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Question
If z1, z2 and z3 are three complex numbers such that |z1| = 1, |z2| = 2, |z3| = 3 and |z1 + z2 + z3| = 1, show that |9z1z2 + 4z1z3 + z2z3| = 6
Solution
|z1| = 1, |z1| = 2, |z3| = 3
|z1 + z2 + z3| = 1
Now |9z1 z2 + 4z1 z3 + z2 z3|
= `|"z"_1| |"z"_2| |"z"_3| |9/z_3 + 4/z_2 + 1/z_1|`
= `(1)(2)(3) |(9bar(z)_3)/|z_3|^2 + (4bar(z)_2)/|z_2|^2 + (bar(z)_1)/|z_1|^2|`
= `6|(9bar(z)_3)/(3)^2 + (4bar(z)_2)/(2)^2 + (bar(z)_1)/(1)^2|`
= `6|bar(z)_3 + bar(z)_2 + bar(z)_1|`
= `6|bar(z_3 + z_2 + z_1)|`
= `6|z_1 + z_2 + z_3|`
= 6(1)
= 6
Hence proved.
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