Question

find the locus of a complex number z=x+iy satisfying z+i by z+i = 1by root 2

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Answer 1

Answer:

Answer

∣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)