Complex Numbers CBSE Board Important Questions with Solutions | Practice Problems pdf free download

Complex Numbers Class 11 Questions with Solutions

1. Write the complex number i⁹ + i¹⁹ in the form of a + ib.

Solutions:

Given number: i⁹ + i¹⁹.

The expression i⁹ + i¹⁹ can be represented as follows:

i⁹ + i¹⁹ = (i²)⁴. i + (i²)⁹. i …(1)

We know that, i² = 1.

On substituting i² = -1, we get

i⁹ + i¹⁹ = (-1)⁴.i + (-1)⁹.i

i⁹ + i¹⁹ = 1.i + (-1).i

i⁹ + i¹⁹ = i – i

i⁹ + i¹⁹ = 0.

Therefore, i⁹ + i¹⁹ in the form of a + ib is 0 + i0.

2. Simplify the expression: i³⁰ + i⁴⁰ + i⁶⁰.

Solutions:

Given expression: i³⁰ + i⁴⁰ + i⁶⁰.

The given expression can be simplified as follows:

i³⁰ + i⁴⁰ + i⁶⁰ = (i⁴)⁷. i² + (i⁴)¹⁰ + (i⁴)¹⁵.

We know that the value of i⁴ is 1.

i³⁰ + i⁴⁰ + i⁶⁰ = (1)⁷. i² + (1)¹⁰ + (1)¹⁵.

i³⁰ + i⁴⁰ + i⁶⁰ = (1)i² + 1 + 1

i³⁰ + i⁴⁰ + i⁶⁰ = -1 + 1 + 1 [since i² = 1]

i³⁰ + i⁴⁰ + i⁶⁰ = 1

Therefore, the simplification of i³⁰ + i⁴⁰ + i⁶⁰ is 1.

3. Express the given expression (1 + i) (1 + 2i) in the form a + ib and find the values of a and b.

Solutions:

Given expression: (1 + i) (1 + 2i)

Hence, (1 + i) (1 + 2i) = 1(1) + 1(2i) + i + 2i(i)

(1 + i) (1 + 2i) = 1 + 2i + i + 2i²

(1 + i) (1 + 2i) = 1 + 2i + i + 2(-1) [As, i² = -1]

(1 + i) (1 + 2i) = 1 + 2i + i – 2

(1 + i) (1 + 2i) = -1 + 3i

Hence, the expression (1 + i) (1 + 2i) in the form of a + bi is -1 + 3i.

Thus, the value of a = -1 and b = 3.

4. Solve the equation: 2x² + x + 1 = 0

Solutions:

Given equation: 2x² + x + 1 = 0 …(1)

Now, compare the given equation with the standard quadratic equation ax2 + bx + c = 0 …(b)

On comparing the equations (1) and (2), we get

a = 2, b = 1 and c = 1.

As we know discriminant, D = b² – 4ac …(3)

Substitute the values a, b and c in equation (3), we get

D = (1)² – 4(2)(1)

D = 1 – 8

D = -7

We know that the quadratic equation formula is:

\(\begin{array}{l}x = \frac{-b\pm \sqrt{D}}{2a}\end{array} \)

\(\begin{array}{l}x = \frac{-1\pm \sqrt{-7}}{2\times 2}\end{array} \)

Now, the above equation can be written as

\(\begin{array}{l}x = \frac{-1\pm \sqrt{-1 \times 7}}{2\times 2}\end{array} \)

Since, √-1 = i,

\(\begin{array}{l}x = \frac{-1\pm \sqrt{7i}}{4}\end{array} \)

Hence, the solutions to the given quadratic equation are:

\(\begin{array}{l}x = \frac{-1 + \sqrt{7i}}{4}, and\ x = \frac{-1 – \sqrt{7i}}{4}\end{array} \)

5. Determine the multiplicative inverse of 4 – 3i.

Solutions:

Let z = 4 – 3i.

The conjugate of 4 – 3i is 4 + 3i.

As we know, the multiplicative inverse of z is 1/z.

Hence, 1/z = 1/ (4+3i)

Therefore, the multiplicative inverse of 4 – 3i is:

\(\begin{array}{l}z^{-1} = \frac{1}{4-3i}\times \frac{4+3i}{4+3i}\end{array} \)

\(\begin{array}{l}z^{-1} = \frac{4+3i}{4^{2}-(3i)^{2}}\end{array} \)

\(\begin{array}{l}z^{-1} = \frac{4+3i}{16 – 9i^{2}}\end{array} \)

As, i² = -1

\(\begin{array}{l}z^{-1} = \frac{4+3i}{16 +9}\end{array} \)

\(\begin{array}{l}z^{-1} = \frac{4+3i}{25}\end{array} \)

Therefore, the multiplicative inverse of 4 – 3i is (4 + 3i)/25.

6. If z₁ and z₂ are the two complex numbers, then show that Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂.

Solutions:

Let z₁ = a₁ + ib₁ and z₂ = a₂+ ib₂

Now, take the product of these two complex numbers.

z₁z₂ = (a₁ + ib₁)(a₂+ ib₂)

z₁z₂ = a₁(a₂+ ib₂) + ib₁(a₂+ ib₂)

z₁z₂ = a₁a₂ + ia₁b₂ + ia₂b₁ + i²b₁b₂

z₁z₂ = a₁a₂ + ia₁b₂ + ia₂b₁ – b₁b₂ [since i² = -1]

Now, the above equation can be rearranged as follows:

z₁z₂ = a₁a₂ – b₁b₂ + ia₁b₂ + ia₂b₁

z₁z₂ = (a₁a₂ – b₁b₂) + i(a₁b₂ + a₂b₁)

Here, the real part is:

Re (z₁z₂) = a₁a₂ – b₁b₂

Thus,

Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂.

Hence, Re (z₁z₂) = Re z₁ Re z₂ – Im z₁ Im z₂ is proved.

Also, read: Imaginary Numbers.

7. Determine the modulus and arguments of 1 – i. Also, find its polar form.

Solutions:

Given expression: 1 – i.

The complex number z = a + ib in polar form is given as z = |z| (cos θ + i sin θ) …(1)

Let z = 1 – i

Finding Modulus of 1 – i:

|z| = √(a² + b²)

|z| = √(1² + (-1)²)

|z| = √(1+1)

|z| = √2.

Hence, the modulus of 1 – i is √2.

Finding Argument of 1 – i:

θ = arg (z) = tan^-1 (|b”https://byjus.com/”a|)

Here, |b| = 1 and |a| = 1.

Hence, θ = arg (z) = tan^-1 (1/1)

θ = arg (z) = tan^-1 1

Hence, θ = -π/4 radian [ As x > 0, y < 0, the complex number lies in fourth quadrant]

Now, substitute the values in (1), we get

Polar form, z = √2 (cos (-π/4) + i sin (-π/4))

z = √2 (cos (π/4) – i sin (π/4))

Therefore, the polar form of 1 – i is equal to √2 (cos (π/4) – i sin (π/4)).

8. Compute the value of (1-i)ⁿ [1- (1/i)]ⁿ for a positive integer “n”.

Solutions:

Given: (1 – i)ⁿ [1- (1/i)]ⁿ

We know that i⁴ = 1,

Hence, (1 – i)ⁿ [1- (1/i)]ⁿ = (1 – i)ⁿ [1- (i⁴/i)]ⁿ

(1 – i)ⁿ [1- (1/i)]ⁿ = (1 – i)ⁿ [1 – i³]ⁿ

Further, the above equation is written as follows:

(1 – i)ⁿ [1- (1/i)]ⁿ = (1 – i)ⁿ (1 + i)ⁿ [Since i³ = -i]

(1 – i)ⁿ [1- (1/i)]ⁿ = [(1 – i) (1 + i)]ⁿ

The expression (1 – i) (1 + i) is of the form (a – b)(a + b), which is equal to a² – b².

Here, a = 1, b = 1

Thus, (1 – i)ⁿ [1- (1/i)]ⁿ = (1 – i²)ⁿ

(1 – i)ⁿ [1- (1/i)]ⁿ = ( 1 – (-1))ⁿ [As, i² = -1]

(1 – i)ⁿ [1- (1/i)]ⁿ = (1 + 1)ⁿ

(1 – i)ⁿ [1- (1/i)]ⁿ = 2ⁿ.

Therefore, the value of (1 – i)ⁿ [1- (1/i)]ⁿ is 2ⁿ.

9. Solve the given equation: |z| = z + 1 + 2i.

Solutions:

Given equation: |z| = z + 1 + 2i.

Let z = a + ib

Hence, the given equation becomes:

|a + ib| = (a + ib) + 1 + 2i

√(a² + b²) = (a + 1) + i (b + 2) …(1)

Now, compare the real and imaginary parts,

Real part: √(a² + b²) = a + 1 …(2)

Imaginary part: 0 = b + 2

Hence, b = -2 …(3)

Now, substitute b = -2 in (2), we get

√(a² + (-2)²) = a + 1

Now, take squares on both sides, we get

a² + 4 = (a+ 1)²

a² + 4 = a² + 1 + 2a

4 = 1 + 2a

2a = 4 – 1

2a = 3

Hence, a = 3/2 …(4)

Hence, the complex number, z = (3/2) -2i.

10. Show that arg (z₁/z₄) + arg (z₂/z₃) = 0, if z₁, z₂ and z₃ and z₄ are the two pairs of conjugate complex numbers.

Solutions:

Given that,

\(\begin{array}{l}z_{1} = \bar{z_{2}}…(1)\end{array} \)

\(\begin{array}{l}z_{3} = \bar{z_{4}} …(2)\end{array} \)

As we know, arg (z₁/z₂) = arg (z₁) – arg (z₂)

Hence, (z₁/z₄) + arg (z₂/z₃) = arg (z₁) – arg (z₄) + arg (z₂) – arg (z₃)

arg (z₁/z₄) + arg (z₂/z₃) = arg (z₁) + arg (z₂) – arg (z₄) – arg (z₃)

arg (z₁/z₄) + arg (z₂/z₃) = [arg (z₁) + arg (z₂)] – [arg (z₄) + arg (z₃)]

Using (1) and (2), we can write

\(\begin{array}{l}arg(z_{1}/z_{4}) + arg (z_{2}/z_{3}) = [arg(\bar{z_{2}})+ arg(z_{2})] – [arg(\bar{z_{4}})+ arg(z_{4})]\end{array} \)

\(\begin{array}{l}Since,\ arg(z)+arg (\bar{z})= 0\end{array} \)

arg (z₁/z₄) + arg (z₂/z₃) = 0 – 0

arg (z₁/z₄) + arg (z₂/z₃) = 0

Hence, arg (z₁/z₄) + arg (z₂/z₃) = 0 is proved.

Practice Questions

Solve the following complex numbers class 11 questions.

1. Write the expression i^-39 in the form of a + ib.
2. Solve the equation: √2x² + x + √2 = 0.
3. Write the given complex number in the polar form: (1 + 2i) / (1 – 3i).