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Question

# If z=x+iy such that |z+1|=|z−1| and amp(z−1z+1)=π4 then

A
x=2+1;y=0
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B
x=0;y=2+1
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C
x=0;y=21
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D
x=21;y=0
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Solution

## The correct option is B x=0;y=√2+1Let z=x+iyGiven that |z+1|=|z−1|$\Rightarrow |(x+1)+iy|=|(x-1)+iy|$⇒√(x+1)2+y2=√(x−1)2+y2⇒(x+1)2+y2=(x−1)2+y2⇒x2+1+2x=x2+1−2x⇒x=0Given that amp(z−1z+1)=π4⇒amp(z−1z+1)=amp(x+iy−1x+iy+1)But x=0⇒amp(z−1z+1)=amp(−1+iy1+iy)⇒amp(−1+iy1+iy)=π4⇒amp(−1+iy1+iy×1−iy1−iy)=π4⇒amp(−1+iy+iy+y21+y2)=π4⇒amp(y2−1+2iy1+y2)=π4⇒tan−1(2yy2−1)=π4⇒2yy2−1=1⇒y2−2y−1=0On solving we get,⇒y=1±√2Since θ lies in Q1⇒y=√2+1 and x=0

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