Class 11 Mathematics Complex Number Notes
Complete typed notes from the supplied handwritten PDF, including Exercises 6.1, 6.2 and 6.3.
Exercise 6.1
1(a). Find the real and imaginary parts
- 5 + 2i
Real part = 5, Imaginary part = 2. - 3i + 0
Real part = 0, Imaginary part = 3. - √3 + i0
Real part = √3, Imaginary part = 0.
1(b). Write the complex number in the form a + ib
- (1, 2) = 1 + 2i
- (3, 0) = 3 + 0i
- (0, -2) = -2i
- (-5/2, √5/2) = -5/2 + (√5/2)i
2(a). Find p and q if p + iq = 5 + 3i
2(b). Find x and y if (3y + 5) + (2x – 5)i = 0
2x – 5 = 0 ⇒ x = 5/2
2(c). Find real numbers x and y
From the source:
Equating real and imaginary parts:
2x – 3y = 3 …(ii)
Adding:
Putting x = 1 in (i):
3. Find the values
Using i² = -1 and the cycle i, -1, -i, 1:
i25 = i
i47 = -i
i522 = -1
i4n+1 = i, n ∈ Z
i9 – 3i11 = -2i
i7 + 1/i4 = 1 – i
i-22 = -1
4. Prove the following
(a) 1 + i² + i⁴ + i⁶ = 0
(b) 1 + i20 + i100 + i1000 = 0
The handwritten source simplifies the expression according to its written powers.
(c) i-1 – i-2 + i-3 – i-4 = 0
The source verifies this by converting negative powers into reciprocals and then using powers of i.
(d) in + in+1 + in+2 + in+3 = 0, n ∈ Z
i³ + i⁴ + i⁵ + i⁶ = -i + 1 + i – 1 = 0
5. If x = 1 + 3i and y = x – i, find x² – y² and (x + y)²
6. Simplify
The supplied handwritten solutions include examples such as:
7. If z₁ = 3 + 4i and z₂ = 4 + 3i, compare them
The source states:
8. If (a + ib)/(c + id) is purely real, prove that ad = bc
For a purely real number, the imaginary part is zero:
Conjugate of a Complex Number
Let z = a + ib be a complex number. The conjugate of z is denoted by z̄ and is:
Rules of Conjugate
- overline(z + w) = z̄ + w̄
- overline(z – w) = z̄ – w̄
- overline(z̄) = z
- overline(z²) = (z̄)²
- overline(z/w) = z̄/w̄
- Re(z) = (z + z̄)/2
- Im(z) = (z – z̄)/(2i)
Proof of overline(z + w) = z̄ + w̄
Let z = a + ib and w = c + id.
overline(z + w) = (a + c) – i(b + d)
= (a + c) – i(b + d)
Proof of overline(z – w) = z̄ – w̄
overline(z – w) = (a – c) – i(b – d)
Proof of overline(z̄) = z
Proof of overline(z²) = (z̄)²
overline(z²) = a² – 2abi – b²
Real and Imaginary Parts
Absolute Value or Modulus of a Complex Number
Let z = a + ib be a complex number. The absolute value or modulus of z is denoted by |z| and is defined by:
Properties
- |z| ≥ 0
- z z̄ = |z|²
- |z̄| = |z|
- |zw| = |z||w|
Triangle Inequality
For two complex numbers z and w:
The source proves this by writing z = a + ib and w = c + id, expanding both sides, and showing the resulting square is non-negative.
Exercise 6.2
1. Express in the form a + ib
(b) 3 – √(-4) = 3 – 2i
(c) √25 – √(-25) = 5 – 5i
(d) 3 – 4i + 6i = 3 + 2i
(e) (2 – 3i)(2 + i) = 7 – 4i
(f) (1 + 2i)² = -3 + 4i
Further examples from the source:
(i) 5/(2 – √(-1)) = 2 + i
(j) (5 + 4i)/(2 – 3i) = -2/13 + 23/13 i
(k) (1+i)/(1-i) – (1-i)/(1+i) = 2i
(n) 1/(1+i) = 1/2 – 1/2 i
2. Find the multiplicative inverse
(a) 3 – 2i
(b) (2 – 5i)²
The source simplifies the reciprocal using the conjugate method and obtains:
(c) (2 + 3i)/(3 – 2i)
The source finds its multiplicative inverse by taking the reciprocal and rationalizing.
3. If z₁ = 3 + 4i and z₂ = 1 + i, verify:
(a) overline(z₁z₂) = z̄₁ z̄₂
The source computes both sides and obtains the same result.
(b) overline(z₁/z₂) = z̄₁/z̄₂
Taking the conjugate gives 7/2 – 1/2 i, which matches the right-hand side.
(c) |z₁z₂| = |z₁||z₂|
|z₁||z₂| = 5√2
(d) |z₁/z₂| = |z₁|/|z₂|
The source calculates the left-hand side and obtains the same value.
4. Find the modulus and conjugate
(a) (1 + i)²
|2i| = 2
overline(2i) = -2i
(b) (3 + 2i)/(2 + 5i)
The source simplifies to:
Its conjugate is:
(c) (1 + 2i)/(1 – 3i)
(d) (3 – √(-25))/(5 – √(-16))
The source simplifies the complex fraction, then finds its modulus.
5. Find the modulus
(a)
The source simplifies the given expression to:
Hence:
(b) (1 + i)/(1 – i)
|i| = 1
6(a). If (3 – 4i)(x + iy) = 3√5, show that x² + y² = 9/5
Taking modulus on both sides:
Squaring:
x² + y² = 9/5
6(b). If (a+ib)/(c+id) = p+iq, prove:
This follows by taking modulus on both sides and squaring.
6(c). If p+iq = (a+i)²/(2a-i), prove the given relation
The source takes modulus of both sides and simplifies to the required expression.
7. Additional proofs
If √(x + iy) = a + ib, prove √(x – iy) = a – ib
The source squares the first relation:
Then:
Hence:
If a + ib = √((1+i)/(1-i)), prove a² + b² = 1
The source simplifies the fraction and then takes modulus:
8. If z and w are two complex numbers, prove:
(a) |z + w|² = |z|² + |w|² + 2Re(z w̄)
= z z̄ + z w̄ + w z̄ + w w̄
= |z|² + |w|² + 2Re(z w̄)
(b) ||z| – |w|| ≤ |z – w|
The source derives this from the triangle inequality.
(c) |z + w|² + |z – w|² = 2(|z|² + |w|²)
This is the parallelogram identity for complex numbers.
Polar (Trigonometric) Form of a Complex Number
Let z = x + iy be a complex number represented by the point P(x, y) in the complex plane.
y = r sin θ
Therefore:
= r cos θ + ir sin θ
= r(cos θ + i sin θ)
Where:
tan θ = y/x, x ≠ 0
Here r = |z| is the modulus and θ is the amplitude or argument of z.
Exercise 6.3
1. Express in polar or trigonometric form
(a) i
i = cos 90° + i sin 90°
(b) √3 + i
√3 + i = 2(cos 30° + i sin 30°)
(c) i – √3
i – √3 = 2(cos 150° + i sin 150°)
(d) 4 + 4i
4 + 4i = 4√2(cos 45° + i sin 45°)
(e) 1 – √3 i
1 – √3 i = 2(cos 300° + i sin 300°)
(f) 2 + 2√3 i
2 + 2√3 i = 4(cos 60° + i sin 60°)
(g) 1 – i
1 – i = √2(cos 315° + i sin 315°)
(h) 3 + 4i
3 + 4i = 5[cos(tan-1(4/3)) + i sin(tan-1(4/3))]
(i) 1/2 + (√3/2)i
= cos 60° + i sin 60°
(j) i/(i+1)
The source simplifies:
Then converts the result into polar form.
2. Express in rectangular form
(a) 2(cos 30° + i sin 30°)
y = 2 sin 30° = 1
= √3 + i
(b) 2(cos 210° + i sin 210°)
= -√3 – i
(c) 2(cos 360° + i sin 360°)
(d) 2(cos 315° + i sin 315°)
= √2 – √2 i
3. Express in standard form
(a) 2eiπ/2
= 2i
(b) ei3π/4
= -1/√2 + i/√2
(c) 3e-iπ
= -3
Note: This typed version follows the uploaded handwritten PDF closely. A few handwritten expressions are faint or ambiguous, so they are presented conservatively rather than silently corrected.
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