# Match the statements of Column A and Column B. (a) Given z = i + √3

So, |z| = |i + √3| = 2

Also, z lies in the first quadrant. = π/6

The polar form of z is (b) Given z = -1 + √-3

= -1 + i √3

Here z lies in the second quadrant.

arg (z) = amp (z) = π – tan-1 √3

= π - π/3

= 2π/3

(c) Given |z + 2| = |z – 2|

|x + 2 + iy| = |x – 2 + iy|

(x + 2)2 + y2 = (x – 2)2 + y2

x2 + 4x + 4 = x2 – 4x + 4

8x = 0

x = 0

It is a straight line which is a perpendicular bisector of segment joining the points (-2, 0) and (2, 0).

(d) Given |z + 2i| = |z – 2i|

|x + i(y + 2)| = |x + i(y - 2)|

(x)2 + (y + 2)2 = (x)2 + (y – 2)2

4y = 0

y = 0

It is a straight line which is a perpendicular bisector of segment joining the points (0, -2) and (0, 2).

(e) Given |z + 4i| ≥ 3

|x + iy + 4i| ≤ 3

|x + i(y + 4)|≤ 3 (x)2 + (y + 4)2 ≤ 9

x2 + y2 + 8y + 16 ≤ 9

x2 + y2 + 8y + 7 ≤ 0

This represents the region on or outside circle having centre (0, -4) and radius 3.

(f) Given |z + 4| ≤ 3

|x + iy + 4| ≤ 3

|x + 4 + iy|≤ 3 (x + 4)2 + y2 ≤ 9

x2 + 8x + 16 + y2 ≤ 9

x2 + 8x + y2 + 7 ≤ 0

This represents the region on or inside circle having centre (-4, 0) and radius 3.

(g) Given     Hence z̅ lies in the third quadrant.

(h) Given z̅ = 1 - i

,outr inside circle having centre (-4, 0) and radius 3.nts (-2, x numbers is not possible or has no = 1/2 (1 + i)

Reciprocal of z lies in first quadrant.

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