Question:

Let $a, b$ be two real numbers such that $a b<$ If the complex number $\frac{1+ai}{b+i}$ is of unit modulus and $a$ $+i b$ lies on the circle $|z-I|=|2 z|$, then a possible value of $\frac{1+[a ]}{4 b}$, where $[t]$ is greatest integer function, is :

Updated On: Apr 28, 2025
  • $\frac{1}{2}$
  • $-\frac{1}{2}$
  • $1$
  • $-1$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Approach Solution - 1













or
Similarly it is not matching with
Was this answer helpful?
0
0
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

Provided conditions are:
1. ab < 0,
2. 1 + ai / b + i = 1,
3. a + ib lies on the circle |z − 1| = |2z|.

Step 1: Simplify the Unit Modulus Condition

For 1 + ai / b + i = 1, we have:

|1 + ai| = |b + i|.

Squaring both sides:

a² + 1 = b² + 1.

a² = b².

a = ±b.

Given ab < 0, we deduce: b = −a.

Step 2: Simplify the Circle Condition

The point a + ib lies on the circle |z − 1| = |2z|. Substituting z = a + ib, we have:

|a + ib − 1| = |2(a + ib)|.

Simplify each term:

|a + ib − 1| = |(a − 1) + ib| = √[(a − 1)² + b²].

|2(a + ib)| = 2|a + ib| = 2√(a² + b²).

Equating the two:

√[(a − 1)² + b²] = 2√(a² + b²).

Squaring both sides:

(a − 1)² + b² = 4(a² + b²).

Substitute b = −a:

(a − 1)² + (−a)² = 4(a² + (−a)²).

(a − 1)² + a² = 8a².

a² − 2a + 1 + a² = 8a².

2a² − 2a + 1 = 8a².

6a² + 2a − 1 = 0. (1)

Step 3: Solve for a and b

Solve the quadratic equation 6a² + 2a − 1 = 0 using the quadratic formula:

a = −2 ± √(2² − 4(6)(−1)) / 2(6).

a = −2 ± √(4 + 24) / 12.

a = −2 ± √28 / 12.

a = −2 ± 2√7 / 12.

a = −1 ± √7 / 6.

Since b = −a, we have:

b = 1 ∓ √7 / 6.

Step 4: Calculate 1 + ⌊a⌋ 4b

Substitute a = −1 + √7 / 6:

⌊a⌋ = 0 (since −1 < a < 0).

1 + ⌊a⌋ 4b = 1 / 4b.

Substitute b = 1 − √7 / 6:

1 / 4b = 1 / 4 · 1 − √7 / 6 = 6 / 4(1 − √7).

Rationalize the denominator:

6 / 4(1 − √7) · (1 + √7) / (1 + √7) = 6(1 + √7) / 4(1 − 7) = 6(1 + √7) / −24.

1 / 4b = −1 + √7 / 4.

For a = −1 − √7 / 6, a similar calculation shows that no option matches the result.

Conclusive Answer: No option matches the calculated values. This question was marked as dropped by NTA.

Was this answer helpful?
0
0

Concepts Used:

Algebra of Complex Numbers

Algebra of complex numbers

1. Addition of two complex numbers:

Consider z1 and z2 are two complex numbers. 

For example, z1 = 3+4i and z2 = 4+3i

Here a=3, b=4, c=4, d=3

∴z1+ z2 = (a+c)+(b+d)i 

⇒z1 + z2 = (3+4)+(4+3)i 

⇒z1 + z2 = 7+7i 

Properties of addition of complex numbers

  • Closure law: While adding two complex numbers the resulting number is also a complex number.
  • Commutative law: For the complex numbers z1 and z2 , the commutation can be z1+ z2 = z2+z1
  • Associative law: While considering three complex numbers, (z1+ z2) + z?3 = z1 + (z2 + z3)
  • Additive identity: An Additive identity is nothing but zero complex numbers that go as 0+i0. For every complex number z, z+0 = z.
  • Additive inverse: Every complex number has an additive inverse denoted as -z.

2. Difference between two complex numbers

It is similar to the addition of complex numbers, such that, z1 - z= z1 + ( -z2

For example: (5+3i) - (2+1i) = (5-2) + (-2-1i) = 3 - 3i

3. Multiplication of complex numbers

Considering the same value of z1 and z2 , the product of the complex numbers are

z* z2 = (ac-bd) + (ad+bc) i

For example: (5+6i) (2+3i) = (5×2) + (6×3)i = 10+18i

 

 

Properties of Multiplication of complex numbers

Note: The properties of multiplication of complex numbers are similar to the properties we discussed in addition to complex numbers.

  • Closure law: When two complex numbers are multiplied the result is also a complex number.
  • Commutative law: z1* z2 = z2 * z1

Associative law: Considering three complex numbers, (z1 z2) z3 = z1 (z2 z3)

  • Multiplicative identity: 1+0i is always denoted as 1. This is multiplicative identity. This means that z.1 = z for every complex number z.
  • Distributive law: Considering three complex numbers, z1 (z2 + z3) =z1 z2 + z1 z3 and (z1+ z2) z3 = z1 z2 + z2 z3.

Read More: Complex Numbers and Quadratic Equations

4. Division of complex numbers

If z1 / z2 of a complex number is asked, simplify it as z1 (1/z2 )

For example: z1 = 4+2i and z2 = 2 - i

z1 / z2 =(4+2i)×1/(2 - i) = (4+i2)(2/(2²+(-1)² ) + i (-1)/(2²+(-1)² )) 

=(4+i2) ((2+i)/5) = 1/5 [8+4i + 2(-1)+1] = 1/5 [8-2+1+41] = 1/5 [7+4i]