Match the statements of column A and Column B.

Explanation

(i) Given that, `z= - i+sqrt(-3) = r (cos theta + sintheta)`
`:' rcostheta = sqrt(3), rsintheta = 1`
`rArr r^(2)= = 1+3 = 4 rArr r = 2 [ :' r gt 0]`
`rArr tan alpha = |(rsintheta)/(rconstheta) | = (1)/(sqrt(3))`
`rArr tan alpha = (1)/(sqrt(3)) rArr alpha = (pi)/(6)`
`:' x gt 0, y gt 0`
and arg `(z) = theta = (pi)/(6)`
So tha polor form of z is `2("cos"(pi)/(6) + isin (pi)/(6))`.
(ii) Given that, ` z -1 + sqrt(-3) = -1 + isqrt(3)`
`:. tan alpha = |(sqrt(3))/(-1) | = sqrt(3)`
`rArr tan theta alpha = "tan" (pi)/(3) rArr alpha =(pi)/(3) `

`:' x lt 0, y gt 0`
`theta = pi - alpha = pi = (pi)/(3) = (2pi)/(3)`
(iii) Given that, `| z+ 2| = | z -2|`
`rArr |x + 2 + iy| = |x -2 + iy|`
`rArr (x + 2)^(2) + y^(2) = (x- 2)^(2) y^(2)`
`rArr x^(2) + 4x + 4 = x^(2) - 4x + 4 rArr 8x = 0 `
`:. x = 0 `
"It is a traight line which is a perpendicular biscector of segment joining the point" `(-2, 0) "and" (2,0)`.
(iv) Given that, `|z + 2i | = | z - 2i |` ltbtgt `rArr | x + i (y+2)| = | x + i(y - 2)|`
`rArr x^(2) + (y = 2) ^(2) = x^(2) + (y - 2)^(2)`
`rArr 4y = 0 rArr y = 0 `
"It is a straight line, which is a perpendicular bisector of segment joining" `(0, - 2) "and" (0, 2)`
(v) Given that, `| z + 4i|ge 3 = |x + iy + 4i | ge 3`
`rArr = |x + i (y + 4 )|ge3 `
`rArr =sqrt(x^(2)+ (y+ 4) ^(2)) ge 3`
`rArr x^(2) + (y+4)^(2) ge 9`
`rArr = x^(2) + y^(2) + 8y + 16 ge 9`
` = x^(2) + y^(2) + 8y + 7 ge 0`
which represent a circle. On or outside having centre (0. -4) and radius 3.
(vi) Given that, `| z+ 3 |le 3`
`rArr | x+ iy + 4| le 3`
`rArr |x + 4 + iy | le3`
`rAtt sqrt ((x + 4 )^(2) + y^(2)) le 3`
`rArr (x + 4) ^(2) + y^(2) le 9`
`rArr x^(2) + 8x + 16 + y^(2) le 9`
`rArr x^(2) + 8x + y^(2) + 7 le 0`
It represent the region which is on or inside the circle having the centre (-4, 0) and radius 3 .
(vii) Given that," " ` z =(1 + 2i)/(1 -i) = ((1 + 2i)(1 + i))/((1 -i)(1 + i))`
= `(1 + 2i + i 2i)^(2)/(1 - i^(2)) = (1 -2 - 3i)/(1+1) = (-1 + 3i)/(2)`
` :." " barz = (-1)/(2) - (3i)/(2)`
Hence, `((-1)/(2), (-3)/(2))` "lies in third quadrant".
(viii) Given that, z = 1-i
Hence, " " `((1)/(2),(1)/(2))`
"Hence, the correct matches are`(a) rarr(v), (b)rarr(iii), (c)rarr(i), (d)rarr(iv), (e)rarr(ii), (f)rarr (vi), (g)rarr(viii), (h)rarr(vii)`