Q26
jee_main_2024_31_jan_morning
Properties of Modulus and Argument
If alpha denotes the number of solutions of |1 - i|^x = 2^x and beta = left(frac|z|arg(z)right), where z = fracpi4 (1 + i)^4 left(frac1 - sqrtpi isqrtpi + i + fracsqrtpi - i1 + sqrtpi iright), i = sqrt-1, then the distance of the point (alpha, beta) from the line 4x - 3y = 7 is
Numerical Answer. Answer: 3 to 3
Solution
### Core Logic
|1 - i|^x = 2^x implies (sqrt2)^x = 2^x implies 2^x/2 = 2^x
This implies fracx2 = x implies x = 0.
There is exactly 1 solution, so alpha = 1.
### Step 1: Simplify complex number z
(1+i)^4 = ((1+i)^2)^2 = (1 + i^2 + 2i)^2 = (2i)^2 = -4
Thus, z = -pi left( frac(1-sqrtpii)(sqrtpi-i)pi + 1 + frac(sqrtpi-i)(1-sqrtpii)1 + pi right)
### Step 2: Simplify Bracket
Let's expand the terms directly:
z = fracpi4(-4) left[ fracsqrtpi - pi i - i - sqrtpipi + 1 + fracsqrtpi - i - pi i - sqrtpi1 + pi right]
= -pi left[ frac-i(pi+1)pi+1 + frac-i(pi+1)pi+1 right]
= -pi [ -i - i ] = 2pi i
### Step 3: Find beta
For z = 2pi i:
|z| = 2pi and arg(z) = fracpi2.
beta = frac|z|arg(z) = frac2pipi/2 = 4
### Step 4: Distance from Line
Distance of point (alpha, beta) = (1, 4) from the line 4x - 3y - 7 = 0:
D = frac|4(1) - 3(4) - 7|sqrt4^2 + (-3)^2 = frac|4 - 12 - 7|5 = frac|-15|5 = 3
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Complex Numbers and Quadratic Equations
Class 11 Maths: Straight Lines
