CBSE Class 11 Applied Mathematics Complex Numbers Notes Set 01

Imaginary Numbers

Concept of Imaginary Numbers and Iota

If the square of a given number is negative, then such a number is called an imaginary number.
For example, \( \sqrt{-1}, \sqrt{-2} \) etc. are imaginary numbers. We denote \( \sqrt{-1} \) by the Greek letter iota 'i'. Thus \( \sqrt{-4} = 2i,\ \sqrt{-9} = 3i \) etc.

Powers of i

We have i⁰ = 1, i¹ = i, i² = −1, i³ = −i, i⁴ = 1.
Let us consider iⁿ, where n is a positive integer and n > 4. On dividing n by 4, let the quotient be m and the remainder be r. Then
n = 4m + r, where 0 ≤ r < 4
∴ \( i^n = i^{4m + r} = i^{4m} \times i^r = (i^4)^m \times i^r = i^r \)

Example 1. Evaluate: (i) i⁹⁹⁸ (ii) i⁻⁷¹ (iii) i⁻¹
Answer:
(i) \( i^{998} = i^{4 \times 249 + 2} = i^2 = -1 \)
(ii) \( i^{-71} = \frac{1}{i^{71}} = \frac{1}{i^{71}} \times \frac{i}{i} = \frac{i}{i^{72}} = \frac{i}{(i^4)^{18}} = \frac{i}{1} = i \)
(iii) \( i^{-1} = \frac{1}{i} = \frac{1}{i} \times \frac{i^3}{i^3} = \frac{i^3}{i^4} = \frac{-i}{1} = -i \)

Example 2. Show that the sum (1 + i² + i⁴ + … + i²ⁿ) is 0 when n is odd and 1 when n is even.
Answer:
Let S = 1 + i² + i⁴ + … + i²ⁿ
This is clearly a GP having (n + 1) terms with a = 1 and r = i² = −1.
∴ \( S = \frac{a(1 - r^{n+1})}{1 - r} = \frac{1 \times [1 - (i^2)^{n+1}]}{1 - i^2} = \frac{1 - (-1)^{n+1}}{1 - (-1)} = \frac{1 - (-1)^{n+1}}{2} \)
\( = \frac{1}{2}(1 - 1) = 0 \), when n is odd
\( = \frac{1}{2}(1 + 1) = 1 \), when n is even

Note: For any two real numbers a and b, the result \( \sqrt{a} \times \sqrt{b} = \sqrt{ab} \) is true only when at least one of the given numbers is either zero or positive.
Thus \( \sqrt{-2} \times \sqrt{-3} = \sqrt{(-2) \times (-3)} = \sqrt{6} \) is wrong.
In fact, \( \sqrt{-2} \times \sqrt{-3} = i\sqrt{2} \times i\sqrt{3} = i^2 \times \sqrt{6} = -\sqrt{6} \).

Example 3. Evaluate: \( \sqrt{-16} + 3\sqrt{-25} + \sqrt{-36} - \sqrt{-625} \)
Answer:
\( \sqrt{-16} + 3\sqrt{-25} + \sqrt{-36} - \sqrt{-625} = 4i + 3 \times 5i + 6i - 25i \)
= 4i + 15i + 6i − 25i = 0

Complex Numbers

Complex Number

The numbers of the form (a + ib), where a and b are real numbers and \( i = \sqrt{-1} \), are known as complex numbers. The set of all complex numbers is denoted by C.
∴ C = {(a + ib) : a, b ∈ R}
For a complex number z = (a + ib), we have a = real part of z, written as Re (z), and b = imaginary part of z, written as Im (z).

Purely Real and Purely Imaginary Numbers

A complex number z is said to be
(i) purely real, if Im (z) = 0
(ii) purely imaginary, if Re (z) = 0
Thus each of the numbers 2, −7, \( \sqrt{3} \) is purely real and each of the numbers 2i, \( \sqrt{3}\,i \), \( \frac{3}{2}i \) is purely imaginary.

Conjugate of a Complex Number

The conjugate of a complex number z = (a + ib) is defined as \( \bar{z} = a - ib \).

Algebra of Complex Numbers

Addition of Two Complex Numbers

Let z₁ = a + ib and z₂ = c + id be any two complex numbers. Then the sum z₁ + z₂ is defined as follows:
z₁ + z₂ = (a + c) + i(b + d), which is again a complex number.
The addition of complex numbers satisfies the following properties:
(i) The closure law: The sum of two complex numbers is a complex number, i.e. z₁ + z₂ is a complex number for all complex numbers z₁ and z₂.
(ii) The commutative law: For any two complex numbers z₁ and z₂, z₁ + z₂ = z₂ + z₁.
(iii) The associative law: For any three complex numbers z₁, z₂, z₃, (z₁ + z₂) + z₃ = z₁ + (z₂ + z₃).
(iv) The existence of additive identity: There exists the complex number 0 + i0 (denoted as 0), called the additive identity or the zero complex number, such that for every complex number z, z + 0 = z.
(v) The existence of additive inverse: To every complex number z = a + ib, we have the complex number −a + i(−b) (denoted as −z), called the additive inverse or negative of z. We observe that z + (−z) = 0.

Difference of Two Complex Numbers

Given any two complex numbers z₁ and z₂, the difference z₁ − z₂ is defined as follows:
z₁ − z₂ = z₁ + (−z₂)

Multiplication of Two Complex Numbers

Let z₁ = a + ib and z₂ = c + id be any two complex numbers. Then the product z₁z₂ is defined as follows:
z₁z₂ = (ac − bd) + i(ad + bc)
The multiplication of complex numbers possesses the following properties:
(i) The closure law: The product of two complex numbers is a complex number, i.e. the product z₁z₂ is a complex number for all complex numbers z₁ and z₂.
(ii) The commutative law: For any two complex numbers z₁ and z₂, z₁z₂ = z₂z₁.
(iii) The associative law: For any three complex numbers z₁, z₂, z₃, (z₁z₂)z₃ = z₁(z₂z₃).
(iv) The existence of multiplicative identity: There exists the complex number 1 + i0 (denoted as 1), called the multiplicative identity, such that z · 1 = z for every complex number z.
(v) The existence of multiplicative inverse: For every non-zero complex number z = a + ib (a ≠ 0 or b ≠ 0), we have the complex number \( \frac{a}{a^2 + b^2} + i\frac{-b}{a^2 + b^2} \) (denoted by \( \frac{1}{z} \) or z⁻¹), called the multiplicative inverse of z, such that \( z \cdot \frac{1}{z} = 1 \).
(vi) The distributive law: For any three complex numbers z₁, z₂, z₃,
(a) z₁(z₂ + z₃) = z₁z₂ + z₁z₃
(b) (z₁ + z₂)z₃ = z₁z₃ + z₂z₃

Division of Two Complex Numbers

Given two complex numbers z₁ and z₂, where z₂ ≠ 0, the quotient \( \frac{z_1}{z_2} = z_1 \cdot \frac{1}{z_2} \).
For example, let z₁ = 6 + 3i and z₂ = 2 − i. Then
\( \frac{z_1}{z_2} = (6 + 3i) \times \frac{1}{2 - i} = (6 + 3i)\left\{\frac{2}{2^2 + (-1)^2} + i\frac{-(-1)}{2^2 + (-1)^2}\right\} = (6 + 3i)\left(\frac{2 + i}{5}\right) \)
\( = \frac{1}{5}[12 - 3 + i(6 + 6)] = \frac{1}{5}(9 + 12i) \)

Identities

For any complex numbers z₁ and z₂, the following identities can be proved to be true:
(i) (z₁ + z₂)² = z₁² + z₂² + 2z₁z₂
(ii) (z₁ − z₂)² = z₁² − 2z₁z₂ + z₂²
(iii) (z₁ + z₂)³ = z₁³ + 3z₁²z₂ + 3z₁z₂² + z₂³
(iv) (z₁ − z₂)³ = z₁³ − 3z₁²z₂ + 3z₁z₂² − z₂³
(v) z₁² − z₂² = (z₁ + z₂)(z₁ − z₂)

Modulus and Conjugate

The Modulus and the Conjugate of a Complex Number

Let z = a + ib be a complex number. Then the modulus of z, denoted by |z|, is defined to be the non-negative real number \( \sqrt{a^2 + b^2} \), i.e. \( |z| = \sqrt{a^2 + b^2} \), and the conjugate of z, denoted as \( \bar{z} \), is the complex number a − ib, i.e. \( \bar{z} = a - ib \).
For example, \( |3 + i| = \sqrt{3^2 + 1^2} = \sqrt{10} \), \( \overline{3 + i} = 3 - i \) and \( \overline{3i - 5} = -3i - 5 \).
Observe that the multiplicative inverse of the non-zero complex number z is given by
\( z^{-1} = \frac{1}{a + ib} = \frac{a - ib}{(a + ib)(a - ib)} = \frac{a - ib}{a^2 + b^2} = \frac{\bar{z}}{|z|^2} \)
or \( z \cdot z^{-1} = \frac{z\bar{z}}{|z|^2} \), i.e. \( 1 = \frac{z\bar{z}}{|z|^2} \), or \( z\bar{z} = |z|^2 \).
Furthermore, the following results can easily be derived. For any two complex numbers z₁ and z₂, we have:
(i) |z₁z₂| = |z₁| |z₂|
(ii) \( \left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \), provided |z₂| ≠ 0
(iii) \( \overline{z_1 z_2} = \bar{z}_1 \bar{z}_2 \)
(iv) \( \overline{z_1 \pm z_2} = \bar{z}_1 \pm \bar{z}_2 \)
(v) \( \overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z}_1}{\bar{z}_2} \), provided z₂ ≠ 0

Example 4. Find the multiplicative inverse of 2 − 3i.
Answer:
Let z = 2 − 3i
Then \( \bar{z} = 2 + 3i \) and |z|² = 2² + (−3)² = 13
Therefore, the multiplicative inverse of 2 − 3i is given by \( z^{-1} = \frac{\bar{z}}{|z|^2} = \frac{2 + 3i}{13} = \frac{2}{13} + \frac{3}{13}i \)
The above working can be reproduced in the following manner also:
\( z^{-1} = \frac{1}{2 - 3i} = \frac{2 + 3i}{(2 - 3i)(2 + 3i)} = \frac{2 + 3i}{2^2 + 3^2} = \frac{2 + 3i}{13} = \frac{2}{13} + \frac{3}{13}i \)

Example 5. Express the following in the form a + ib: (i) \( \frac{5 + \sqrt{2}\,i}{1 - \sqrt{2}\,i} \) (ii) i⁻³⁵
Answer:
(i) We have \( \frac{5 + \sqrt{2}\,i}{1 - \sqrt{2}\,i} = \frac{5 + \sqrt{2}\,i}{1 - \sqrt{2}\,i} \times \frac{1 + \sqrt{2}\,i}{1 + \sqrt{2}\,i} = \frac{5 + 5\sqrt{2}\,i + \sqrt{2}\,i - 2}{1 - (\sqrt{2}\,i)^2} \)
\( = \frac{3 + 6\sqrt{2}\,i}{1 + 2} = \frac{3(1 + 2\sqrt{2}\,i)}{3} = 1 + 2\sqrt{2}\,i \)
(ii) \( i^{-35} = \frac{1}{i^{35}} = \frac{1}{i^{4 \times 8 + 3}} = \frac{1}{i^3} = \frac{1}{-i} = \frac{1}{-i} \times \frac{i}{i} = \frac{i}{-i^2} = i \)

Example 6. If \( a + ib = \frac{(x + i)^2}{2x^2 + 1} \), then prove that \( a^2 + b^2 = \frac{(x^2 + 1)^2}{(2x^2 + 1)^2} \).
Answer:
We have \( a + ib = \frac{(x + i)^2}{2x^2 + 1} = \frac{x^2 + i^2 + 2ix}{2x^2 + 1} = \frac{(x^2 - 1) + i(2x)}{2x^2 + 1} \)
⇒ \( a + ib = \frac{x^2 - 1}{2x^2 + 1} + i \cdot \frac{2x}{2x^2 + 1} \)
⇒ \( |a + ib|^2 = \left|\frac{x^2 - 1}{2x^2 + 1} + i \cdot \frac{2x}{2x^2 + 1}\right|^2 \)
⇒ \( a^2 + b^2 = \frac{(x^2 - 1)^2}{(2x^2 + 1)^2} + \frac{4x^2}{(2x^2 + 1)^2} = \frac{(x^2 - 1)^2 + 4x^2}{(2x^2 + 1)^2} = \frac{(x^2 + 1)^2}{(2x^2 + 1)^2} \)
Hence \( a^2 + b^2 = \frac{(x^2 + 1)^2}{(2x^2 + 1)^2} \).

Example 7. If |z₁| = |z₂| = |z₃| = … = |z_n| = 1, then prove that \( \left|\frac{1}{z_1} + \frac{1}{z_2} + \ldots + \frac{1}{z_n}\right| = |z_1 + z_2 + \ldots + z_n| \).
Answer:
We have |z₁| = |z₂| = … = |z_n| = 1
⇒ |z₁|² = |z₂|² = … = |z_n|² = 1
⇒ \( z_1\bar{z}_1 = 1,\ z_2\bar{z}_2 = 1, \ldots, z_n\bar{z}_n = 1 \)
⇒ \( \frac{1}{z_1} = \bar{z}_1,\ \frac{1}{z_2} = \bar{z}_2, \ldots, \frac{1}{z_n} = \bar{z}_n \)
⇒ \( \left|\frac{1}{z_1} + \frac{1}{z_2} + \ldots + \frac{1}{z_n}\right| = |\bar{z}_1 + \bar{z}_2 + \ldots + \bar{z}_n| = \left|\overline{z_1 + z_2 + \ldots + z_n}\right| \)
\( = |z_1 + z_2 + \ldots + z_n| \) [∵ \( |\bar{z}| = |z| \)]

Square Root of a Complex Number

To Evaluate \( \sqrt{a + ib} \)

Let \( \sqrt{a + ib} = x + iy \) … (i)
On squaring both sides of (i), we get a + ib = (x + iy)²
⇒ a + ib = (x² − y²) + i(2xy) … (ii)
On comparing real parts and imaginary parts on both sides of (ii), we get x² − y² = a and 2xy = b
⇒ \( x^2 + y^2 = \sqrt{(x^2 - y^2)^2 + 4x^2y^2} = \sqrt{a^2 + b^2} \)
∴ x² − y² = a … (iii) and \( x^2 + y^2 = \sqrt{a^2 + b^2} \) … (iv)
On solving (iii) and (iv), we get \( x^2 = \frac{1}{2}\left(\sqrt{a^2 + b^2} + a\right) \) and \( y^2 = \frac{1}{2}\left(\sqrt{a^2 + b^2} - a\right) \)
Hence \( \sqrt{a + ib} = \pm\left[\sqrt{\frac{1}{2}\left(\sqrt{a^2 + b^2} + a\right)} + i\sqrt{\frac{1}{2}\left(\sqrt{a^2 + b^2} - a\right)}\right] \), when b > 0 (for b < 0, x and y have opposite signs).
Remark: Similarly, by assuming that \( \sqrt{a - ib} = x - iy \), we may find \( \sqrt{a - ib} \).

Example 8. Evaluate: \( \sqrt{-i} \)
Answer:
Let \( \sqrt{-i} = x - iy \) … (i)
On squaring both sides of (i), we get −i = (x − iy)²
⇒ −i = (x² − y²) − i(2xy) … (ii)
On comparing real parts and imaginary parts on both sides of (ii), we get x² − y² = 0 and 2xy = 1
∴ \( x^2 + y^2 = \sqrt{(x^2 - y^2)^2 + 4x^2y^2} = \sqrt{0^2 + 1^2} = 1 \)
Thus x² − y² = 0 … (iii) and x² + y² = 1 … (iv). On solving (iii) and (iv), we get \( x^2 = \frac{1}{2} \) and \( y^2 = \frac{1}{2} \)
∴ \( x = \pm\frac{1}{\sqrt{2}} \) and \( y = \pm\frac{1}{\sqrt{2}} \)
Since xy > 0, x and y are of the same sign.
∴ \( \left(x = \frac{1}{\sqrt{2}},\ y = \frac{1}{\sqrt{2}}\right) \) or \( \left(x = -\frac{1}{\sqrt{2}},\ y = -\frac{1}{\sqrt{2}}\right) \)
Hence \( \sqrt{-i} = \left(\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\right) \) or \( \left(-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right) \)