CBSE Class 11 Mathematics Complex Numbers and Quadratic Equations Notes Set 01

Important Notes and Formulas

Definition

A number of the form a + ib, where a, b ∈ R and \( i = \sqrt{-1} \), is called a complex number and is denoted by z.
z = a + ib, where a = Re(z) is the real part and b = Im(z) is the imaginary part of z.

Conjugate of a Complex Number

For a given complex number z = a + ib, its conjugate \( \bar{z} \) is defined as \( \bar{z} = a - ib \).

Algebra of Complex Numbers

Let z₁ = a + ib and z₂ = c + id be two complex numbers, where a, b, c, d ∈ R and \( i = \sqrt{-1} \).
1. Addition: z₁ + z₂ = (a + bi) + (c + di) = (a + c) + (b + d)i
2. Subtraction: z₁ − z₂ = (a + bi) − (c + di) = (a − c) + (b − d)i
3. Multiplication: z₁ · z₂ = (a + bi)(c + di) = a(c + di) + bi(c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i   (∵ i² = −1)
4. Division: \( \dfrac{z_1}{z_2} = \dfrac{a + bi}{c + di} = \dfrac{a + bi}{c + di} \cdot \dfrac{c - di}{c - di} = \left( \dfrac{ac + bd}{c^2 + d^2} \right) + \left( \dfrac{bc - ad}{c^2 + d^2} \right) i \)

Note:
1. a + ib = c + id ⇔ a = c and b = d
2. Powers of i: \( i^{4k + r} = \begin{cases} 1, & r = 0 \\ i, & r = 1 \\ -1, & r = 2 \\ -i, & r = 3 \end{cases} \)
3. \( \sqrt{a}\,\sqrt{b} = \sqrt{ab} \) only if at least one of a or b is non-negative.

Argand Plane

A complex number z = a + ib can be represented by a unique point P(a, b) in the Argand plane. The horizontal axis is the real axis Re(z) and the vertical axis is the imaginary axis Im(z).

z = a + ib is represented by the point P(a, b).

Modulus of a Complex Number

If z = a + ib is a complex number represented by the point P(a, b), then the distance of P from the origin O is called the modulus of z.

It is denoted by \( r = |z| = \sqrt{a^2 + b^2} \).
The distance between the points representing z₁ and z₂ is |z₂ − z₁|.
Note: z₁ > z₂ or z₁ < z₂ has no meaning, but |z₁| > |z₂| or |z₁| < |z₂| holds meaning.

Some Important Properties

1. \( \overline{(\bar{z})} = z \)
2. \( z + \bar{z} = 2\,\text{Re}(z) \)
3. \( z - \bar{z} = 2i\,\text{Im}(z) \)
4. \( \overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2 \)
5. \( \overline{z_1 z_2} = \bar{z}_1 \, \bar{z}_2 \)
6. |z| = 0 ⇒ z = 0
7. \( z\bar{z} = |z|^2 \)
8. \( |z_1 z_2| = |z_1|\,|z_2| \); \( \left| \dfrac{z_1}{z_2} \right| = \dfrac{|z_1|}{|z_2|} \)
9. \( |\bar{z}| = |z| = |-z| \)
10. \( |z_1 \pm z_2|^2 = |z_1|^2 + |z_2|^2 \pm 2\,\text{Re}(z_1 \bar{z}_2) \)
11. \( |z_1 + z_2| \le |z_1| + |z_2| \)   (Triangle Inequality)
12. \( |z_1 - z_2| \ge \big|\, |z_1| - |z_2| \,\big| \)
13. \( |az_1 - bz_2|^2 + |bz_1 + az_2|^2 = (a^2 + b^2)(|z_1|^2 + |z_2|^2) \), where a, b ∈ R
14. Useful factorisation: x² + y² = (x + iy)(x − iy)

Solved Examples

Example 1. Express the following in the form of a + ib, a, b ∈ R, \( i = \sqrt{-1} \). State the values of a and b.
(i) \( \dfrac{i(4 + 3i)}{(1 - i)} \)   (ii) \( \dfrac{(2 + i)}{(3 - i)(1 + 2i)} \)   (iii) \( \dfrac{4i^8 - 3i^9 + 3}{3i^{11} - 4i^{10} - 2} \)   (iv) (1 + i)⁶ + (1 − i)³
Answer:
(i) \( z = \dfrac{i(4 + 3i)}{(1 - i)} = \dfrac{4i + 3i^2}{1 - i} \times \dfrac{1 + i}{1 + i} = \dfrac{(4i - 3)(1 + i)}{1 - i^2} = \dfrac{4i + 4i^2 - 3 - 3i}{1 + 1} = \dfrac{-7 + i}{2} = -\dfrac{7}{2} + \dfrac{1}{2}i \)
Here, \( a = -\dfrac{7}{2},\ b = \dfrac{1}{2} \)
(ii) \( z = \dfrac{(2 + i)}{(3 - i)(1 + 2i)} = \dfrac{2 + i}{3 + 6i - i - 2i^2} = \dfrac{2 + i}{5 + 5i} \times \dfrac{5 - 5i}{5 - 5i} = \dfrac{10 - 10i + 5i - 5i^2}{25 - 25i^2} = \dfrac{15 - 5i}{50} = \dfrac{3}{10} - \dfrac{1}{10}i \)
Here, \( a = \dfrac{3}{10},\ b = -\dfrac{1}{10} \)
(iii) \( z = \dfrac{4i^8 - 3i^9 + 3}{3i^{11} - 4i^{10} - 2} = \dfrac{4 \cdot i^4 \cdot i^4 - 3i^4 \cdot i^4 \cdot i + 3}{3 \cdot i^4 \cdot i^4 \cdot i^3 - 4i^4 \cdot i^4 \cdot i^2 - 2} = \dfrac{4 - 3i + 3}{-3i + 4 - 2} = \dfrac{7 - 3i}{2 - 3i} \times \dfrac{2 + 3i}{2 + 3i} = \dfrac{14 + 21i - 6i - 9i^2}{4 - 9i^2} = \dfrac{23}{13} + \dfrac{15}{13}i \)
Here, \( a = \dfrac{23}{13},\ b = \dfrac{15}{13} \)
(iv) (1 + i)⁶ = {(1 + i)²}³ = (1 + i² + 2i)³ = (1 − 1 + 2i)³ = 8i³ = −8i
and (1 − i)³ = 1 − i³ − 3i + 3i² = 1 + i − 3i − 3 = −2 − 2i
Therefore, (1 + i)⁶ + (1 − i)³ = −8i − 2 − 2i = −2 − 10i
Here, a = −2, b = −10

Example 2. If \( x + iy = \sqrt{\dfrac{a + ib}{c + id}} \), prove that \( (x^2 + y^2)^2 = \dfrac{a^2 + b^2}{c^2 + d^2} \).
Answer:
\( x + iy = \sqrt{\dfrac{a + ib}{c + id}} \)
∴ \( x - iy = \sqrt{\dfrac{a - ib}{c - id}} \)   (Taking complex conjugate)
∴ \( (x + iy)(x - iy) = \sqrt{\dfrac{a + ib}{c + id} \times \dfrac{a - ib}{c - id}} \)
\( x^2 - i^2y^2 = \sqrt{\dfrac{a^2 - i^2b^2}{c^2 - i^2d^2}} \)
∴ \( x^2 + y^2 = \sqrt{\dfrac{a^2 + b^2}{c^2 + d^2}} \)
∴ \( (x^2 + y^2)^2 = \dfrac{a^2 + b^2}{c^2 + d^2} \)

Example 3. Find the value of x³ + x² − x + 22 if x = 1 + 2i.
Answer:
x = 1 + 2i
(x − 1)² = (2i)²
x² − 2x + 5 = 0
Now, x³ + x² − x + 22 = (x² − 2x + 5)(x + 3) + 7
So, putting x = 1 + 2i, we get:
x³ + x² − x + 22 = 0 + 7 = 7

Example 4. Find the modulus of the following complex numbers.
(i) \( \sqrt{3} + \sqrt{2}\,i \)   (ii) 1 + i
Answer:
(i) \( z = \sqrt{3} + \sqrt{2}\,i \), here \( a = \sqrt{3},\ b = \sqrt{2} \)
∴ \( |z| = \sqrt{a^2 + b^2} = \sqrt{3 + 2} = \sqrt{5} \)
(ii) z = 1 + i, here a = 1, b = 1
∴ \( |z| = \sqrt{a^2 + b^2} = \sqrt{1 + 1} = \sqrt{2} \)

Example 5. For z = 2 + 3i verify the following:
(i) \( \overline{(\bar{z})} = z \)   (ii) \( z\bar{z} = |z|^2 \)   (iii) \( (z + \bar{z}) \) is real   (iv) \( z - \bar{z} = 6i \)
Answer:
z = 2 + 3i, \( \bar{z} = 2 - 3i \)
(i) \( \overline{(\bar{z})} = 2 + 3i \). Hence, \( \overline{(\bar{z})} = z \)
(ii) \( z\bar{z} = (2 + 3i)(2 - 3i) = 4 - 9i^2 = 13 \)
\( |z| = \sqrt{4 + 9} = \sqrt{13} \). Hence, \( z\bar{z} = |z|^2 \)
(iii) \( z + \bar{z} = 2 + 3i + 2 - 3i = 4 \). Hence, \( z + \bar{z} \) is a real number.
(iv) \( z - \bar{z} = (2 + 3i) - (2 - 3i) = 2 + 3i - 2 + 3i = 6i \). Hence, \( z - \bar{z} \) is an imaginary number.

Example 6. Solve the equation \( z^2 = \bar{z} \), where z = x + iy.
Answer:
\( z^2 = \bar{z} \) ⇒ x² − y² + i2xy = x − iy
Therefore, x² − y² = x   ... (1)
and 2xy = −y   ... (2)
From (2), we have y = 0 or \( x = -\dfrac{1}{2} \)
When y = 0, from (1), we get x² − x = 0, i.e., x = 0 or x = 1.
When \( x = -\dfrac{1}{2} \), from (1), we get \( y^2 = \dfrac{1}{4} + \dfrac{1}{2} \) or \( y^2 = \dfrac{3}{4} \), i.e., \( y = \pm\dfrac{\sqrt{3}}{2} \)
Hence, the solutions of the given equation are \( 0 + i0,\ 1 + i0,\ -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2},\ -\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} \)

Example 7. (i) If |z| = 1, prove that \( \dfrac{z - 1}{z + 1} \) (z ≠ −1) is a purely imaginary number.
(ii) If the number \( \dfrac{z - 1}{z + 1} \) is purely imaginary, then prove that |z| = 1.
Answer:
(i) Let \( w = \dfrac{z - 1}{z + 1} \). Since |z| = 1, \( z\bar{z} = 1 \), i.e. \( \bar{z} = \dfrac{1}{z} \).
\( \text{Re}(w) = \dfrac{w + \bar{w}}{2} = \dfrac{1}{2}\left[ \dfrac{z - 1}{z + 1} + \dfrac{\bar{z} - 1}{\bar{z} + 1} \right] = \dfrac{1}{2}\left[ \dfrac{z - 1}{z + 1} + \dfrac{\frac{1}{z} - 1}{\frac{1}{z} + 1} \right] \)
\( = \dfrac{1}{2}\left[ \dfrac{z - 1}{z + 1} + \dfrac{1 - z}{1 + z} \right] = \dfrac{1}{2}\left[ \dfrac{z - 1}{z + 1} - \dfrac{z - 1}{z + 1} \right] = 0 \)
⇒ Re(w) = 0 ⇒ w is a purely imaginary number.
(ii) \( w = \dfrac{z - 1}{z + 1} \). As w is purely imaginary, Re(w) = 0
⇒ \( \dfrac{w + \bar{w}}{2} = 0 \) ⇒ \( \dfrac{z - 1}{z + 1} + \dfrac{\bar{z} - 1}{\bar{z} + 1} = 0 \)
⇒ \( (z - 1)(\bar{z} + 1) + (\bar{z} - 1)(z + 1) = 0 \) ⇒ \( 2z\bar{z} - 2 = 0 \)
⇒ \( z\bar{z} = 1 \) ⇒ |z|² = 1 ⇒ |z| = 1.

Example 8. If iz³ + z² − z + i = 0, then show |z| = 1.
Answer:
iz³ + z² − z + i = 0
By substituting z = i in the equation, we get 0 = 0. Hence z − i is a factor of iz³ + z² − z + i
⇒ iz²(z − i) − 1(z − i) = 0 ⇒ (iz² − 1)(z − i) = 0
Either iz² − 1 = 0 or z − i = 0
When z − i = 0, z = i ∴ \( |z| = |0 + i \cdot 1| = \sqrt{0^2 + 1^2} = 1 \)
When iz² − 1 = 0, \( z^2 = \dfrac{1}{i} = -i \)
∴ \( |z^2| = |0 - 1 \cdot i| = \sqrt{0^2 + (-1)^2} = 1 \) ⇒ |z|² = 1 or |z| = 1
∴ In any case we have |z| = 1

Example 9. Prove that: x⁴ + 4 = (x + 1 + i)(x + 1 − i)(x − 1 + i)(x − 1 − i).
Answer:
Consider R.H.S.
= [(x + 1 + i)(x + 1 − i)] [(x − 1 + i)(x − 1 − i)]
= [(x + 1)² − i²] [(x − 1)² − i²]
= (x² + 2x + 1 + 1)(x² − 2x + 1 + 1) = [(x² + 2) + 2x] [(x² + 2) − 2x]
= (x² + 2)² − (2x)² = x⁴ + 4x² + 4 − 4x² = x⁴ + 4 = L.H.S.

Example 10. Find all complex numbers z which satisfy the following equations:
(i) \( z = \bar{z} \)   (ii) \( z = -\bar{z} \)   (iii) \( \bar{z} = 4 - z \)   (iv) \( z^2 = -\bar{z} \)
Answer:
Let z = x + iy. Then \( \bar{z} = x - iy \).
(i) The equation \( z = \bar{z} \) becomes x + iy = x − iy, or 2iy = 0, which gives y = 0.
Hence z = x, i.e., all the real numbers constitute the solutions of the given equation.
(ii) The equation \( z = -\bar{z} \) is equivalent to x + iy = −(x − iy), or 2x = 0, or x = 0.
Hence z = iy, i.e., the solutions of the given equation are all purely imaginary numbers.
(iii) \( \bar{z} = 4 - z \) or x − iy = 4 − x − iy or x = 4 − x. This gives x = 2. Hence z = 2 + iy.
∴ The given equation is satisfied by all complex numbers whose real part is 2.
(iv) \( z^2 = -\bar{z} \) or (x + iy)² = −(x − iy) or x² − y² + 2ixy = −x + iy
Equating real and imaginary parts, we get x² − y² = −x   ... (i)
and 2xy = y or y(2x − 1) = 0   ... (ii)
From (ii), either y = 0 or 2x − 1 = 0, i.e., \( x = \dfrac{1}{2} \).
When y = 0, (i) gives x² = −x or x(x + 1) = 0, which gives x = 0, −1.
Thus we get two sets of solutions x = 0, y = 0 and x = −1, y = 0.
When \( x = \dfrac{1}{2} \), (i) gives \( y^2 = \dfrac{3}{4} \), which gives \( y = \pm\dfrac{\sqrt{3}}{2} \).
Thus we get two more sets of solutions \( x = \dfrac{1}{2},\ y = \dfrac{\sqrt{3}}{2} \) and \( x = \dfrac{1}{2},\ y = -\dfrac{\sqrt{3}}{2} \).
Hence the given equation has in all the following four solutions: z₁ = 0 + i0, z₂ = −1 + 0i = −1, \( z_3 = \dfrac{1}{2} + i\dfrac{\sqrt{3}}{2},\ z_4 = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} \)