Q. 275.0( 2 Votes )

# Match the statements of Column A and Column B.

Answer :

(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} + y^{2} = (x – 2)^{2} + y^{2}

⇒ x^{2} + 4x + 4 = x^{2} – 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

⇒ x^{2} + y^{2} + 8y + 16 ≤ 9

⇒ x^{2} + y^{2} + 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} + y^{2} ≤ 9

⇒ x^{2} + 8x + 16 + y^{2} ≤ 9

⇒ x^{2} + 8x + y^{2} + 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

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.

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