Question
The number of complex numbers z such that |z−1|=|z+1|=|z−i| equals
Answer: Option A
:
A
Let z = x + iy
|z−1|=|z+1|⇒(x−1)2+y2=(x+1)2+y2⇒Re(z)=0⇒x=0|z−1|=|z−i|⇒(x−1)2+y2=x2+(y−1)2⇒x=y|z+1|=|z−i|⇒(x+1)2+y2=x2+(y−1)2
Only (0, 0) satisfies all conditions.
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:
A
Let z = x + iy
|z−1|=|z+1|⇒(x−1)2+y2=(x+1)2+y2⇒Re(z)=0⇒x=0|z−1|=|z−i|⇒(x−1)2+y2=x2+(y−1)2⇒x=y|z+1|=|z−i|⇒(x+1)2+y2=x2+(y−1)2
Only (0, 0) satisfies all conditions.
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