Buy BOOKS at Discounted Price

Complex Numbers And Quadratic Equations

Class 11th Mathematics CBSE Solution
Exercise 5.1
  1. Express each of the complex number given in the Exercises 1 to 10 in the form a…
  2. i^9 + i^19 Express each of the complex number given in the Exercises 1 to 10 in…
  3. i-39 Express each of the complex number given in the Exercises 1 to 10 in the…
  4. 3(7 + i7) + i(7 + i7) Express each of the complex number given in the Exercises…
  5. (1-i) - (-1 + i6) Express each of the complex number given in the Exercises 1 to…
  6. (1/5 + i 2/5) - (4+i 5/2) . Express each of the complex number given in the…
  7. [(1/3 + i 7/3) + (4+i 1/3)] - (- 4/3 + i) Express each of the complex number…
  8. (1 - i)^4 Express each of the complex number given in the Exercises 1 to 10 in…
  9. (1/3 + 3i)^3 Express each of the complex number given in the Exercises 1 to 10…
  10. (- 2 - 1/3 i)^3 Express each of the complex number given in the Exercises 1 to…
  11. Find the multiplicative inverse of the complex number (4-3i)
  12. Find the multiplicative inverse of √5 + 3i
  13. Find the multiplicative inverse of -i
  14. Express the following expression in the form of a + ib: (3+i root 5) (3-i root…
Exercise 5.2
  1. z = -1 - i√ 3 Find the modulus and the arguments of each of the complex numbers…
  2. z = - √3 + i Find the modulus and the arguments of each of the complex numbers…
  3. 1 - i Convert each of the complex numbers given in Exercises 3 to 8 in the polar…
  4. - 1 + i Convert each of the complex numbers given in Exercises 3 to 8 in the…
  5. - 1 - i Convert each of the complex numbers given in Exercises 3 to 8 in the…
  6. - 3 Convert each of the complex numbers given in Exercises 3 to 8 in the polar…
  7. √2 + i Convert each of the complex numbers given in Exercises 3 to 8 in the…
  8. i Convert each of the complex numbers given in Exercises 3 to 8 in the polar…
Miscellaneous Exercise
  1. Evaluate:
  2. For any two complex numbers z1 and z2, prove that Re (z1 z2) = Re z1 Re z2 -…
  3. Reduce (1/1-4i - 2/1+i) (3-4i/5+i) to the standard form.
  4. If x-iy = root a-ib/c-id prove that (x^2 + y^2)^2 = a^2 + b^2/c^2 + d^2…
  5. 1+7i/(2-i)^2 Convert the following in the polar form:
  6. 1+3i/1-2i Convert the following in the polar form:
  7. 3x^2 - 4x + 20/3 = 0 Solve each of the equation in Exercises 6 to 9:…
  8. x^2 - 2x + 3/2 = 0 Solve each of the equation in Exercises 6 to 9:…
  9. 27x^2 - 10 x + 1 = 0 Solve each of the equation in Exercises 6 to 9:…
  10. 21x^2 − 28x + 10 = 0 Solve each of the equation in Exercises 6 to 9:…
  11. If z1 = 2 - i, z2 = 1 + i, find | z_1+z_2+1/z_1-z_2+1|
  12. If a+ib = (x+i)^2/2x^2 + 1 prove that a^2 + b^2 = (x^2 + 1)^2/(2x^2 + 1)^2…
  13. Let z1 = 2 - i, z2 = -2 + i. Find: (i) re (z_1z_2/z_1) (ii) im (1/z_1 bar z_1)…
  14. Find the modulus and argument of the complex number 1+2i/1-3i
  15. Find the real numbers x and y if (x - iy) (3 + 5i) is the conjugate of -6 - 24i…
  16. Find the modulus of 1+i/1-i - 1-i/1+i
  17. If (x + iy)^3 = u + iv, then show that: u/x + v/y = 4 (x^2 - y^2)…
  18. If α and β are different complex numbers with |β| = 1 , then find…
  19. Find the number of non-zero integral solutions of the equation |1 - i|x = 2x…
  20. If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that: (a^2 + b^2)…
  21. If (1+i/1-i)^m = 1 , then find the least positive integral value of m.…
Exercise 5.3
  1. x^2 + 3 = 0 Solve each of the following equations:
  2. 2x^2 + x + 1 = 0 Solve each of the following equations:
  3. x^2 + 3x + 9 = 0 Solve each of the following equations:
  4. - x^2 + x - 2 = 0 Solve each of the following equations:
  5. x^2 + 3x + 5 = 0 Solve each of the following equations:
  6. x^2 - x + 2 = 0 Solve each of the following equations:
  7. √2x^2 + x + √2 = 0 Solve each of the following equations:
  8. √3x^2 - √2x + 3√3 = 0 Solve each of the following equations:
  9. x^2 + x + 1/√2 = 0 Solve each of the following equations:
  10. x^2 + x/√2 + 1 = 0 Solve each of the following equations:

Exercise 5.1
Question 1.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.


Answer:

Given Complex Number:



Multiplying it,



⇒ - 3 i2

We know that i2 = -1

So,

⇒ - 3 × -1

= 3



Question 2.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

i9 + i19


Answer:



⇒ 1×i + 1×(-i)


⇒ i + (-i)


⇒ 0


Question 3.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

i–39


Answer:

i -39

Now, - 39 = - 36 - 3

So we have,






Rationalizing it,





Question 4.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

3(7 + i7) + i(7 + i7)


Answer:

Multiplying like-like terms,


⇒ 21 + 21i + 7i +


⇒ 21 + 28i + 7× (-1)


⇒ 14 + 28i



Question 5.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

(1-i) – (-1 + i6)


Answer:

(1-i) – (-1 + i6)

⇒ 1-i + 1-i6

⇒ 2-7i


Question 6.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

.


Answer:

Given complex Number:



Now to Express the complex number in the form a + i b, we need to separate the real parts and imaginary parts for both the complex numbers.









Question 7.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.



Answer:


Separating like–like terms,





Question 8.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.

(1 – i)4


Answer:






⇒ -4



Question 9.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.



Answer:

Given Complex number:

Applying the formula: (a + b)3 = a3 + b3 + 3a2b + 3ab2

Therefore,





Question 10.

Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.



Answer:

Taking -1 outside the bracket,










Question 11.

Find the multiplicative inverse of the complex number (4-3i)


Answer:

Let z = 4 - 3 i

We know that multiplicative inverse of z is 1/z.

So the multiplicative inverse of given complex number will be,


Now rationalizing the number we get,



As i 2 = -1




Question 12.

Find the multiplicative inverse of √5 + 3i


Answer:

To Find: Multiplicative Inverse of √5 + 3 i

Concept: Multiplicative inverse of x is 1/x

Therefore,


Rationalizing numerator and denominator by multiplying and dividing by √5 - 3 i


Ans.

Question 13.

Find the multiplicative inverse of –i


Answer:

Multiplicative inverse of a number x is given by 1/x.

To Find: Multiplicative inverse of - i

Therefore,


Multiplying and dividing the expression by i we get,


we know that, i2 = -1

Putting this value we get,



Hence, multiplicative inverse of - i is i.


Question 14.

Express the following expression in the form of a + ib:



Answer:


Use formula: (a + b)(a - b) = a2 - b2 in the numerator we get,






Rationalizing it,







Or,





Exercise 5.2
Question 1.

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

z = –1 – i√ 3


Answer:

Given: z = –1 – i√ 3

Formulas:

Modulus of a complex number z = x + i y is given by,

|z| = √(x2 + y2)

So modulus of given z is,

|z| = √[(-1)2 + (√3)2]

|z| = √4 = 2

Now for argument let us find out the quadrant in which the complex number is present.
As real and imaginary parts are negative, z lies in third quadrant.

Thus Argument of z in third quadrant is,

arg(z) = α - π

And where,





α = π/3

arg(z) = π/3 - π


= - 2π/3



Question 2.

Find the modulus and the arguments of each of the complex numbers in Exercises 1 to 2.

z = – √3 + i


Answer:

As we know that the polar representation of a complex number z = x + iy is


z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.


So, now, let –√3 = r cos θ and 1 = r sin θ ……….(i)


Squaring both sides, we get


3 = r2 cos2 θ and 1 = r2 sin2 θ


Adding both the equations, we get


3 + 1 = r2 cos2 θ + r2 sin2 θ


⇒ 4 = r2 (cos2 θ + sin2 θ)


⇒ 4 = r2 or r2 = 4 [∵ sin2 θ + cos2 θ = 1]


⇒ r = √4


r = 2 (conventionally, r>0) ……….(ii)


Substituting r = 2 in (i), we get


–√3 = 2 cos θ and 1 = 2 sin θ




∵ We know that the complex number –√3 + i lies in the second quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.



……….(iii)


From (ii) and (iii), we have


r = 2 and


Question 3.

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

1 – i


Answer:

Let 1 = r cos θ and –1 = r sin θ ……….(i)


Squaring both sides, we get


1 = r2 cos2 θ and 1 = r2 sin2 θ


Adding both the equations, we get


1 + 1 = r2 cos2 θ + r2 sin2 θ


⇒ 2 = r2 (cos2 θ + sin2 θ)


⇒ 2 = r2 or r2 = 2 [∵ sin2 θ + cos2 θ = 1]


⇒ r = √2


⇒ r = √2 (conventionally, r>0)


Substituting r = √2 in (i), we get


1 = √2 cos θ and – 1 = √2 sin θ




∵ We know that the complex number 1 – i lies in the third quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.




As we know that the polar representation of a complex number z = x + iy is


z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.


So, the required polar form is .



Question 4.

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

– 1 + i


Answer:

Let – 1 = r cos θ and 1 = r sin θ ……….(i)


Squaring both sides, we get


1= r2 cos2 θ and 1 = r2 sin2 θ


Adding both the equations, we get


1 + 1 = r2 cos2 θ + r2 sin2 θ


⇒ 2 = r2 (cos2 θ + sin2 θ)


⇒ 2 = r2 or r2 = 2 [∵ sin2 θ + cos2 θ = 1]


⇒ r = √2


⇒ r = √2 (conventionally, r>0)


Substituting r = √2 in (i), we get


– 1 = √2 cos θ and 1 = √2 sin θ




∵ We know that the complex number – 1 + i lies in the second quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.




As we know that the polar representation of a complex number z = x + iy is


z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.


So, the required polar form is .



Question 5.

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

– 1 – i


Answer:

Let – 1 = r cos θ and – 1 = r sin θ ……….(i)


Squaring both sides, we get


1= r2 cos2 θ and 1 = r2 sin2 θ


Adding both the equations, we get


1 + 1 = r2 cos2 θ + r2 sin2 θ


⇒ 2 = r2 (cos2 θ + sin2 θ)


⇒ 2 = r2 or r2 = 2 [∵ sin2 θ + cos2 θ = 1]


⇒ r = √2


⇒ r = √2 (conventionally, r>0)


Substituting r = √2 in (i), we get


– 1 = √2 cos θ and – 1 = √2 sin θ




∵ We know that the complex number –1 – i lies in the third quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.




As we know that the polar representation of a complex number z = x + iy is


z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.


So, the required polar form is .



Question 6.

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

– 3


Answer:

Let – 3 = r cos θ and 0 = r sin θ ……….(i)


Squaring both sides, we get


9= r2 cos2 θ and 0 = r2 sin2 θ


Adding both the equations, we get


9 + 0 = r2 cos2 θ + r2 sin2 θ


⇒ 9 = r2 (cos2 θ + sin2 θ)


⇒ 9 = r2 or r2 = 9 [∵ sin2 θ + cos2 θ = 1]


⇒ r = √9


⇒ r = 3 (conventionally, r>0)


Substituting r = √2 in (i), we get


– 3 = 3 cos θ and 0 = 3 sin θ


⇒ cos θ = – 1 and sin θ = 0


⇒ θ = cos-1 (-1) and θ = sin-1 0


∵ We know that the complex number – 3 lies on the real axis (x-axis) and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.


∴ θ = cos-1 cos π and θ = sin-1 sin π


⇒ θ = π


As we know that the polar representation of a complex number z = x + iy is


z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.


So, the required polar form is z = 3 (cos π + i sin π)



Question 7.

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

√2 + i


Answer:

Let √3 = r cos θ and 1 = r sin θ ……….(i)


Squaring both sides, we get


3= r2 cos2 θ and 1 = r2 sin2 θ


Adding both the equations, we get


3 + 1 = r2 cos2 θ + r2 sin2 θ


⇒ 4 = r2 (cos2 θ + sin2 θ)


⇒ 4 = r2 or r2 = 4 [∵ sin2 θ + cos2 θ = 1]


⇒ r = √4


⇒ r = 2 (conventionally, r>0)


Substituting r = 2 in (i), we get


√3 = 2 cos θ and 1 = 2 sin θ




∵ We know that the complex number √3 + i lies in the first quadrant and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.




As we know that the polar representation of a complex number z = x + iy is


z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.


So, the required polar form is .



Question 8.

Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

i


Answer:

Let 0 = r cos θ and 1 = r sin θ……….(i)


Squaring both sides, we get


0= r2 cos2 θ and 1 = r2 sin2 θ


Adding both the equations, we get


0 + 1 = r2 cos2 θ + r2 sin2 θ


⇒ 1 = r2 (cos2 θ + sin2 θ)


⇒ 1 = r2 or r2 = 1 [∵ sin2 θ + cos2 θ = 1]


⇒ r = √1


⇒ r = 1 (conventionally, r>0)


Substituting r = √2 in (i), we get


0 = cos θ and 1 = sin θ


⇒ cos θ = 0 and sin θ = 1


⇒ θ = cos-1 0 and θ = sin-1 1


∵ We know that the complex number i lies on the imaginary axis (y-axis) and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.




As we know that the polar representation of a complex number z = x + iy is


z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.


So, the required polar form is .




Miscellaneous Exercise
Question 1.

Evaluate:


Answer:

It is given in the question that,

=


=


=


=


=


= [- 1 – i]3


= (- 1)3 [1 + i]3


= - [13 + i3 + 3 × 1 × i (1 + i)]


= - [1 + i3 + 3i + 3i2]


= - [1 – i + 3i – 3]


= - [- 2 + 2i]


= 2 – 2i


Question 2.

For any two complex numbers z1 and z2, prove that

Re (z1 z2) = Re z1 Re z2 – Imz1 IMz2


Answer:

It is given in the question that,

z1 and z2 are two complex numbers and we have to prove that:


Re (z1z2) = Rez1 Rez2 – Imz1 Imz2


For this, firstly let z1 = x1 + iy1 and z2 = x2 + iy2


Thus, z1z2 = (x1 + iy1) (x2 + iy2)


= x1 (x2 + iy2) + iy1 (x2 + iy2)


= x1x2 + ix2y2 + iy1x2 + i2y1y2


= x1x2 + ix2y2 + iy1x2 – y1y2 (i2 = - 1)


= (x1x2 – y1y2) + i (x1y2 + y1x2)


Re (z1z2) = x1x2 – y1y2


∴ Re (z1z2) = Rez1 Rez2 – Imz1 Imz2


Hence, proved


Question 3.

Reduce to the standard form.


Answer:

We have,



=


=


=


=


=


=


Now by multiplying numerator and denominator by (14 + 5i), we get:


=


=


=


=


=


Hence, this is the required standard form



Question 4.

If prove that


Answer:

It is given in the question that,


We have to prove that,



Proof:


Now by multiplying the numerator and denominator by (c + id), we get:




Thus,



Now, by comparing the real and the imaginary parts we get:





=


=


=


=


=


=


Hence, proved


Question 5.

Convert the following in the polar form:



Answer:

We have,



=


Now, by multiplying numerator and denominator by 3 + 4i we get:


=


=


=


=


= - 1 + i


Let us assume and


Now by squaring and adding the given values, we get:



We know that,


∴ r2 = 2



As conventionally r > 0, thus


Also,


As lies in the 2nd quadrant



Hence,


=


=


This is the required polar form of the given equation



Question 6.

Convert the following in the polar form:



Answer:

We have,


Now, by multiplying numerator and denominator by 1 + 2i we get:


=


=


=


= - 1 + i


Let us assume and


Now by squaring and adding the given values, we get:



We know that,


∴ r2 = 2



As conventionally r > 0, thus


Also,


As lies in the 2nd quadrant



Hence,


=


=


This is the required polar form of the given equation



Question 7.

Solve each of the equation in Exercises 6 to 9:



Answer:

We have,


This equation can be rewritten as follows:


9x2 – 12x + 20 = 0


Now, in order to solve the equation we have to compare this with ax2 + bx + c = 0 where a = 9, b = - 12 and c = 20


Thus, the discriminant of the above given equation is:


D = b2 – 4ac


= (- 12)2 – 4 × 9 × 20


= 144 – 720


= - 576


Hence, the required solutions are:



=


=


=


=


=



Question 8.

Solve each of the equation in Exercises 6 to 9:



Answer:

We have,


This equation can be rewritten as follows:


2x2 – 4x + 3 = 0


Now, in order to solve the equation we have to compare this with ax2 + bx + c = 0 where a = 2, b = - 4 and c = 3


Thus, the discriminant of the above given equation is:


D = b2 – 4ac


= (- 4)2 – 4 × 2 × 3


= 16 – 24


= - 8


Hence, the required solutions are:



=


=


=



Question 9.

Solve each of the equation in Exercises 6 to 9:

27x2 – 10 x + 1 = 0


Answer:

We have,


Now, in order to solve the equation we have to compare this with ax2 + bx + c = 0 where a = 27, b = - 10 and c = 1


Thus, the discriminant of the above given equation is:


D = b2 – 4ac


= (- 10)2 – 4 × 27 × 1


= 100 – 108


= - 8


Hence, the required solutions are:



=


=


=



Question 10.

Solve each of the equation in Exercises 6 to 9:

21x2 − 28x + 10 = 0


Answer:

We have,


Now, in order to solve the equation we have to compare this with ax2 + bx + c = 0 where a = 21, b = - 28 and c = 10


Thus, the discriminant of the above given equation is:


D = b2 – 4ac


= (- 28)2 – 4 × 21 × 10


= 784 – 840


= - 56


Hence, the required solutions are:



=


=


=


=



Question 11.

If z1 = 2 – i, z2 = 1 + i, find


Answer:

It is given in the question that,

z1 = 2 – i


z2 = 1 + i



=


=


=


=


=


=


=


=


=


Hence, the value of is .


Question 12.

If prove that


Answer:

It is given in the question that,


=


=


=


Now, by comparing the real and the imaginary parts we obtain:




=


=


=



Hence, proved



Question 13.

Let z1 = 2 – i, z2 = –2 + i. Find:

(i)

(ii)


Answer:

(i) It is given in the question that,


z1 = 2 – 1 and z2 = - 2 + i


Now, we have:


z1z2 = (2 – i) (- 2 + i)


= - 4 + 2i + 2i – i^2


= - 4 + 4i – (- 1)


= - 3 + 4i


Now, by multiplying numerator and denominator by (2 – i) we get:



=


=


=


=


Now, on comparing the real parts we get:



(ii) We have,



=


=


Now, on comparing the imaginary parts we get:



Question 14.

Find the modulus and argument of the complex number


Answer:

Let us assume,

Now multiplying the numerator and denominator by 1 + 3i, we get:



=


=


=


=


Let us now assume,



Now, by squaring the both sides we get:




As conventionally r > 0, thus




Hence,


Thus, modulus of the given complex number =


And, argument of given complex number =



Question 15.

Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of –6 – 24i


Answer:

Let us assume, z = (x – iy) (3 + 5i)

z = 3x + 5xi – 3yi – 5yi2


= 2x + 5xi – 3yi + 5y


= (3x + 5y) + i (5x – 3y)


= (3x + 5y) – i (5x – 3y)


It is given in the question that, = - 6 – 24i


Now by equating the real and imaginary parts of the equation, we get:


3x + 5y = - 6 (i)


5x – 3y = 24 (ii)


Now, by multiply equation (i) by 3 and equation (ii) by 5 and then by adding them both we get:


34x = 102



= 3


Now putting the value of x in (i), we get:


3 × 3 + 5y = - 6


5y = - 15



Hence, x = 3 and y = - 3



Question 16.

Find the modulus of


Answer:

Given in the question,


Thus, in order to find the modulus we will follow following steps:







Question 17.

If (x + iy)3 = u + iv, then show that:


Answer:

It is given in the question that,

(x + iy)3 = u + iv

we know that, (x + y)3 = x3 + y3 + 3xy(x + y)

x3 + i3y3 + 3 × x × iy (x + iy) = u + iv


x3 + i3y3 + 3x2yi + 3xi2y2= u + iv

Now we know that, i3 = - i and i2 = -1

x3 – iy3 + 3x2yi – 3xy2 = u + iv


(x3 – 3xy2) + i(-y3 + 3x2y)= u + iv


Now, equating the imaginary and real part we will get,


u = (x3 – 3xy2), v = (-y3 + 3x2y)


According to question,




= x2 – 3y2 + 3x2 – y2


= 4x2 – 4y2


= 4(x2 – y2)


Thus,


Hence, proved


Question 18.

If α and β are different complex numbers with |β| = 1 , then find


Answer:

It is given thatand β are different complex numbers,

And, |β|=1


= 1


Now,





=


= 1


Question 19.

Find the number of non-zero integral solutions of the equation |1 – i|x = 2x


Answer:

We have equation as:

|1-i|x = 2x

|z| denotes the modulus function

If z = x + i y, then |z| = √(x2 + y2)

For 1 - i, |1 - i| is given by,

|1 - i| = √(12 + (-1)2 ) = √2


2 x/2 = 2 x


On equating powers, we get:



x = 2x


2x - x = 0


x = 0


Hence, for the given solution 0 is the only integral solution.


Question 20.

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that:

(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2


Answer:

It is given in the question that,

(a + ib) (c + id) (e + if) (g + ih) = A + iB


Using, [|z1z2|=|z1||z2|]


|(a + ib)| × | (c + id) | × |(e + if) | × |(g + ih)| = |A + iB|



Now, we will square on both sides,


(a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2


Hence, proved



Question 21.

If , then find the least positive integral value of m.


Answer:

It is given in the question that,







im = 1


We can also write,


im = i4k


On equating the powers,


Thus, m = 4k, Where k is some integer.


∴ 1 is the least positive integer.


Least positive integral value of m is 4 × 1 = 4



Exercise 5.3
Question 1.

Solve each of the following equations:

x2 + 3 = 0


Answer:

It is given in the question that,

x2 + 3 = 0



Hence,



Question 2.

Solve each of the following equations:

2x2 + x + 1 = 0


Answer:

It is given in the question that,

2x2 + x + 1 = 0


Now, comparing the given value with ax2 + bx + c = 0 where a = 2, b = 1 and c = 1


Thus,


=


=


=


Hence, the value of x should be and



Question 3.

Solve each of the following equations:

x2 + 3x + 9 = 0


Answer:

It is given in the question that,

x2 + 3x + 9 = 0


Now, comparing the given value with ax2 + bx + c = 0 where a = 1, b = 3 and c = 9


Thus,


=


=


=


Hence, the value of x should be and



Question 4.

Solve each of the following equations:

– x2 + x – 2 = 0


Answer:

It is given in the question that,

- x2 + x - 2 = 0


Now, comparing the given value with ax2 + bx + c = 0 where a = - 1, b = 1 and c = - 2


Thus,


=


=


=


Hence, the value of x should be and



Question 5.

Solve each of the following equations:

x2 + 3x + 5 = 0


Answer:

It is given in the question that,

x2 + 3x + 5 = 0


Now, comparing the given value with ax2 + bx + c = 0 where a = 1, b = 3 and c = 5


Thus,


=


=


=


Hence, the value of x should be and



Question 6.

Solve each of the following equations:

x2 – x + 2 = 0


Answer:

It is given in the question that,

x2 - x + 2 = 0


Now, comparing the given value with ax2 + bx + c = 0 where a = 1, b = - 1 and c = 2


Thus,


=


=


=


Hence, the value of x should be and



Question 7.

Solve each of the following equations:

√2x2 + x + √2 = 0


Answer:

It is given in the question that,


Now, comparing the given value with ax2 + bx + c = 0 where , b = 1 and


Thus,


=


=


=


Hence, the value of x should be and



Question 8.

Solve each of the following equations:

√3x2 – √2x + 3√3 = 0


Answer:

It is given in the question that,


Now, comparing the given value with ax2 + bx + c = 0 where , and


Thus,


=


=


=


Hence, the value of x should be and



Question 9.

Solve each of the following equations:

x2 + x + 1/√2 = 0


Answer:

It is given in the question that,


Now, comparing the given value with ax2 + bx + c = 0 where , and


Thus,


=


=


=


Hence, the value of x should be and



Question 10.

Solve each of the following equations:

x2 + x/√2 + 1 = 0


Answer:

It is given in the question that,


Now, comparing the given value with ax2 + bx + c = 0 where , and


Thus,


=


=


=


=


Hence, the value of x should be and