Complex Numbers Class 11

Complex Numbers Class 11 – A number that can be represented in form p + iq is defined as a complex number. Here, p and q are real numbers and \(i=\sqrt{-1}\). For a complex number z = p + iq, p is known as the real part, represented by Re z and q is known as the imaginary part, it is represented by Im z of complex number z.

Complex Numbers Class 11 Concepts

The topics and subtopics covered in Complex numbers class 11 are:

• Introduction
• Complex Numbers
• Algebra of Complex Numbers
  • Addition of two complex Numbers
  • Difference of two complex Numbers
  • Multiplication of two complex Numbers
  • Division of two complex Numbers
  • Power of i
  • The square root of a negative real number
  • Identities
• The Modulus and the Conjugate of a Complex Numbers
• Argand Plane and Polar Representation
  • Polar Representation of a Complex Numbers

Complex Numbers Class 11 Notes

The notes for class 11 gives a detailed knowledge of all the concepts involved in complex numbers. Some important concepts like the algebra of complex number, what is the principal argument, modulus and argument of complex numbers, and so on. To learn more, click here: Complex Numbers

Consider two complex numbers z₁ = p + iq and z₂ = m + in.

z₁ + z₂ = (p + m) + i(q + n)

z₁ – z₂ = (p – m) + i(q – n)

z₁ . z₂ = = (pm – nq) + i(pn + mq)

\(\frac{z_1}{z_2}=\frac{p+iq}{m+in}=\frac{p+iq}{m+in}\times \frac{m-in}{m-in}=\frac{pm+qn+i(qm-pn)}{m^2+n^2}\)

Identities for complex numbers z₁ and z₂ are given by:

(z₁ + z₂)² = z₁² + z₂² + 2z₁z₂

(z₁ – z₂)² = z₁² + z₂² – 2z₁z₂

(z₁² – z₂²) = (z₁ + z₂)(z₁ – z₂)

(z₁ + z₂)³ = z₁³ + z₂³ + 3z₁z₂(z₁ + z₂)

(z₁ – z₂)³ = z₁³ – z₂3 – 3z₁z₂(z₁ – z₂)

For any integer k,

i^4k = 1

i^{4k + 1} = i

i^{4k + 2} = -1

i^{4k + 3} = -i

The conjugate of complex number z = p + iq is given by \(\bar{z}\) = p – iq.

Complex Numbers Class 11 Examples

Example 1: If 4x + i(3x – y) = 3 + i (– 6), where x and y are real numbers, then find the values of x and y.

Solution: Given,

4x + i (3x – y) = 3 + i (–6) ….(1)

By equating the real and the imaginary parts of equation (1), 

4x = 3, 3x – y = –6,

Now, 4x = 3

⇒ x = 3/4

And 3x – y = -6

⇒ y = 3x + 6

Substituting the value of x,

⇒ y = 3(3/4) + 6

⇒ y = 33/4

Therefore, x = 3/4 and y = 33/4.

Example 2: Express (-√3 + √-2)(2√3 – i) in the form of a + ib.

Solution: We know that i² = -1

(-√3 + √-2)(2√3 – i) = (-√3 + i√2)(2√3 – i)

= (-√3)(2√3) + (i√3) + i(√2)(2√3) – i²√2

= -6 + i√3(1 + 2√2) + √2

= (-6 + √2) + i√3(1 + 2√2)

This is of the form a + ib, where a = -6 + √2 and b = √3(1 + √2).

Example 3: Find the multiplicative inverse of 2 – 3i.

Solution: Let z = 2 – 3i

\(\bar{z}\) = 2 + 3i

|z|² = (2)² + (-3)² = +  = 13

We know that the multiplicative inverse of z is given by the formula:

\(z^{-1}=\frac{\bar{z}}{|z|^2}\)

= (2 + 3i)/13

= (2/13) + i(3/13)

Alternatively,

Multiplicative inverse of z is:

z^-1 = 1/(2 – 3i)

By rationalising the denominator we get,

= (2 + 3i)/(4 + 9)

= (2 + 3i)/ 13

= (2/13) + i(3/13)

Example 4:  Represent the complex number z = 1 + i√3 in the polar form.

Solution: Given, z = 1 + i√3

Let 1 = r cos θ, √3 = r sin θ

By squaring and adding, we get

r²(cos²θ + sin²θ) = 4

r² = 4

r = 2 (as r > 0)

Therefore, cos θ = 1/2 and sin θ = √3/2

This is possible when θ = π/3.

Thus, the required polar form is z = 2[cos π/3 + i sin π/3].

Hence, the complex number z = 1 + i√3 is represented as shown in the below figure.

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