If $\alpha$ denotes the number of solutions of $|1-i|^{x}=2^{x}$ and $\beta=\left(\frac{|z|}{\arg(z)}\right)$, where $z=\frac{\pi}{4}(1+i)^{4}\left(\frac{1-\sqrt{\pi}i}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi}i}\right), i=\sqrt{-1}$, then the distance of the point $(\alpha, \beta)$ from the line $4x-3y=7$ is ____

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Complex Numbers

α

denotes the number of solutions of

∣1 - i ∣^{x} = 2^{x}

and

β = (\frac{∣ z ∣}{a r g ( z )})

, where

z = \frac{π}{4} (1 + i)^{4} (\frac{1 - π i}{π + i} + \frac{π - i}{1 + π i}), i = - 1

, then the distance of the point

(α, β)

from the line

4 x - 3 y = 7

is ____

Answer: 3

∣1 - i ∣ = 2

, so

∣1 - i ∣^{x} = 2^{x}

becomes

(2)^{x} = 2^{x}

, i.e.

2^{x /2} = 2^{x}

. Equating exponents:

x /2 = x ⟹ x = 0

. There is exactly one solution, so

α = 1

.

For

z

(1 + i)^{4} = ((1 + i)^{2})^{2} = (2 i)^{2} = - 4

, so

\frac{π}{4} (1 + i)^{4} = - π

.

For the bracketed sum, with

a = π

\frac{1 - ai}{a + i} = \frac{( 1 - ai ) ( a - i )}{a ^{2} + 1} = \frac{a - i - a ^{2} i - a}{a ^{2} + 1} = - i, \frac{a - i}{1 + ai} = \frac{( a - i ) ( 1 - ai )}{1 + a ^{2}} = - i

So the bracket sums to

- 2 i

, giving

z = (- π) (- 2 i) = 2 π i

∣ z ∣ = 2 π

ar g (z) = \frac{π}{2}

, so

β = \frac{2 π}{π /2} = 4

.

Distance of

(α, β) = (1, 4)

from

4 x - 3 y = 7

\frac{∣4 ( 1 ) - 3 ( 4 ) - 7∣}{16 + 9} = \frac{∣ - 15∣}{5} = 3

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer