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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 and z2 = 2 − 3i
∴ `bar("z"_1)` = 1 − i and `bar("z"_2)` = 2 + 3i
`"z"_1/"z"_2 = (1 + "i")/(2 -3"i")`
= `(1 + "i")/(2 - 3"i") xx (2 + 3"i")/(2 + 3"i")`
= `(2 + 3"i" + 2"i" + 3"i"^2)/(4 - 9"i"^2)`
=`(2 + 5"i" - 3)/(4 + 9)` ...[∵ i2 = – 1]
=`(-1 + 5"i")/13`
=`-1/13 + 5/13"i"`
∴ `bar(("z"_1/"z"_2)) = -1/13 - 5/13"i"` .......(1)
`bar("z"_1)/(bar("z")_2) = (1 - "i")/(2 + 3"i")`
= `(1 - "i")/(2 + 3"i") xx (2 - 3"i")/(2 - 3"i")`
= `(2- 3"i" - 2"i" + 3"i"^2)/(4 - 9"i"^2)`
= `(2 - 5"i" - 3)/(4 + 9)` ...[∵ i2 = – 1]
= `(-1 - 5"i")/13`
= `-1/13 - 5/13"i"` ........(2)
From (1) and (2), we get,
`bar(("z"_1/"z"_2))=bar("z"_1)/bar("z"_2)`
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