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Question
For any two complex numbers z1 and z2, such that |z1| = |z2| = 1 and z1 z2 ≠ -1, then show that `(z_1 + z_2)/(1 + z_1 z_2)` is real number
Solution
Given |z1| = |z2| = 1 and z1z2 ≠ 1
|z1|2 = 1
|z2|2 = 1
`"z"_1 bar("z")_1` = 1
Similarly `"z"_2 bar("z")_2` = 1
z1 = `1/bar(z)_1`
z2 = `1/bar(z)_2`
`(z_1 + z_2)/(1 + z_1z_2) = (1/bar(z)_1 + 1/bar(z)_2)/(1 + 1/bar(z)_1 1/bar(z)_2)`
= `((bar(z)_2 + bar(z)_1)/(bar(z)_1 + bar(z)_2))/((bar(z)_1 bar(z)_2 + 1)/(bar(z)_1 bar(z)_2))`
= `(bar(z)_1 + bar(z)_2)/(1 + bar(z)_1 bar(z)_2)`
= `(bar(z_1 + z_2))/(bar(1 + z_1 + z_2))`
= `bar(((z_1 + z_2)/(1 + z_1 + z_2))`
∴ `(z_1 + z_2)/(1 + z_1 z_2) = bar(((z_1 + z_2)/(1 + z_1 + z_2))`
∴ `(z_1 + z_2)/(1 + z_1 z_2)` is real.
Since z = `bar(z)` [z is real]
Since z = `bar(z)`, it is a real number.
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