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Q.

If Re(Z2) = 2. Re(Z – 1) then the locus of Z is

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a

x2−y2+2x−2=0

b

x2−y2−2x+2=0

c

x2+y2−2x+2=0

d

−x2−y2−2x−2=0

answer is C.

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Detailed Solution

Re[x2−y2+2ixy]=2Re[(x−1)+iy]⇒x2−y2=2x−2
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