If z=x+iy and |z|=1 then find the locus of z
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Substitucting z=x+iy
⇒∣x+iy−ai∣=∣x+iy+ai∣
⇒x
2
+(y−a)
2
=x
2
+(y+a)
2
⇒2ay=0
⇒y=0
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