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z=x+iy satisfies arg(z+2)=arg(z+i) then ...

`z=x+iy` satisfies `arg(z+2)=arg(z+i)` then (A) `x+2y+1=0` (B) `x+2y+2=0` (C) `x-2y+1=0` (D) `x-2y-2=0`

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