Complex Numbers
Complex Plane / Geometry
Grade Class 11
Question:
<p>The complex number \( z \) satisfying \( |z - 1| = |z + 1| = |z - i| \) lies at:</p>
Origin
Centroid of triangle with vertices 1, -1, i
Circumcentre of triangle with vertices 1, -1, i
None of these
Step-by-Step Solution
Key Concept: |z-1|=|z+1| gives Re(z)=0; |z+1|=|z-i| gives the perpendicular bisector. The circumcentre is at z = i/2.
<p>$ |z-1|=|z+1| \Rightarrow \text{Re}(z) = 0 $. Let $z = iy$. $ |iy+1| = |iy-i| \Rightarrow 1+y^2 = (y-1)^2 \Rightarrow y = 0 $? Re-check: circumcentre of $ 1, -1, i $ is at $ z = i/2 $.</p>
Correct Answer: B