Question

Match the statements of Column A and Column B.

Column A	Column B
(a) The polar form of `i + sqrt(3)` is	(i) Perpendicular bisector of segment joining (–2, 0) and (2, 0).
(b) The amplitude of `-1 + sqrt(-3)` is	(ii) On or outside the circle having centre at (0, –4) and radius 3.
(c) If |z + 2| = |z − 2|, then locus of z is	(iii) `(2pi)/3`
(d) If |z + 2i| = |z − 2i|, then locus of z is	(iv) Perpendicular bisector of segment joining (0, –2) and (0, 2).
(e) Region represented by |z + 4i| ≥ 3 is	(v) `2(cos  pi/6 + i sin  pi/6)`
(f) Region represented by |z + 4| ≤ 3 is	(vi) On or inside the circle having centre (–4, 0) and radius 3 units.
(g) Conjugate of `(1 + 2i)/(1 - i)` lies in	(vii) First quadrant
(h) Reciprocal of 1 – i lies in	(viii) Third quadrant

Answer 1

Column A	Answer
(a) The polar form of `i + sqrt(3)` is	(v) `2(cos  pi/6 + i sin  pi/6)`
(b) The amplitude of `-1 + sqrt(-3)` is	(iii)  `(2pi)/3`
(c) If |z + 2| = |z − 2|, then locus of z is	(i) Perpendicular bisector of segment joining (–2, 0) and (2, 0)
(d) If |z + 2i| = |z − 2i|, then locus of z is	(iv) Perpendicular bisector of segment joining (0, –2) and (0, 2).
(e) Region represented by |z + 4i| ≥ 3 is	(ii) On or outside the circle having centre at (0, –4) and radius 3.
(f) Region represented by |z + 4i| ≤ 3 is	(vi) On or inside the circle having centre (–4, 0) and radius 3 units.
(g) Conjugate of `(1 + 2i)/(1 - i)` lies in	(viii) Third quadrant
(h) Reciprocal of 1 – i lies in	(vii) First quadrant

Explanation:

(a) Given that z = `i + sqrt(3)`

Polar form of z = `r[cos theta + i sin theta]`

⇒ `sqrt(3) + i = r cos theta + ri sin theta`

⇒ r = `sqrt((sqrt(3))^2 + (1)^2)` = 2

And `tan alpha = 1/sqrt(3)`

⇒ `alpha = pi/6`

Since x > 0, y > 0

∴ Polar form of z = `2[cos  pi/6 + "i" sin  pi/6]`

(b) Given that z = `-1 + sqrt(-3)`

= `1 + sqrt(3)"i"`

Here argument (z) = `tan^-1 |sqrt(3)//4|`

= `tan^-1 |sqrt(3)|`

= `pi/3`

So, `alpha = pi/3`

Since x < 0 and y > 0

Then θ = π – α

= `pi - pi/3`

= `(2pi)/3`

(c) Given that: |z + 2| = |z − 2|

Let z = x + yi

∴ |x + yi + 2| = |x + yi − 2|

⇒ |(x + 2) + yi| = |(x − 2) + yi|

⇒ `sqrt((x + 2)^2 + y^2) = sqrt((x - 2)^2 + y^2)`

⇒ (x + 2)² + y² = (x – 2)² + y²

⇒ (x + 2)² = (x – 2)²

⇒ x² + 4 + 4x = x² + 4 – 4x

⇒ 8x = 0

⇒ x = 0

Which represent equation of y-axis and it is perpendicular to the line joining the points (–2, 0) and (2, 0).

(d) |z + 2i| = |z − 2i|

Let z = x + yi

∴ |x + yi + 2i| = |x + yi – 2i|

⇒ |x + (y + 2)i| = |x + (y – 2)i|

⇒ `sqrt(x^2 + (y + 2)^2) = sqrt(x^2 + (y -2)^2)`

⇒ x² + (y + 2)² = x² + (y – 2)² 

⇒ (y + 2)² = (y – 2)²

⇒ y² + 4 + 4y = y² + 4 – 4y

⇒ 8y = 0

⇒ y = 0.

Which is the equation of x-axis and it is perpendicular to the line segment joining (0, –2) and (0, 2).

(e) Given that: |z + 4i| ≥ 3

Let z = x + yi

∴ |x + yi + 4i| ≥ 3 

⇒ |x + (y + 4)i| ≥ 3 

⇒ `sqrt(x^2 + (y + 4)^2` ≥ 3

⇒ x² + (y + 4)² ≥ 9

⇒ x² + y² + 8y + 16 ≥ 9

⇒ x² + y² + 8y + 7 ≥ 0

⇒ r = `sqrt((4)^2 - 7)` = 3

Which represents a circle on or outside having centre (0, –4) and radius 3.

(f) |z + 4| ≤ 3

Let z = x + yi

Then |x + yi + 4| ≤ 3

⇒ |(x + 4) + yi| ≤ 3

⇒ `sqrt((x + 4)^2 + y^2)` ≤ 3

⇒ x² + 8x + 16 + y² ≤ 9

⇒ x² + y² + 8x + 7 ≤ 0

Which is a circle having centre (–4, 0) and r = `sqrt((4)^2 - 7) = sqrt(9)` = 3 and is on or inside the circle.

(g) Let z = `(1 + 2i)/(1 - i)`

= `(1 + 2i)/(1 - i) xx (1 + i)/(1 + i)`

= `(1 + i + 2i + 2i^2)/(1 - i^2)`

= `(1 + i + 2i - 2)/(1 + 1)`

= `(-1 + 3i)/2`

= `- 1/2 + 3/2 i`

∴ `barz = - 1/2 - 3/2 i` which lies in third quadrant.
(h) Given that: z = 1 – i

Reciprocal of z = `1/z`

= `1/(1 - i) xx (1 + i)/(1 + i)`

= `(1 + i)/(1 - i^2)`

= `(1 + i)/2`

= `1/2 + 1/2 i`

Which lies in first quadrant.