If denotes the number of solutions of | 1 - i | x = 2 x and = ( | z | arg ( z ) ) , where , then the distance of the point from the line 4 x - 3 y = 7 is __________ [2024]

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Solution

Q.
If $α$ denotes the number of solutions of $| 1 - i |^{x}$ $= 2^{x}$ and $β = (\frac{| z |}{\arg (z)}),$ where $z = \frac{π}{4} {(1 + i)}^{4}$ $[\frac{1 - \sqrt{π} i}{\sqrt{π} + i} + \frac{\sqrt{π} - i}{1 + \sqrt{π} i}],$ $i = \sqrt{- 1},$ then the distance of the point $(α, β)$ from the line $4 x - 3 y = 7$ is __________ [2024]

Ans.
(3)

We have, $| 1 - i |^{x}$ $= 2^{x}$

$= ({\sqrt{{(1)}^{2} + {(- 1)}^{2}})}^{x} = 2^{x} \Rightarrow {(\sqrt{2})}^{x} = 2^{x}$

$2^{x / 2} = 2^{x} \Rightarrow \frac{x}{2} = x \Rightarrow 2 x - x = 0 \Rightarrow x = 0$

$\Rightarrow α = 1$

and $β = (\frac{| z |}{a r g (z)}),$ where $z = \frac{π}{4} {(1 + i)}^{4}$ $[\frac{1 - \sqrt{π} i}{\sqrt{π} + i} + \frac{\sqrt{π} - i}{1 + \sqrt{π} i}]$

$z = 2 π i \Rightarrow \arg (z) = \frac{π}{2}$

and $| z | = 2 π$

$β = \frac{| z |}{\arg (z)} = \frac{2 π}{π / 2} = 4 \Rightarrow β = 4$

Now, the distance of point $(α, β)$ from the line $4 x - 3 y = 7$ is given by

$d = \frac{| 4 \times 1 - 3 \times 4 - 7 |}{\sqrt{16 + 9}} = \frac{| 4 - 12 - 7 |}{5} = \frac{15}{5} = 3$

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