Go To Index
Unit 1: Real And Complex Numbers
Explanation Of Exercise 1.6
Operations on Complex Numbers:
At complex number all basic mathematical operations are performed. i.e. Carry out all Basic Operations (Addition, Subtraction, Multiplication and Division).(1) Addition of complex number:
Real part addition in real and imaginary with imaginary.
Let
z1 = a + ib and z2 = c + id be any two complex numbers.
∀ a, b, c, d ∈ R. then, their sum,
z1 + z2 = (a + ib) + (c + id)
z1 + z2 = (a + c) + i(b + d) = (a + c, b + d)
Remember that: (a + c) + (b + d) = (a + c, b + d)
Example: If z1 = 6 + 9i and z2 = -1 + 2i, find z1 + z2
Solution:
Given that
z1 = 6 + 9i = (6, 9) and
z2 = -1 + 2i = (-1, 2)
We know that z1 + z2 = (a + c) + i(b + d) = (a + c, b + d)
∴ z1 + z2 = (6, 9) + (-1, 2)
⇒ z1 + z2 = (6 - 1, 9 + 2)
⇒ z1 + z2 = (5, 11) Ans.
(2) Subtraction of complex number:
Real part subtraction in real and imaginary with imaginary.
Let
z1 = a + ib and z2 = c + id ∀ a, b, c, d ∈ R.
then,
z1 - z2 = (a + ib) - (c + id)
z1 - z2 = (a - c) + i.(b - d) = (a - c, b - d)
Remember that (a, b) - (c, d) = (a - c, b - d)
Example: If z1 = -7 + 2i and z2 = 4 - 9i, find z1 - z2
Solution:
Given that
z1 = -7 + 2i = (-7, 2) and
z2 = 4 - 9i = (4, -9)
We know that z1 + z2 = (a - c, b - d)
∴ z1 + z2 = (6, 9) + (-1, 2)
⇒ z1 - z2 = (-7 - 4, 2 + 9)
⇒ z1 + z2 = (-11, 11) Ans.
(3) Multiplication of complex number:
Normal multiplication applied and removed i2 by substituting its value.
Let
z1 = a + ib and z2 = c + id, be any two complex number.
∀a, b, c, d ∈ R
z1 . z2 = (a + ib).(c + id)
z1z2 = c(a + ib) + id(a + ib)
z1z2 = ac + bci + adi + bdi2
z1z2 = ac + (ad + bc)i + (-1)bd (∴ i2 = - 1)
z1z2 = (ac - bd) + (ad + bc)i = (ac -bd, ad + bc)
Remember that: (a, b)(c, d) = (ac - bd, ad + bc)
Example: If z1 = 3 + 4i = (3, 4) and z2 = -3 - 4i = (-3, -4), find the product z1z2
Solution:
Given that
z1 = 3 + 4i = (3, 4) and
z2 = -3 - 4i = (-3, -4)
We know that z1 + z2 = (ac - bd, ad + bc)
∴ z1 . z2 = (3, 4) . (-3, -4)
⇒ z1 . z2 = (-9 + 16, -12 -12)
⇒ z1 . z2 = (7, -24) Ans.
(3) Division Of complex number:
In this we solve by the multiply and divide by the denominator of complex number conjugate and solve.
Let z1 = a + ib = (a, b) and z2 = c + id =(c, d), z2 ≠ 0
No comments:
Post a Comment