The equation z bar z + a bar z+b=0 , b∈R represents a circle if (1) |a|^2 = b (2) |a|^2 > b (3) |a|^2 < b(4) None of these

Solution:

\(z\bar{z}+a\bar{z}+\bar{a}z+b = 0\) \(a\bar{a}+z\bar{z}+a\bar{z}+b = 0\) \((z+a)(\bar{z}+\bar{a})= a\bar{a}-b\)

|z+a|2 = |a|2-b

This will represent a circle if |a|2-b > 0 or |a|2 >b

Hence option (1) is the answer.

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