Question

find the locus of a complex number z=x+iy satisfying |(z+i)|/|(z-i)| =1/√2​

Answer 1

Answer:

x²+y²+6y+3=0

Step-by-step explanation:

z+i = x+iy +i = x +i(y+1)

|z+i| = √{x²+(y+1)²}

z-i = x+i(y-1)

|z-i| = √{x²+(y-1)²}

√{x²+(y+1)²}/√{x²+(y-1)²} = 1/√2

or, {x²+(y+1)²}/{x²+(y-1)²} = 1/2

or, 2{x²+(y+1)²}={x²+(y-1)²}

or, x²+y²+6y+3=0