State ture or false for the following. (i) The order relation is defined on the set of complex numbers. (ii) Multiplication of a non-zero complex number by -i rotates the point about origin through a right angle in the anti-clockwise direction.(iiI) For any complex number z, the minimum value of `|z|+|z-1|` is 1. (iv) The locus represent by `|z-1|= |z-i|` is a line perpendicular to the join of the points `(1,0) and (0,1)`. (v) If z is a complex number such that `zne 0" and" Re (z) = 0, then Im (z^(2)) = 0 . (vi) The inequality`|z-4|lt |z-2|` represents the region given by `xgt3. (vii) Let `z_(1) "and" z_(2)`be two complex numbers such that `|z_(1)+z_(2)|= |z_(1)+z_(2)|`,then arg`(z_(1)-z_(2))=0`. 2 is not a complex number.

Explanation

(i) False
we can compare two complex numbers when they are purely real. Otherwise comperison of complex number is not possible .
(ii) False ` (x,y){:[(0),(-1)]:}={:[(0),(-1):}],` which is false
(iii) Ture
Let `z =x + iy`
`|z|+|z-1|=sqrt(x^(2)+y^(2)) + sqrt((x-1)^(2)+y^(2)`
If x 0, y = 0, then the value of `|z|+|z-1| =1`
(iv) Ture
Let `z =x + iy`
`|z-1|+|z-i|`
Then, `|x-1+iy|=|x-i(1-y)|`
`x^(2)-2x+1+y^(2) =x^(2)+1+y^(2) - 2y`
`-2x+1=1-2y`
`-2x+2y=0` ,brgt `x-y=0 ...(i) `
Equation of a line through the point (1,0) and (0,1),
`y-0=(1-0)/(0-1)(x-1)`
`rArr y= -(x-1) rArr x+ y=1 ...(ii)`
which is perpendicular to the line x-y =0.
(v) False
Let `z =x + iy, zne0 "and "Re (z) =0`
i.e., x= 0
`:. z =iy`
Im`(m^(2)) = I^(2) Y^(2) = - Y^(2)` which is real.
(vi) Ture
Given inequality, `|z-4|lt|z-2|`
`Let z = x + iy`
`:. z = x + iy`
`:. |x - 4 + iy|lt|x-2-iy|`
`rArr sqrt((x-4)^(2) + y^(2))ltsqrt((x-2)^(2)+y^(2))`
`rArr (x-4)^(2) +y^(2)lt(x-2)^(2)_y^(2)`
`rArr x^(2)-8x+16 +y^(2)ltx^(2)-4x + 4+y^(2)`
`rArr -8x+16 larr4x + 4`
`rArr -8x larr4x -12`
`rArr -4x lt -12`
`rArr 4x gt 12`
`rArr xgt3`
(vii) False
Let ` z_(1)=x_(1)+iy_(1)" and " z_(2)=x_(2)+iy_(1)`
Given that, `|z_(1)+z_(2)|=|z_(1)|+|z_(2)|`
`|x_(1)+iy_(1)+x_(2)+iy2|=|x_(1)+iy_(1)|+|x_(2)+iy_(2)|`
`rArr sqrt((x_(1)+x_(2))^(2) +(y_(2)+y_(2))^(2))=sqrt((x_(1)^(2)+y_(1)^(2)))+sqrt((x_(2)^(2)+y_(2)^(2)))`
On squaring both sides, we get
`(x_(1)+x_(2))^(2) +(y_(1)+y_(2))^(2)= (x_(1)^(2)+y_(1)^(2)) +(x_(2)^(2)+y_(2)^(2))+2sqrt((x_(1)^(2)+x_(1)^(2))(x_(2)^(2)+x_(2)^(2))) `
`rArr x_(1)^(2)+x_(2)^(2) +2x_(1)x_(2)+y_(1)^(2)+y_(2)^(2)+2y_(1)y_(2) = x_(1)^(2)+y_(1)^(2) +x_(2)^(2)+y_(2)^(2)+2sqrt((x_(1)^(2)+x_(1)^(2))(x_(2)^(2)+x_(2)^(2))) `
`rArr 2x_(1)x_(2)+2y_(1)y_(2) =2sqrt((x_(1)^(2)+x_(1)^(2))(x_(2)^(2)+x_(2)^(2))) `
`rArr x_(1)x_(2)+y_(1)y_(2) =2sqrt((x_(1)^(2)+x_(1)^(2))(x_(2)^(2)+x_(2)^(2))) `
On squaring both sides, we get
2x_(1)^(2)x_(2)^(2)+2y_(1)^(2)y_(2)^(2)
`rArr (x_(1)y_(2)-x_(2)y_(1))^(2)=0`
`rArr x_(1)y_(2)=x_(2)y_(1)`
`rArr (y_(1))/(x_(1))=(y_(2))/(y_(2))`
`rArr ((y_(1))/(x_(1)))-((y_(2))/(y_(2)))=0`
`rArr arg(z_(1)- arg(z_(2) = 0`
(viii) We know that , 2 is a real number.
Since, 2 is not a complex number.