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.
Let \( K \) be an algebraically closed field containing a finite field \( F \). Let \( L \) be the subfield of \( K \) consisting of elements of \( K \) that are algebraic over \( F \).
Consider the following statements:
S1: \( L \) is algebraically closed.
S2: \( L \) is infinite.
Then, which one of the following is correct?
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).

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
It is similar to the addition of complex numbers, such that, z1 - z2 = z1 + ( -z2)
For example: (5+3i) - (2+1i) = (5-2) + (-2-1i) = 3 - 3i
Considering the same value of z1 and z2 , the product of the complex numbers are
z1 * 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.
Associative law: Considering three complex numbers, (z1 z2) z3 = z1 (z2 z3)
Read More: Complex Numbers and Quadratic Equations
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]