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If arg(z-1)=arg(z+3i), then find (x-1):y...

If `arg(z-1)=arg(z+3i)`, then find `(x-1):y`, where `z=x+iy`.

Text Solution

Verified by Experts

Given that, arg `(z - 1) = arg (z + 3i)`
and , let `z = x + iy`
Now, `arg (z -1) = arg(z + 3i)`
`rArr arg (x + iy - 1) = arg (x + iy + 3i)`
`rArr arg (x - 1 + iy) = arg [x + i (y + 3)]`
`rArr tan^(-1).(y)/(x -1) = tan^(-1).(y + 3)/ (x)`
`rArr (y)/(x -1) = (y + 3)/ (x) rArr xy = (x - 1)(y +3)`
` rArr xy = xy - y 3x - 3 rArr 3x - 3 = y`
`rArr (3(x - 1))/(y ) =1 rArr (x - 1)/(y) =(1)/(3)`
`:. (x -1) : y = 1 : 3`
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