Solved Excerxise 1.3 - Class 10 Mathematics

 Exercise 1.3 – Complex Numbers

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🌟 Introduction

In this exercise, you will learn how to:
- Solve quadratic equations that have complex roots.
- Check whether a complex number is a solution to a given equation.
- Factorize expressions involving imaginary numbers.
- Solve linear systems involving complex numbers.
- Use the formula E = IZ (used in electrical circuits) where E is voltage, I is current, and Z is impedance.
- Perform arithmetic using complex numbers.

This exercise helps you become familiar with complex number operations and their applications in mathematics and real-life electrical circuits.

✏️ Step-by-Step Solutions

1. Solve: x² + 7 = 0

Step 1: Move constant to other side
x² = -7
Step 2: Take square root
x = ±√(-7)
Step 3: Convert to complex number
x = ±√7 i
Answer: x = ±√7 i

2. Solve: x² + 9 = 0

Step 1: x² = -9
Step 2: x = ±√(-9)

x= ±3i
Answer: x = ±3i

3. Solve: x² + 100 = 0

Step 1: x² = -100
Step 2: x = ±√(-100)

x= ±10i
Answer: x = ±10i

4. Check if 1 + 2i is a solution of x² - 2x + 5 = 0

Let x = 1 + 2i;
Putting the value of x in given equation:

= (1 + 2i)² - 2(1 + 2i) + 5
= -3 + 4i - 2 - 4i + 5

= 0
Answer: Yes, it is a solution.

5. Check if 1 - 2i is a solution of x² - 2x + 5 = 0

Let x = 1 - 2i;

Putting the value of x in given equation:
= (1 - 2i)² - 2(1 - 2i) + 5
= -3 - 4i - 2 + 4i + 5

= 0
Answer: Yes, it is a solution.

6. Check if 1 - i is a solution of x² + 2x + 2 = 0

Let x = 1 – I;

Putting the value of x in given equation:
= (1 - i)² + 2(1 - i) + 2
= -2i + 2 - 2i + 2 = 4 - 4i

≠ 0
Answer: No, it is not a solution.

7. Check if i is a solution of x² + 1 = 0

Let x = I;

Putting the value of x in given equation:
= i² + 1 = -1 + 1

= 0
Answer: Yes, it is a solution.

8. Factor: x² + 16

x² + 16

= x² - (-16)

= x² - (4i)²
= (x + 4i)(x - 4i)
Answer: (x + 4i)(x - 4i)

9. Factor: a² + b²

a² + b²

= a² - (-b²)

= (a + bi)(a - bi)
Answer: (a + bi)(a - bi)

10. Factor: x² + 25y²

x² + 25y²

= x² - (-25y²)

= (x + 5iy)(x - 5iy)
Answer: (x + 5iy)(x - 5iy)

11. Solve the system:
    z - 4w = 3i
    2z + 3w = 11 - 5i

Step 1: Multiply first equation by 2:

2z - 8w = 6i
Step 2: Subtract Second Eq. by First:

2z + 3w - (2z - 8w)  = 11 - 5i - 6i
11w = 11 - 11i

→ w = 1 – i

Step 3: Plug w into first:

z - 4(1 - i) = 3i

→ z = 4 - i
Answer: z = 4 - i, w = 1 – i

12. Solve the system:
    3z + (2 + i)w = 11 - i
    (2 - i)z - w = -1 + i

Step 1: Use substitution or elimination method.
Let’s solve the second equation for w:

(2 - i)z - w = -1 + i

→ w = (2 - i)z + 1 - i
Step 2: Substitute this value of w into the first equation:

3z + (2 + i)[(2 - i)z + 1 - i] = 11 - i
Step 3: Expand:

(2 + i)(2 - i)z + (2 + i)(1 - i)

= (4 + 1)z + (2 - 2i + i - i²)

= 5z + (3 - i)
So: 3z + 5z + 3 - i = 11 - i → 8z + 3 = 11 → 8z = 8 → z = 1
Step 4: Put z = 1 in w equation:

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

w = 2 - i + 1 - i = 3 - 2i
Answer: z = 1, w = 3 - 2i

13. (a) Find I, given the values:

Use the formula: I = E / Z
(i) E = 70 + 220i, Z = 16 + 8i
Multiply numerator and denominator by conjugate of denominator:
I = (70 + 220i)(16 - 8i) / ((16 + 8i)(16 - 8i))
Denominator = 256 + 64 = 320
Numerator = Use FOIL: 70×16 - 70×8i + 220i×16 - 220i×8i
= 1120 - 560i + 3520i - 1760i²

= 1120 + 2960i + 1760
= 2880 + 2960i
I = (2880 + 2960i) / 320 = 9 + 9.25i
Answer: I = 9 + 9.25i amps

(ii) E = 85 + 110i, Z = 3 - 4i
I = (85 + 110i)(3 + 4i) / ((3 - 4i)(3 + 4i))
Denominator = 9 + 16 = 25
Numerator = 255 + 340i + 330i + 440i² = 255 + 670i - 440
= -185 + 670i
I = (-185 + 670i)/25 = -7.4 + 26.8i
Answer: I = -7.4 + 26.8i amps

13. (b) Find Z, given the values:

Use: Z = E / I
(i) E = -50 + 100i, I = -6 - 2i
Z = (-50 + 100i)(-6 + 2i) / ((-6 - 2i)(-6 + 2i))
Denominator = 36 + 4 = 40
Numerator = 300 - 100i + 600i - 200i² = 300 + 500i + 200
= 500 + 500i
Z = (500 + 500i)/40 = 12.5 + 12.5i
Answer: Z = 12.5 + 12.5i ohms

(ii) E = 100 + 10i, I = -8 + 3i
Z = (100 + 10i)(-8 - 3i) / ((-8 + 3i)(-8 - 3i))
Denominator = 64 + 9 = 73
Numerator = -800 - 300i - 80i - 30i² = -800 - 380i + 30
= -770 - 380i
Z = (-770 - 380i)/73 ≈ -10.55 - 5.21i
Answer: Z ≈ -10.55 - 5.21i ohms

13. (c) Evaluate: 1 / (z - z²), when z = (1 - i)/10

Step 1: Find z²:
z = (1 - i)/10 → z² = [(1 - i)²]/100 = (-2i)/100 = -i/50

Step 2: z - z² = (1 - i)/10 + i/50 = (5(1 - i) + i)/50 = (5 - 5i + i)/50 = (5 - 4i)/50

Step 3: Invert:
1 / [(5 - 4i)/50] = 50 / (5 - 4i)
= 50(5 + 4i)/(25 + 16) = (250 + 200i)/41 ≈ 6.1 + 4.88i
Answer: ≈ 6.1 + 4.88i