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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 |
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Solution
| 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.
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