Exercise 1.2 – Complex Numbers
Class 10th Mathematics | Solved Exercises
Watch the Complete video: Click Here
Introduction;
In mathematics, complex numbers extend the concept of the
one-dimensional number line to the two-dimensional complex plane. A complex
number is in the form:
z = a + ib
Where:
- a is the real part
- b is the imaginary part
- i is the imaginary unit, with the property i² = -1
This exercise covers foundational concepts such as:
- Additive and multiplicative inverses
- Conjugates
- Modulus (absolute value)
- Graphical representation
- Separation of real and imaginary parts
- Basic algebraic operations
- Inequalities involving complex numbers
Question 1: Find the additive inverse of each complex number.
1.
(a) 4 + 5i
·
Additive inverse is the
negative of the number:
·
→ -4 - 5i
2.
(b) -3 - 5i
·
→ 3 + 5i
3.
(c) 5 - 5i
·
→ -5 + 5i
4.
(d) 4i
·
→ 0 - 4i
Question 2: Show that each pair of complex numbers are additive inverses of
each other.
I.
2 + 3i and -2 - 3i
·
Sum: (2 + 3i) + (-2 - 3i) =
0
·
Hence, additive inverses.
II.
-5 + 4i and 5 - 4i
·
Sum = 0 ⇒ additive
inverses.
III.
6 - 2i and -6 + 2i
·
Sum = 0 ⇒ additive
inverses.
IV.
4 - 3i and -4 + 3i
·
Sum = 0 ⇒ additive
inverses.
Question 3: Find the multiplicative inverse of each complex number.
a)
2 + i
·
Formula: 1/(a + bi) = (a -
bi)/(a² + b²)
·
→ (2 - i)/(2² + 1²) = (2 -
i)/5
b)
1 + i
·
→ (1 - i)/(1² + 1²) = (1 -
i)/2
c)
10 - 3i
·
→ (10 + 3i)/(10² + 3²) =
(10 + 3i)/109
Question 4: Find the product of each complex number and its conjugate.
1) 2 + 3i
·
→ (2 + 3i)(2 - 3i) = 4 + 9
= 13
2) 1 - 2i
·
→ (1 - 2i)(1 + 2i) = 1 + 4
= 5
3) -4 + i
·
→ (-4 + i)(-4 - i) = 16 + 1
= 17
Question 5: If z₁ = 1 - 2i and z₂ = 2 + i, show that z₁/z₂ = (1 - 5i)/5
Solution:
·
z₁ / z₂ = (1 - 2i)/(2 + i)
·
Multiply numerator and
denominator by conjugate of denominator:
·
→ (1 - 2i)(2 - i)/(2 + i)(2
- i)
·
= (2 - i - 4i + 2i²)/(4 +
1) = (2 - 5i - 2)/5 = -5i/5 = -i
Question 6: Let z = a + ib, then show |z| = √(a² + b²), and |z₁ – z₂| = |z₂
– z₁|
Solution:
·
Let z₁ = 1 + i, z₂ = 2 - i
·
z₁ - z₂ = (1 + i) - (2 - i)
= -1 + 2i
·
|z₁ - z₂| = √((-1)² + 2²) =
√5
·
z₂ - z₁ = (2 - i) - (1 + i)
= 1 - 2i
·
|z₂ - z₁| = √((1)² + (-2)²)
= √5
·
Hence, |z₁ - z₂| = |z₂ -
z₁|
Question 7: Find |z|, |-z|, |z₁ – z₂|, and |z₁| – |z₂|
i.
z = -1 - i
·
|z| = √((-1)² + (-1)²) = √2
·
|-z| = |1 + i| = √2
ii.
z = 1 - 2i
·
|z| = √(1² + (-2)²) = √5
·
|-z| = |-1 + 2i| = √5
iii.
z₁ = 2 - i, z₂ = -3 + 2i
·
z₁ - z₂ = (2 - i) - (-3 +
2i) = 5 - 3i
·
|z₁ - z₂| = √(5² + (-3)²) =
√34
·
|z₁| = √(2² + (-1)²) = √5
·
|z₂| = √((-3)² + 2²) = √13
·
|z₁| - |z₂| = √5 - √13
Question 8: Represent the numbers in the complex plane
(a)
·
3 - i → point (3, -1)
(b)
·
2 + 3i → point (2, 3)
(c)
·
-2 - 3i → point (-2, -3)
(d)
·
4 - 2i → point (4, -2)
Question 9: Separate real and imaginary parts of each complex number
1. -5/(1 + 5i)
·
Multiply numerator and
denominator by conjugate:
·
→ (-5)(1 - 5i)/(1² + 25) =
(-5 + 25i)/26
·
Real part = -5/26,
Imaginary part = 25/26
2. (1 + 2i)/(1 + i)
·
→ (1 + 2i)(1 - i)/(1 + i)(1
- i) = (3 + i)/2
·
Real = 3/2, Imaginary = 1/2
3. (2 + 3i)/(1 - i)
·
→ (2 + 3i)(1 + i)/(1 + 1) =
(-1 + 5i)/2
·
Real = -1/2, Imaginary =
5/2
4. (4 + 3i)/(1 + i)
·
→ (4 + 3i)(1 - i)/2 = (7 -
i)/2
·
Real = 7/2, Imaginary =
-1/2
5. (2 - 5i)/(2 + 3i)
·
→ (2 - 5i)(2 - 3i)/(13) =
(-11 - 16i)/13
·
Real = -11/13, Imaginary =
-16/13
6. (1 - 2i)/(2 + i)
·
→ (1 - 2i)(2 - i)/5 = -i
·
Real = 0, Imaginary = -1
7. (2 + 3i)/(3 - 4i)
·
→ (2 + 3i)(3 + 4i)/25 = (-6
+ 17i)/25
·
Real = -6/25, Imaginary =
17/25
Question 10: Show properties using z and its conjugate
Ø Part (i)
·
z + z̄ = 2a → real number
Ø Part (ii)
·
z - z̄ = 2ib → imaginary
number
Ø Part (iii)
·
z * z̄ = a² + b² → real
number
Ø Part (iv)
·
z / z̄ = (a + ib)/(a - ib)
→ generally not real or imaginary
Ø Part (v)
·
Conjugate of conjugate: z̄̄
= z
Question 11: Represent the sum and difference of complex numbers
graphically
v z₁ = 3 + 2i, z₂ = 2 + 3i
·
z₁ + z₂ = 5 + 5i
·
z₁ - z₂ = 1 - i
v z₁ = 2 - i, z₂ = -3 + i
·
z₁ + z₂ = -1
·
z₁ - z₂ = 5 - 2i
Question 12: Represent z₁ + z₂ and show |z₁ + z₂|² ≤ (|z₁| + |z₂|)²
Solution:
Example:- z₁ = 1 + 2i, z₂ = 2 + 3i
·
z₁ + z₂ = 3 + 5i → |z₁ +
z₂|² = 9 + 25 = 34
·
|z₁| = √5 ≈ 2.24, |z₂| =
√13 ≈ 3.61
·
(|z₁| + |z₂|)² ≈ (5.85)² =
34.22
·
Hence, 34 ≤ 34.22 ⇒ verified
0 Comments