find the locus of a complex number z=x+iy satisfying z+i by z+i = 1by root 2
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∣z+i∣=∣z+2∣
assume z=x+iy
x
2
+(y+1)
2
=(x+2)
2
+y
2
x
2
+y
2
+2y+1=x
2
+2x+y+y
2
2y+1=2x+1
x=y
So, z=z+ix
=x(1+i)
So z is always multiple pf (1+i)
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