Complex Numbers

1 )

If z = x + iy, x, y ∈ R and 3x + (3x – y) i = 4 – 6 i then z =___

(4 / 3) + i 10

(4 / 3) – i 10

[(– 4) / 3] + i 10

[(– 4) / 3] – i 10

View Solution

Answer :

z = x + iy (1)

3x + (3x – y)i = 4 – 6i

i.e. 3x = 4 and 3x – y = – 6 ----- equating real and imaginary parts

∴ x = (4 / 3) and y = 3x + 6

∴ x = (4 / 3) and y = 10

∴ from (2), z = (4 / 3) + 10 i

2 )

Evaluate [i¹⁹ + (1 / i)²⁵]²

-4

View Solution

Answer :

we know i² = – 1

∴ i¹⁹ = (i²)⁹ ∙ i

i¹⁹ = (– 1)⁹ i = – i

also (1 / i) = – i

and (1 / i)²⁵ = (– i)²⁵

Hence

[i¹⁹ + (1 / i)²⁵]² = [– i + (– i)²⁵]²

= i² + 2(– i) (– i)²⁵ + [(– i)²⁵]²

= i² + 2 i²⁶ + (– i)⁵⁰

= i² + 2(i²)¹³ + (i²)²⁵

= – 1 + 2(– 1)¹³ + (– 1)²⁵

= – 1 – 2 – 1

= – 4

3 )

i¹ + i² + i³ + i⁴ +…….. i¹⁰⁰⁰ =__________

-1

-2

View Solution

Answer :

i¹ + i² + i³ + i⁴ + -------- i¹⁰⁰⁰

This is G. P with first term a = i

and common ratio = r = i

Sum of n terms in G. P is given by

S_n = [{a(1 – rⁿ)} / (1 – r)]

= [{(i) (1 – i¹⁰⁰⁰)} / (1 – i)]

= [{(i) (1 – (i²)⁵⁰⁰)} / (1 – i)]

= [{(i) (1 – ( – 1))⁵⁰⁰} / (1 – i)]

= [i / (1 – i)] [1 – 1] = 0

4 )

The expression of complex number [1 / (1 + cos θ – i sin θ)] in the form a + ib is_________

[(sin θ) / {2(1 + cos θ)}] + i(1/2)

(1/2) – i[(sin θ) / {2(1 + cos θ)}]

(1/2) + i(1/2)tan (θ/2)

(1/2) tan (θ/2) – i(1/2)

View Solution

Answer :

z = [1 / (1 + cos θ – i sin θ)]

= [{(1 + cos θ) + i sin θ} / {(1 + cos θ – i sin θ) (1 + cos θ + i sin θ)]

= [{2 cos² (θ/2) + i 2 sin (θ/2) ∙ cos (θ/2)} / {(1 + cos θ)² + sin² θ}]

= [{2 cos² (θ/2) + i ∙ 2 ∙ sin (θ/2) ∙ cos (θ/2)} / {1 + cos² θ + 2 cos θ + sin² θ}]

= [{2 cos (θ/2) {cos (θ/2) + i sin (θ/2)}} / {2(1 + cos θ)}]

= [{cos (θ/2) {cos (θ/2) + i sin (θ/2)}} / {2 cos² (θ/2)}]

= (1/2) [{{cos (θ/2)} / {cos (θ/2)}} + i {{sin (θ/2)} / {cos (θ/2)}}]

= (1/2) + (i/2) tan (θ/2)

5 )

If [{(1 + i)x – 2i} / (3 + i)] + [{(2 – 3i)y + i} / (3 – i)}] = i then (x, y) =___________

(3, 1)

(3, – 1)

(-3, 1)

(-3, -1)

View Solution

Answer :

[{(1 + i)x – 2i} / (3 + i)] + [{(2– 3i)y + i} / (3 – i)] = i

∴ (3 – i) (1 + i)x – 2i(3 – i) + (3 + i) (2 – 3i)y + (3 + i) (i) – i(3 + i) (3 – i)

∴ (3 + 3i – i – i²)x – (6i – 2i²) + (6 – 9i + 2i – 3i²)y + 3i + i² = i(9 – i²)

put i² = – 1 we get

(2i + 4)x – (6i + 2) + (9 – 7i)y + 3i – 1 = i(10)

∴ (4 + 2i)x + (9 – 7i)y – 3i – 3 = 10i

∴ (4x + 9y – 3) + i(2x – 7y – 3) = 10i

∴ equating real and imaginary part,

4x + 9y – 3 = 0 (1)

2x – 7y – 3 = 10

2x – 7y – 13 = 0 (2)

Solving (1) & (2) we get x = 3, y = – 1 hence (x, y) = (3, – 1)

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