Complex Numbers Class 11 notes are provided here in detail. The concepts covered in complex numbers class 11 Maths are important for students, as it will help them to solve the problems in the higher class. Go through the notes provided below and practice the problems as well to score good marks in the examination.
Complex Numbers: A number that can be represented in form p + iq is defined as a complex number. Here, p and q are real numbers and
Complex Numbers Class 11 Concepts
The topics and subtopics covered in Complex numbers class 11 are:
- 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
- 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)
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,
i4k = 1
i4k + 1 = i
i4k + 2 = -1
i4k + 3 = -i
The conjugate of complex number z = p + iq is given by
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.
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
|z|2 = (2)2 + (-3)2 = + = 13
We know that the multiplicative inverse of z is given by the formula:
= (2 + 3i)/13
= (2/13) + i(3/13)
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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