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.
Solve for \( x \):
\( \log_{10}(x^2) = 2 \).
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?
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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]