Important Questions for Class 11 Maths Chapter 5

Important Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Here is the Important Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations-

Question 1:

Write the given complex number (1 – i) – ( –1 + i6) in the form a + ib

Solution:

Given Complex number: (1 – i) – ( –1 + i6) Multiply (-) by the term inside the second bracket ( –1 + i6) = 1 – i +1 – i6 = 2 – 7i, which is of the form a + ib.

Question 2:

Express the given complex number (-3) in the polar form.

Solution:

Given, complex number is -3. Let r cos θ = -3 …(1) and r sin θ = 0 …(2) Squaring and adding (1) and (2), we get r ² cos ² θ + r ² sin ² θ = (-3) ² Take r ² outside from L.H.S, we get r ² (cos ² θ + sin ² θ) = 9 We know that, cos ² θ + sin ² θ = 1, then the above equation becomes, r ² = 9 r = 3 (Conventionally, r > 0) Now, subsbtitute the value of r in (1) and (2) 3 cos θ = -3 and 3 sin θ = 0 cos θ = -1 and sin θ = 0 Therefore, θ = π Hence, the polar representation is, -3 = r cos θ + i r sin θ 3 cos π + 3 sin π = 3(cos π + i sin π) Thus, the required polar form is 3 cos π+ 3i sin π = 3(cos π+i sin π)

Question 3:

Solve the given quadratic equation 2x ² + x + 1 = 0.

Solution:

Given quadratic equation: 2x ² + x + 1 = 0 Now, compare the given quadratic equation with the general form ax ² + bx + c = 0 On comparing, we get a = 2, b = 1 and c = 1 Therefore, the discriminant of the equation is: D = b ² – 4ac Now, substitute the values in the above formula D = (1) ² – 4(2)(1) D = 1- 8 D = -7 Therefore, the required solution for the given quadratic equation is x =[-b ± √D]/2a x = [-1 ± √-7]/2(2) We know that, √-1 = i x = [-1 ± √7i] / 4 Hence, the solution for the given quadratic equation is (-1 ± √7i) / 4.

Question 4:

For any two complex numbers z ₁ and z ₂ , show that Re(z ₁ z ₂ ) = Rez ₁ Rez ₂ – Imz ₁ Imz ₂

Solution:

Given: z ₁ and z ₂ are the two complex numbers To prove: Re(z ₁ z ₂ ) = Rez ₁ Rez ₂ – Imz ₁ Imz ₂ Let z ₁ = x ₁ +iy ₁ and z ₂ = x ₂ +iy ₂ Now, z ₁ z ₂ =(x ₁ +iy ₁ )(x ₂ +iy ₂ ) Now, split the real part and the imaginary part from the above equation: ⇒ x ₁ (x ₂ +iy ₂ ) +iy ₁ (x ₂ +iy ₂ ) Now, multiply the terms: = x ₁ x ₂ +ix ₁ y ₂ +ix ₂ y ₁ +i ² y ₁ y ₂ We know that, i ² = -1, then we get = x ₁ x ₂ +ix ₁ y ₂ +ix ₂ y ₁ -y ₁ y ₂ Now, again seperate the real and the imaginary part: = (x ₁ x ₂ -y ₁ y ₂ ) +i (x ₁ y ₂ +x ₂ y ₁ ) From the above equation, take only the real part: ⇒ Re (z ₁ z ₂ ) =(x ₁ x ₂ -y ₁ y ₂ ) It means that, ⇒ Re(z ₁ z ₂ ) = Rez ₁ Rez ₂ – Imz ₁ Imz ₂ Hence, the given statement is proved.

Question 5:

Find the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)]

Solution:

Given: [(1+i)/(1-i)] – [(1-i)/(1+i)] Simplify the given expression, we get: [(1+i)/(1-i)] – [(1-i)/(1+i)] = [(1+i) ² – (1-i) ² ]/ [(1+i)(1-i)] = (1+i ² +2i-1-i ² +2i)) / (1 ² +1 ² ) Now, cancel out the terms, = 4i/2 = 2i Now, take the modulus, | [(1+i)/(1-i)] – [(1-i)/(1+i)]| =|2i| = √2 ² = 2 Therefore, the modulus of [(1+i)/(1-i)] – [(1-i)/(1+i)] is 2.

Benefits of Solving Important Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations

Here are the benefits of solving Important Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations:

Enhances Understanding : Deepens knowledge of complex numbers, their properties and quadratic equations with complex roots.

Builds Problem-Solving Skills : Provides practice with different question types, improving analytical and problem-solving abilities.

Prepares for Exams : Familiarizes students with exam-style questions, boosting readiness for Class 11 exams.

Improves Accuracy and Speed : Regular practice increases precision and helps solve problems faster.

Strengthens Foundation for Advanced Math : Establishes a strong base for more complex topics in higher grades and competitive exams.

Boosts Confidence : Tackling a variety of questions increases confidence in handling complex number and quadratic equation problems.