If z = x + iy lies in III quadrant, then `bar z / z` also lies in III quadrant If:

Explanation

Given that, z = x + iy lies in a third quadrant
`rArr x le 0 and y le 0`
Now, `(barz)/(z) =(x-iy)/(x+iy)= ((x-iy)(x-iy))/((x+iy)(x-iy))`
`= (x^(2) -y^(2)-2ixy)/(x^(2)+y^(2)) = (x^(2)-y^(2))/(x^(2) +y^(2))-(2ixy)/(x^(2) +y^(2))`
For `(barz)/(z)` to lie the third quadrant, we must have
`(x^(2) -y^(2))/(x^(2)+y^(2)) lt 0 and (-2xy)/(x^(2) + y^(2))lt 0`
`rArr " "x^(2) - y^(2) and -2xy lt 0`
`rArr " " x^(2) lt y^(2) and xy gt 0`
But `x,y gt0`
`therefore " "y lt x lt0`