To multiply the complex numbers, we can use the distributive multiplication method.
Since i2 = -1
= ac + adi + bci + bd(-1)
= ac + adi + bci – bd
= ac – bd + (ad + bc)i
Write the product in standard form :
Example 1 :
(2 + 3i)(2 – i)
Solution :
(2 + 3i)(2 – i) = 2(2) – 2(i) + 3i(2) – 3i(i)
= 4 – 2i + 6i – 3i2
= 4 + 4i – 3(-1)
= 4 + 4i + 3
= 7 + 4i
So, the standard form is 7 + 4i
Example 2 :
(2 - i)(1 + 3i)
Solution :
(2 - i)(1 + 3i) = 2(1) + 2(3i) - i(1) – i(3i)
= 2 + 6i - i – 3i2
= 2 + 5i – 3(-1)
= 2 + 5i + 3
= 5 + 5i
So, the standard form is 5 + 5i
Example 3 :
(1 - 4i)(3 - 2i)
Solution :
(1 - 4i)(3 - 2i) = 1(3) – 1(2i) - 4i(3) + 4i(2i)
= 3 - 2i - 12i + 8i2
= 3 - 14i + 8(-1)
= 3 - 14i - 8
= -5 - 14i
So, the standard form is -5 - 14i
Example 4 :
(5i - 3)(2i + 1)
Solution :
(5i - 3)(2i + 1) = 5i(2i) + 5i(1) - 3(2i) - 3(1)
= 10i2 + 5i - 6i - 3
= 10(-1) + 5i – 6i - 3
= -10 + 5i – 6i - 3
= -13 - i
So, the standard form is -13 – i
Example 5 :
(7i - 3)(2 + 6i)
Solution :
(7i - 3)(2 + 6i) = 7i(2) + 7i(6i) - 3(2) - 3(6i)
= 14i + 42i2 - 6 – 18i
= 14i + 42(-1) – 6 – 18i
= -4i - 42 – 6
= -4i – 48
= -48 – 4i
So, the standard form is -48 – 4i
Example 6 :
(√-4 + i)(6 - 5i)
Solution :
(√-4 + i)(6 - 5i) = √-4(6) - √-4(5i) + i(6) - i(5i)
= 6√-4 – 5i√-4 + 6i – 5i2
Since i = √-1
= 6i(2) - 5i(i)(2) + 6i – 5(-1)
= 12i – 10i2 + 6i + 5
= 18i – 10(-1) + 5
= 18i + 10 + 5
= 18i + 15
= 15 + 18i
So, the standard form is 15 + 18i
Example 7 :
(-3 - 4i)(1 + 2i)
Solution :
(-3 - 4i)(1 + 2i) = -3(1) - 3(2i) - 4i(1) – 4i(2i)
= -3 - 6i - 4i – 8i2
= -3 - 10i – 8(-1)
= -3 - 10i + 8
= 5 - 10i
So, the standard form is 5 - 10i
Example 8 :
(√-2 + 2i)(6 + 5i)
Solution :
(√-2 + 2i)(6 + 5i) = √-2(6) + √-2(5i) + 2i(6) + 2i(5i)
= 6√-2 + 5i√-2 + 12i + 10i2
Since i = √-1
= 6i√2 + 5i(i)√2 + 12i + 10(-1)
= 6√2i + 5√2i2 + 12i - 10
= 6√2i + 5√2(-1) + 12i - 10
= 6√2i - 5√2 + 12i – 10
By combining like terms, we get
= 6√2i + 12i - 5√2 – 10
= -5√2 – 10 + (6√2 + 12)i
So, the standard form is -5√2 – 10 + (6√2 + 12)i
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