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Question
z1 = 1 + i, z2 = 2 − 3i. Verify the following :
`bar("z"_1."z"_2) = bar("z"_1).bar("z"_2)`
Solution
z1 = 1 + i, z2 = 2 − 3i
∴ `bar("z"_1)` = 1 − i, `bar("z"_2)` = 2 + 3i
`("z"_1."z"_2)` = (1 + i) (2 – 3i)
= 2 – 3i + 2i – 3i2
= 2 – i – 3 (– 1) ...[∵ i2 = – 1]
= 5 – i
∴ `bar("z"_1."z"_2)` = 5 + i
`bar("z"_1).bar("z"_2)` = (1 – i) (2 + 3i)
= 2 + 3i – 2i – 3i2
= 2 + i – 3(–1) ...[∵ i2 = – 1]
= 5 + i
∴ `bar("z"_1."z"_2) = bar("z"_1).bar("z"_2)`
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