_{2} [cos θ_{2} + i sin θ_{2}]
[∴ r_{1} = r_{2}]…(ii)
From eqn. (i) and (ii) we get,
Hence proved.
Q.17. If then show that the real part of z_{2} is zero.
Ans.
Let z_{1} = x + yi
⇒ x^{2} + y^{2} = 1 ...(i)
Now
Hence, the real part of z_{2} is 0.
Q.18. If z_{1}, z_{2} and z_{3}, z_{4} are two pairs of conjugate complex numbers, then find
Ans.
Let the polar form of z_{1} = r_{1} (cos θ_{1} + i sin θ_{1})
∴ = r_{1} (cos θ_{1} – i sin θ_{1}) = r_{1} [cos (– θ_{1}) + i sin (– θ_{1})]
Similarly, z_{3} = r_{2} (cos θ_{2} + i sin θ_{2})
∴ = r_{2} (cos θ_{2} – i sin θ_{2}) = r_{2} [cos (– θ_{2}) + i sin (– θ_{2})]
= arg (z_{1}) – arg (z_{4}) + arg (z_{2}) – arg (z_{3})
= θ_{1} – (- θ_{2}) + (- θ_{1}) – θ_{2}
= θ_{1} + θ_{2} – θ_{1} – θ_{2} = 0
Hence,
Q.19. If then
show that
Ans.
We have
⇒ ...(i)
⇒
⇒
L.H.S. = R.H.S. Hence proved.
Q.20. If for complex numbers z_{1} and z_{2}, arg (z_{1}) – arg (z_{2}) = 0, then show that
Ans.
Given that for z_{1} and z_{2}, arg (z_{1}) – arg (z_{2}) = 0
Let us represent z_{1} and z_{2} in polar form
z_{1} = r_{1} (cos θ_{1} + i sin θ_{1}) and z_{2} = r_{2} (cos θ_{2} + i sin θ_{2})
arg (z_{1}) = θ_{1} and arg (z_{2}) = θ_{2}
Since arg (z_{1}) – arg (z_{2}) = 0
⇒ θ_{1} – θ_{2} = 0 ⇒ θ_{1} = θ_{2}
Now z_{1} – z_{2} = r_{1} (cos θ_{1} + i sin θ_{1}) – r_{2} (cos θ_{2} + i sin θ_{2})
= r_{1} cos θ_{1} + i r_{1} sin θ_{1} – r_{2} cos θ_{1} – i r_{2} sin θ_{1}
[∴ θ_{1} = θ_{2}]
= (r_{1 }cos θ_{1} – r_{2} cos θ_{1}) + i(r_{1} sin θ_{1} – r_{2} sin θ_{1})
Hence,
Q.21. Solve the system of equations Re (z^{2}) = 0,
Ans.
Given that: Re(z^{2}) = 0 and
Let z = x + yi
∴
⇒ = 2 ⇒ x2 + y2 = 4 ...(i)
Since, z = x + yi
z^{2} = x^{2} + y2i^{2} + 2xyi
⇒ z^{2} = x^{2} – y^{2} + 2xyi
∴ Re(z^{2}) = x^{2} – y^{2}
⇒ x^{2} – y^{2} = 0 ...(ii)
From eqn. (i) and (ii), we get x
x^{2} + y^{2} = 4 ⇒ 2x^{2} = 4 ⇒ x^{2} = 2 ⇒ x =
Hence, z =
Q.22. Find the complex number satisfying the equation
Ans.
Given that:
Let z = x + yi
∴
⇒ x^{2} = 2(x^{2} + 2x + 1 + y^{2})
⇒ x^{2} = 2x^{2} + 4x + 2 + 2y^{2}
⇒ x^{2} + 4x + 2 + 2y^{2} = 0
⇒ x^{2} + 4x + 2 + 2(– 1)^{2} = 0 [∴ y = – 1]
⇒ x^{2} + 4x + 4 = 0
⇒ (x + 2)^{2} = 0
⇒ x + 2 = 0 ⇒ x = – 2
Hence, z = x + yi = – 2 – i.
Q.23. Write the complex number in polar form.
Ans.
Given that:
So r = √2
Now arg(z) =
Hence, the polar is
Q.24. If z and w are two complex numbers such that=1 and arg (z) – arg (w) =then show that
Ans.
Let z = r_{1} (cos θ_{1} + i sin θ_{1}) and w = r_{2} (cos θ_{2} + i sin θ_{2})
zw = r_{1}r_{2} [(cos θ_{1} + i sin θ_{1})] [(cos θ_{2} + i sin θ_{2})]
= r_{1}r_{2} = 1 (given)
Now arg (z) – arg (w) =
= r_{1} (cos θ_{1} – i sin θ_{1}) r_{2} (cos θ_{2} + i sin θ_{2})
= r_{1} r_{2} [cos θ_{1} cos θ_{2} + i cos θ_{1} sin θ_{2} – i sin θ_{1} cos θ_{2} – i_{2} sin θ_{1 }sin θ_{2}]
= r_{1} r_{2} [(cos θ_{1} cos θ_{2} + sin θ_{1} sin θ_{2}) + i(cos θ_{1} sin θ_{2} – sin θ_{1} cos θ_{2})]
= r_{1} r_{2} [cos (θ_{2} – θ_{1}) + i sin (θ_{2} – θ_{1})]
HereHence proved.
Fill in the Blanks
Q.25. (i) For any two complex numbers z_{1}, z_{2} and any real numbers a, b,
(ii) The value of
(iii) The numberis equal to ...............
(iv) The sum of the series i + i^{2} + i^{3} + ... upto 1000 terms is ..........
(v) Multiplicative inverse of 1 + i is ................
(vi) If z_{1} and z_{2} are complex numbers such that z_{1} + z_{2} is a real number, then z_{2 }= ....
(vii) arg (z) + arg
(viii) Ifthen the greatest and least values ofare ............... and ...............
(ix) Ifthen the locus of z is ............
(x) If= 4 and arg (z) =, then z = ............
Ans.
Hence, the value of the filler is
Hence, the value of the filler is – 15.
(iii)
(iv) i + i^{2} + i^{3} + ... upto 1000 terms
= i + i^{2} + i^{3} + ... + i^{1000} = 0
Hence, the value of the filler is 0.
(v) Multiplicative inverse of
Hence, the value of the filler =
(vi) Let z_{1 }= x_{1} + iy_{1} and z_{2}
= x_{2} + iy_{2} z_{1 }+ z_{2}
= (x_{1} + iy_{1}) + (x_{2} + iy_{2}) z_{1} + z_{2}
= (x_{1} + x_{2}) + (y_{1} + y_{2})i
If z_{1} + z_{2} is real then
y_{1} + y_{2} = 0⇒ y_{1} = – y_{2}
∴ z_{2} = x_{2} – iy_{1}
z_{2} = x_{1} – iy_{1} (when x_{1} = x_{2})
So z_{2} =
Hence, the value of the filler is.
(vii) arg ( z) + arg
If arg (z) = θ, then arg ( ) = - θ
So θ + (– θ) = 0
Hence, the value of the filler is 0.
(viii) Given that:
For the greatest value of
Hence, the greatest value ofis 6 and for the least value of= 0.
[∴ The least value of the modulus of complex number is 0] Hence, the value of the filler are 6 and 0.
(ix) Given that:
Let z = x + iy
Which represents are equation of a circle.
Hence, the value of the filler is circle.
(x) Given that:
Let z = x + yi
From eqn. (i) and (ii)
Hence, the value of the filler is
Q.26. State True or False for the following :
(i) The order relation is defined on the set of complex numbers.
(ii) Multiplication of a non zero complex number by – i rotates the point about origin through a right angle in the anti-clockwise direction.
(iii) For any complex number z the minimum value of
(iv) The locus represented by is a line perpendicular to the join of (1, 0) and (0, 1).
(v) If z is a complex number such that z ≠ 0 and Re (z) = 0, then Im (z^{2}) = 0.
(vi) The inequality represents the region given by x > 3.
(vii) Let z1 and z2 be two complex numbers such that , then arg (z1 – z2) = 0.
(viii) 2 is not a complex number.
Ans.
(i) Comparison of two purely imaginary complex numbers is not possible. However, the two purely real complex numbers can be compared.
So it is ‘False’.
(ii) Let z = x + yi
z.i = (x + yi) i = xi – y which rotates at angle of 180°
So, it is ‘False’.
(iii) Let z = x + yi
The value of is minimum when x = 0, y = 0 i.e., 1.
Hence, it is ‘True’.
(iv) Let z = x + yi
Given that:
⇒(x – 1)^{2} + y^{2} = x^{2} + (1 – y)^{2}
⇒ x^{2} – 2x + 1 + y^{2} = x^{2} + 1 + y^{2} – 2y
⇒ – 2x + 2y = 0
⇒ x – y = 0 which is a straight line.
Slope = 1
Now equation of a line through the point (1, 0) and (0, 1)
⇒ y = – x + 1 whose slope = – 1.
Now the multiplication of the slopes of two lines = – 1 x 1 = - 1,
so they are perpendicular.
Hence, it is ‘True’.
(v) Let z = x + yi, z ≠ 0 and Re(z) = 0
Since real part is 0 ⇒ x = 0
∴ z = 0 + yi = yi
∴ lm (z^{2}) = y^{2}i^{2} = - y^{2} which is real.
Hence, it is ‘False’.
(vi) Given that:
Let z = x + yi
⇒ (x – 4)^{2} + y^{2} < (x – 2)^{2} + y^{2}
⇒ (x – 4)^{2} < (x – 2)^{2}
⇒ x^{2} + 16 – 8x < x^{2} + 4 – 4x
⇒ – 8x + 4x < – 16 + 4
⇒ – 4x < – 12 ⇒ x > 3
Hence, it is ‘True’.
(vii) Let z_{1} = x_{1} + y_{1}i and z_{2} = x_{2} + y_{2}i
Squaring both sides, we get
Again squares on both sides, we get
∴ arg (z_{1}) = arg (z_{2})
⇒ arg (z_{1}) – arg (z_{2}) = 0
Hence, it is ‘True’.
(viii) Since 2 has no imaginary part.
So, 2 is not a complex number.
Hence, it is ‘True’.
Q.27. Match the statements of Column A and Column B.
Ans.
(a) Given that z = i + √3
Polar form of z = r [cos θ + i sin θ]
Since x > 0, y > 0
∴ Polar form of z =
Hence, (a) ⇒ 8y = 0 Þ y = 0. Which is the equation of x-axis and it is perpendicular to the line segment joining (0, – 2) and (0, 2).
Hence, (d) ↔ (iv).
(b) Given that
Here argument (z) =
So,
Since x < 0 and y > 0
Hence, (b) ↔ (iii).
(c) Given that:
Let z = x + yi
⇒ 8y = 0 ⇒ y = 0. Which is the equation of x-axis and it is perpendicular to the line segment joining (0, – 2) and (0, 2).
Hence, (d) ↔ (iv).
(e) Given that:
Let z = x + yi
Which represents a circle on or outside having centre (0, – 4) and radius 3.
Hence, (e) ↔ (ii).
(f)
Let z = x + yi
Which is a circle having centre (– 4, 0) and r= and is on or inside the circle.
Hence, (f) ↔ (vi).
(g) Let
∴which lies in third quadrant.
Hence, (g) ↔ (viii).
(h) Given that: z = 1 – i
Which lies in first quadrant.
Hence, (h)↔ (vii).
Hence, the correct matches are (a) ↔ (v), (b) ↔ (iii), (c) ↔ (i), (d) ↔ (iv), (e) ↔ (ii), (f) ↔ (vi), (g) ↔ (viii), (h) ↔ (vii).
Q.28. What is the conjugate of
Ans.
Given that
Q.29. If is it necessary that z_{1} = z_{2}?
Ans.
Let z_{1} = x_{1} + y_{1}i and z_{2} = x_{2} + y_{2}i
⇒ x_{1} = ± x_{2} and y_{1} = ± y_{2}
So z_{1} = x_{1} + y_{1}i and z_{2} = ± x_{2} ± y_{2}i
⇒ z_{1} ≠ z_{2}
Hence, it is not necessary that z_{1} = z_{2}.
Q.30. If = x + iy, what is the value of x^{2} + y^{2}?
Ans.
Given that: = x + iy ...(i)
Taking conjugate on both sides
⇒ = x – iy ...(ii)
Multiplying eqn. (i) and (ii) we have
Hence, the value of x^{2} + y^{2} =
Q.31. Find z if= 4 and arg (z) =
Ans.
Given that:
= 4 ⇒ r = 4
So Polar form of z = r [ cos θ + i sin θ]
= - 2 √3 + 2i
Hence, z = - 2 √3 + 2i .
Q.32. Find
Ans.
= 1
Hence,
Q.33. Find principal argument of (1 + i √3 )^{2} .
Ans.
Given that: (1 + i √3 )^{2} = 1 + i^{2} . 3 + 2 √3i
= 1 - 3 + 2 √3i = - 2 + 2 √3i
⇒
⇒
Now Re(z) < 0 and image (z) > 0.
∴
Hence, the principal arg =
Q.34. Where does z lie, if
Ans.
Given that:
Let z = x + yi
∴
⇒
⇒ x^{2} + (y – 5)^{2} = x^{2} + (y + 5)^{2}
⇒(y – 5)^{2} = (y + 5)^{2}
⇒ y^{2} + 25 – 10y = y^{2} + 25 + 10y
⇒ 20y = 0⇒ y = 0
Hence, z lies on x-axis i.e., real axis.
Choose the correct answer from the given four options
Q.35. sinx + i cos 2x and cos x – i sin 2x are conjugate to each other for:
(A) x = nπ
(C) x = 0
(D) No value of x
Ans.
Let z = sin x + i cos 2x
= sin x – i cos 2x
But we are given that = cos x – i sin 2x
∴ sin x – i cos 2x = cos x – i sin 2x
Comparing the real and imaginary parts, we get
sin x = cos x and cos 2x = sin 2x
⇒ tan x = 1 and tan 2x = 1
⇒
∴
⇒ x = 2x ⇒ 2x - x = 0 ⇒ x = 0
Hence, the correct option is (c).
Q.36. The real value of α for which the expression is purely real is :
(C) n π
(D) None of these, where n ∈N
Ans.
Let
Since, z is purely real, then
So, α = nπ, n ∈ N.
Q.37. If z = x + iy lies in the third quadrant, then also lies in the third quadrant if
(A) x > y > 0
(B) x < y < 0
(C) y < x < 0
(D) y > x > 0
Ans.
Given that: z = x + iy
If z lies in third quadrant.
So x < 0 and y < 0.
When z lies in third quadrant thenwill also be lie in third quadrant
∴
⇒ 2 – y^{2} < 0 and 2xy > 0
⇒ x^{2} < y^{2} and xy > 0
So x < y < 0.
Hence, the correct option is (b)
Q.38. The value of (z + 3)is equivalent to
(C) z^{2} + 3
(D) None of these
Ans.
Given that: ( z + 3)
Let z = x + yi
So ( z + 3)= (x + yi + 3)(x – yi + 3)
= [(x + 3) + yi][(x + 3) – yi]
= (x + 3)^{2} – y^{2}i^{2} = (x + 3)^{2} + y^{2}
Hence, the correct option is (a).
Q.39. Ifthen
(A) x = 2n+1
(B) x = 4n
(C) x = 2n
(D) x = 4n + 1, where n ∈N
Ans.
Given that:
⇒ (i)^{x} = (i)^{4n}
⇒x = 4n, n ∈ N
Hence, the correct option is (b).
Q.40. A real value of x satisfies the equation = α − iβ(α, β ∈ R) if α^{2} + β^{2} =
(A) 1
(B) – 1
(C) 2
(D) – 2
Ans.
Given that:
⇒
⇒
⇒
⇒ ...(i)
⇒ ...(ii)
Multiplying eqn. (i) and (ii) we get
⇒
⇒
⇒
So, α^{2} + β^{2}^{ }= 1
Hence, the correct option is (a).
Q.41. Which of the following is correct for any two complex numbers z_{1 }and z_{2}?
(B) arg (z_{1}z_{2}) = arg (z_{1}). arg (z_{2})
Ans.
Let z_{1} = r_{1} (cos θ_{1} + i sin θ_{1})
∴ = r_{1}
and z_{2} = r_{2} (cos θ_{2} + i sin θ_{2})
∴ = r_{2}
z_{1}z_{2} = r_{1} (cos θ_{1} + i sin θ_{1}) . r_{2} (cos θ_{2} + i sin θ_{2})
= r_{1}r_{2} (cos θ_{1} + i sin θ_{1}) . (cos θ_{2} + i sin θ_{2})
= r_{1}r_{2} (cos θ_{1} cos θ_{2} + i sin θ_{2} cos θ_{1} + i sin θ_{1} cos θ_{2} + i_{2} sin θ_{1} sin θ_{2})
= r_{1}r_{2} [(cos θ_{1} cos θ_{2} – sin θ_{1} sin θ_{2}) + i(sin θ_{1} cos θ_{2} + cos θ_{1} sin θ_{2})]
= r_{1}r_{2} [cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2})]
Hence, the correct option is (a).
Q.42. The point represented by the complex number 2 – i is rotated about origin through an angle in the clockwise direction, the new position of point is:
(A) 1 + 2i
(B) –1 – 2i
(C) 2 + i
(D) –1 + 2 i
Ans.
Given that: z = 2 – i
If z rotated through an angle ofabout the origin in clockwise direction.
Then the new position = z.e^{– (π/2)}
= (2 – i) e^{– (π/2)}
= (2 – i) (0 – i) = – 1 – 2i
Hence, the correct option is (b).
Q.43. Let x, y ∈ R, then x + iy is a non real complex number if:
(A) x = 0
(B) y = 0
(C) x ≠ 0
(D) y ≠ 0
Ans.
x + yi is a non-real complex number if y ≠ 0. If x, y ∈ R.
Hence, the correct option is (d).
Q.44. If a + ib = c + id, then
(A) a^{2} + c^{2} = 0
(B) b^{2} + c^{2} = 0
(C) b^{2} + d^{2} = 0
(D) a^{2} + b^{2} = c^{2} + d^{2}
Ans.
Given that: a + ib = c + id
Squaring both sides, we get a^{2} + b^{2} = c^{2} + d^{2}
Hence, the correct option is (d).
Q.45. The complex number z which satisfies the conditionlies on
(A) circle x^{2} + y^{2} = 1
(B) the x-axis
(C) the y-axis
(D) the line x + y = 1.
Ans.
Given that:
Let z = x + yi
⇒ x^{2} + (y + 1)^{2} = x^{2} + (y – 1)^{2}
⇒ (y + 1)^{2} = (y – 1)^{2}
⇒ y^{2} + 2y + 1 = y^{2} – 2y + 1
⇒ 2y = – 2y
⇒ 4y = 0 ⇒ y = 0 ⇒ x-axis.
Hence, the correct option is (b).
Q.46. If z is a complex number, then
Ans.
Let z = x + yi
= x^{2} + y^{2} ...(i)
Now z^{2} = x^{2} + y^{2}i^{2} + 2xyi
z^{2} = x^{2} – y^{2} + 2xyi
Hence, the correct option is (b).
Q.47. is possible if
(C) arg (z_{1}) = arg (z_{2})
Ans.
Let z_{1} = r_{1} (cos θ_{1} + i sin θ_{1}) and z_{2} = r_{2} (cos θ_{2} + i sin θ_{2})
Since
z_{1} + z_{2} = r_{1} cos θ_{1} + i r_{1} sin θ_{1} + r_{2} cos θ_{2} + i r_{2} sin θ_{2}
Squaring both sides, we get
⇒ 2r_{1}r_{2} – 2r_{1}r_{2} cos (θ_{1} – θ_{2}) = 0
⇒ 1 – cos (θ_{1} – θ_{2}) = 0 ⇒ cos (θ_{1} – θ_{2}) = 1
⇒ θ_{1} – θ_{2} = 0 ⇒ θ_{1} = θ_{2}
So, arg (z_{1}) = arg (z_{2})
Hence, the correct option is (c).
Q.48. The real value of θ for which the expression is a real number is:
(D) none of these.
Ans.
Let
If z is a real number, then
⇒ 3 cos θ = 0 ⇒ cos θ = 0
∴
Hence, the correct option is (c).
Q.49. The value of arg (x) when x < 0 is:
(A) 0
(B) π/2
(C) π
(D) none of these
Ans.
Let z = – x + 0i and x < 0
∴
Since, the point (–x, 0) lies on the negative side of the real axis
(∴ x < 0).
∴ Principal argument (z) = π
Hence, the correct option is (c).
Q.50. If f (z) =where z = 1 + 2i, then
(D) none of these.
Ans.
Given that: z = 1 + 2i
Now
So
Hence, the correct option is (a).