CBSE Class 11 Mathematics Practice Worksheet A on Complex Numbers with various question types and marks distribution.

Let's solve each question from the CBSE Class 11 Mathematics Practice Worksheet A on the topic "Introduction to Complex Numbers" step by step.

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Very Short Answer Questions

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#### Q1: Find the value of $ x $ and $ y $ such that $ 2 + (x + iy) = (3 - i) $

Solution:

We are given:

$$
2 + (x + iy) = 3 - i
$$

Simplify the left-hand side:

$$
(2 + x) + iy = 3 - i
$$

Now, equating real and imaginary parts:

- Real part: $ 2 + x = 3 \Rightarrow x = 1 $
- Imaginary part: $ y = -1 $

✔ Answer: $ x = 1, y = -1 $

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#### Q2: Find the value of $ x $ and $ y $ such that $ (x + iy) = (1 + i)(1 - i) $

Solution:

First compute the right-hand side:

$$
(1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2
$$

So,

$$
x + iy = 2 + 0i
$$

Thus,

- $ x = 2 $
- $ y = 0 $

✔ Answer: $ x = 2, y = 0 $

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#### Q3: If $ z = 1 + 2i $, represent $ z $ on the complex plane.

Solution:

The complex number $ z = 1 + 2i $ has:
- Real part: 1
- Imaginary part: 2

To represent it on the complex plane:
- Plot the point $ (1, 2) $
- The horizontal axis is the real axis.
- The vertical axis is the imaginary axis.

✔ Answer: Plot point at $ (1, 2) $ on the complex plane.

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#### Q4: Find the smallest positive integer $ m $ for which $ \left(\frac{1+i}{1-i}\right)^m = 1 $

Solution:

First simplify $ \frac{1+i}{1-i} $

Multiply numerator and denominator by conjugate of denominator:

$$
\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{(1)^2 - (i)^2} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{1 + 2i - 1}{2} = \frac{2i}{2} = i
$$

So,
$$
\left(\frac{1+i}{1-i}\right)^m = i^m
$$

We want $ i^m = 1 $

We know the powers of $ i $ cycle every 4:
- $ i^1 = i $
- $ i^2 = -1 $
- $ i^3 = -i $
- $ i^4 = 1 $

So the smallest positive $ m $ such that $ i^m = 1 $ is $ m = 4 $

✔ Answer: $ m = 4 $

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#### Q5: Prove that: $ 3(1 - 2i) - (-4 - 5i) + (-8 + 3i) = 2i - 1 $

Solution:

Compute LHS step by step:

1. $ 3(1 - 2i) = 3 - 6i $
2. $ -(-4 - 5i) = +4 + 5i $
3. $ +(-8 + 3i) = -8 + 3i $

Now add them:

Real parts: $ 3 + 4 - 8 = -1 $

Imaginary parts: $ -6i + 5i + 3i = 2i $

So total: $ -1 + 2i $

Which is same as $ 2i - 1 $

✔ Hence proved.

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Short Answer Questions

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#### Q6: Prove that $ i^{107} + i^{112} + i^{117} + i^{122} = 0 $

Solution:

We use the cyclicity of powers of $ i $:

- $ i^1 = i $
- $ i^2 = -1 $
- $ i^3 = -i $
- $ i^4 = 1 $
- Cycle repeats every 4.

So, reduce exponents modulo 4:

- $ 107 \mod 4 = 3 \Rightarrow i^{107} = i^3 = -i $
- $ 112 \mod 4 = 0 \Rightarrow i^{112} = i^0 = 1 $
- $ 117 \mod 4 = 1 \Rightarrow i^{117} = i^1 = i $
- $ 122 \mod 4 = 2 \Rightarrow i^{122} = i^2 = -1 $

Now sum:

$$
(-i) + 1 + i + (-1) = (-i + i) + (1 - 1) = 0 + 0 = 0
$$

✔ Hence proved.

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#### Q7: Find the value of $ x $ and $ y $ such that $ 3 + (x + iy) = (2 + i)^2 $

Solution:

First compute $ (2 + i)^2 $:

$$
(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i
$$

Now:

$$
3 + (x + iy) = 3 + 4i
\Rightarrow (3 + x) + iy = 3 + 4i
$$

Equate real and imaginary parts:

- $ 3 + x = 3 \Rightarrow x = 0 $
- $ y = 4 $

✔ Answer: $ x = 0, y = 4 $

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#### Q8: Find the value of $ x $ and $ y $ such that $ i(x + iy) = (2 + 3i)(3 + 2i) $

Solution:

First compute RHS:

$$
(2 + 3i)(3 + 2i) = 2\cdot3 + 2\cdot2i + 3i\cdot3 + 3i\cdot2i = 6 + 4i + 9i + 6i^2 = 6 + 13i - 6 = 0 + 13i
$$

So RHS = $ 13i $

Now LHS: $ i(x + iy) = ix + i^2y = ix - y = -y + ix $

Set equal:

$$
-y + ix = 0 + 13i
$$

Equate real and imaginary parts:

- $ -y = 0 \Rightarrow y = 0 $
- $ x = 13 $

✔ Answer: $ x = 13, y = 0 $

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#### Q9: Solve for $ x $ and $ y $: $ 3x + (2x - y)i = 6 - 3i $

Solution:

Given:

$$
3x + (2x - y)i = 6 - 3i
$$

Equate real and imaginary parts:

- Real: $ 3x = 6 \Rightarrow x = 2 $
- Imaginary: $ 2x - y = -3 $

Substitute $ x = 2 $:

$ 2(2) - y = -3 \Rightarrow 4 - y = -3 \Rightarrow y = 7 $

✔ Answer: $ x = 2, y = 7 $

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#### Q10: Prove that $ (x+1+i)(x-1+i)(x-1-i)(x+1-i) = x^4 + 4 $

Solution:

Group terms cleverly:

Let’s pair:

- $ (x+1+i)(x+1-i) = [(x+1) + i][(x+1) - i] = (x+1)^2 - i^2 = (x+1)^2 + 1 $
- $ (x-1+i)(x-1-i) = [(x-1) + i][(x-1) - i] = (x-1)^2 - i^2 = (x-1)^2 + 1 $

Now compute both:

- $ (x+1)^2 + 1 = x^2 + 2x + 1 + 1 = x^2 + 2x + 2 $
- $ (x-1)^2 + 1 = x^2 - 2x + 1 + 1 = x^2 - 2x + 2 $

Now multiply:

$$
(x^2 + 2x + 2)(x^2 - 2x + 2)
$$

Use identity: $ (a + b)(a - b) = a^2 - b^2 $, but here it's not exactly that.

Alternatively, treat as:

Let $ A = x^2 + 2 $, then:

$$
(x^2 + 2x + 2)(x^2 - 2x + 2) = [(x^2 + 2) + 2x][(x^2 + 2) - 2x] = (x^2 + 2)^2 - (2x)^2
$$

Compute:

- $ (x^2 + 2)^2 = x^4 + 4x^2 + 4 $
- $ (2x)^2 = 4x^2 $

So:

$$
x^4 + 4x^2 + 4 - 4x^2 = x^4 + 4
$$

✔ Hence proved.

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Long Answer Question

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#### Q11: Find the value of $ x $ and $ y $ such that $ 5x + 3iy + 7 = 3 + 4i $

Solution:

Given:

$$
5x + 3iy + 7 = 3 + 4i
$$

Group real and imaginary parts:

Left-hand side:
- Real part: $ 5x + 7 $
- Imaginary part: $ 3y $

So:

$$
(5x + 7) + (3y)i = 3 + 4i
$$

Equate real and imaginary parts:

1. $ 5x + 7 = 3 \Rightarrow 5x = -4 \Rightarrow x = -\frac{4}{5} $
2. $ 3y = 4 \Rightarrow y = \frac{4}{3} $

✔ Answer: $ x = -\frac{4}{5}, y = \frac{4}{3} $

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✔ Final Answers Summary:

| Q | Answer |
|---|--------|
| Q1 | $ x = 1, y = -1 $ |
| Q2 | $ x = 2, y = 0 $ |
| Q3 | Plot $ (1, 2) $ on complex plane |
| Q4 | $ m = 4 $ |
| Q5 | Verified: $ -1 + 2i = 2i - 1 $ |
| Q6 | Proved using $ i^4 = 1 $, sum = 0 |
| Q7 | $ x = 0, y = 4 $ |
| Q8 | $ x = 13, y = 0 $ |
| Q9 | $ x = 2, y = 7 $ |
| Q10 | Proved: product equals $ x^4 + 4 $ |
| Q11 | $ x = -\frac{4}{5}, y = \frac{4}{3} $ |

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