Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 12 Find the conjugate of ((3 − 2i)(2 + 3i))/((1 + 2i)(2 − i) ) First we calculate ((3 − 2i)(2 + 3i))/((1 + 2i)(2 − i) ) then find its conjugate ((3 −2i)(2+3i))/((1+ 2i)(2−i) ) = (3(2+3i)−2i(2+3i))/(1(2−i)+ 2i(2−i) ) = (3 × 2 +3 × 3𝑖 − 2𝑖 × 2 − 2𝑖 × 3𝑖)/(1 × 2 + 1 ×( −𝑖) + 2𝑖 × 2 − 2𝑖 × 𝑖 ) = (6 + 9i − 4i + 6𝑖2)/(2 − i + 4i − 2i2) = (6 + 5i − 6𝑖2)/(2 + 3i + 2i2) Putting i2 = −1 = (6 + 5𝑖 − 6 ( −1))/(2 + 3𝑖 −2( −1)) = (6 + 5𝑖 + 6)/(2 + 3𝑖 + 2) = (6 + 6 + 5𝑖)/(2 + 2 + 3𝑖) = (12 + 5𝑖)/(4 + 3𝑖) Rationalizing = (12 + 5𝑖)/(4 + 3𝑖) × (4 − 3𝑖)/(4 − 3𝑖) = ((12 + 5𝑖) (4 − 3𝑖))/((4 + 3𝑖) (4 − 3𝑖) ) = (12 × 4 − 12 × 3𝑖 + 5𝑖 × 4 − 5𝑖 × 3𝑖)/((4 + 3𝑖) (4 − 3𝑖)) = (48 − 36𝑖 + 20𝑖 −15𝑖2)/((4 + 3𝑖) (4 − 3𝑖)) Putting i2 = − 1 = (48 − 16𝑖 − 15 (−1))/((4 + 3𝑖) (4 − 3𝑖)) = (48 − 16𝑖 +15)/((4 + 3𝑖) (4 − 3𝑖)) = (63 − 16𝑖)/((4 + 3𝑖) (4 − 3𝑖)) Using (a – b) (a + b) = a2 – b2 = (63 − 16𝑖)/((4)2 − (3𝑖)2 ) = (63 − 16𝑖)/(16 −9𝑖2) Putting i2 = − 1 = (63 −16𝑖)/(16 + 9(−1) ) = (63 − 16𝑖)/(16 + 9) = (63 − 16𝑖)/25 Hence, ((3 −2i)(2+3i))/((1+ 2i)(2−i) ) = 63/25 − 16/25 𝑖 So conjugate of ((3 −2i)(2+3i))/((1+ 2i)(2−i) ) is 63/25 + 16/25 𝑖
Examples
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Example 5
Example 6
Example 7 Deleted for CBSE Board 2022 Exams
Example 8 Deleted for CBSE Board 2022 Exams
Example 9
Example 10
Example 11
Example 12 You are here
Example 13 Deleted for CBSE Board 2022 Exams
Example 14 Important
Example 15
Example 16 Important Deleted for CBSE Board 2022 Exams
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