1.

where does z lie , if `|(z-5i)/(z+5i)|=1?`

Answer» Let z = (x + iy). Then,
`|(z-5i)/(z+5i)|=1 rArr (|z-5i|)/(|z+5i|)=1" "[because |(z_(1))/(z_(2))|=(|z_(1)|)/(|z_(2)|)]`
`rArr" "|z-5i|=|z+5i| rArr |z-5i|^(2)=|z+5i|^(2)`
`rArr" "|(x+iy)-5i|^(2)=|(x+iy)+5i|^(2)" "[because z = (x+iy)]`
`rArr" "|x+i(y-5)|^(2)=|x+i(y+5)|^(2)`
`rArr" "x^(2)+(y-5)^(2)=x^(2)+(y+5)^(2)" "[because |x+iy|^(2)=(x^(2)+y^(2))]`
`rArr" "(y+5)^(2)-(y-5)^(2)= 0 rArr 4 xx y xx 5 = 0 rArr y = 0`.
`therefore" "z=x + i0 rArr z = x`, where x is real.
Hence, z is a real number.


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