If |z−3|=min{|z−1|,|z−5|}, then Re(z) equals to
A
2
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B
52
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C
72
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D
4
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Solution
The correct options are A 2 D 4 Let z=x+iy Hence |z−3|=|z−1| Or (x−3)2+y2=(x−1)2+y2 Or (x−3)2−(x−1)2=0 Or (2x−4)(−2)=0 Hence x=2 ...(i) Else |z−3|=|z−5| (x−3)2=(x−5)2 (x−3)2−(x−5)2=0 (2x−8)(2)=0 x=4 ...(ii) Hence Re(z)=2or4.