1
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Question

# Let z and w be two complex numbers such that |z|≤1, |ω|≤1 and |z+iω|=|z−i¯ω|=2, then z equals

A
1 or i
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B
i or -i
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C
1 or -1
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D
i or -1
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Solution

## The correct option is C 1 or -1Given, |z+iω|=|z−i¯ω|=2⇒|z−(−iω)|=|z−(i¯ω)|=2⇒|z−(−iω)|=|z−(i¯ω)| ∴ z lies on the perpendicular bisector of the line joining −iω and i¯ω. Since i¯ω is the mirror image of −iω in the x-axis, the locus of z is the x-axis. Let z=x+iy and y=0.Now, |z|≤1⇒x2+02≤1⇒−1≤x≤1. ∴ z may take values given in option (c).

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