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Question

If ω is any complex number such that zω=|z|2 and |z¯z|+|ω+¯ω|=4, then as ω varies, then the area bounded by the locus of z is

A
4 sq. units
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B
8 sq. units
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C
16 sq. units
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D
12 sq. units
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Solution

The correct option is C 8 sq. units
zω=|z|2
zω=z¯zω=¯z
|z¯z|+|ω+¯ω|=|z¯z|+|z+¯z|=4
Put z=x+iy
|(x+iyx+iy)|+|x+iy+xiy|=4
|2iy|+|2x|=4
|2y|+|2x|=4
|x|+|y|=2
Now, |x|+|y|=2 has graph of square with side=42
Therefore area=(42)2=162=8 square units

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