Match the statements of Column A and Column B.
(a) Given z = i + √3
So, |z| = |i + √3|
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= 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 ![]()
![]()
![]()
![]()
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Hence z̅ lies in the third quadrant.
(h) Given z̅ = 1 - i
lies in the third quadrant.
,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.
Couldn't generate an explanation.
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