If α denotes the number of solutions of |1 - i| ^ x = 2 ^ x and β...

Given,

${\left|1 - i\right|}^{x} = {2}^{x}$

$⇒ {\left(\sqrt{2}\right)}^{x} = {2}^{x}$

$⇒ x = 0$

Hence, the number of solutions will be one, so $\alpha = 1$

Now, solving

$z = \dfrac{\pi}{4}{\left(1 + i\right)}^{4}\left(\dfrac{1 - \sqrt{\pi} \cdot i}{\sqrt{\pi} + i} + \dfrac{\sqrt{\pi} - i}{1 + \sqrt{\pi} \cdot i}\right)$

$⇒ z = \dfrac{\pi}{4}{\left(1 + i\right)}^{4}\left[\dfrac{\sqrt{\pi} - \pii - i - \sqrt{\pi}}{\pi + 1} + \dfrac{\sqrt{\pi} - i - \pii - \sqrt{\pi}}{1 + \pi}\right]$

$⇒ z = \dfrac{\pi}{4}{\left(1 + i\right)}^{4}\left[\dfrac{- 2\pii - 2i}{\pi + 1}\right]$

$⇒ z = \dfrac{\pii}{2}{\left(1 + i\right)}^{4}$

$⇒ z = - \dfrac{\pii}{2}\left(1 + 4i + 6{i}^{2} + 4{i}^{3} + 1\right)$

$⇒ z = 2\pii$

So, $arg\left(z\right) = \dfrac{\pi}{2}$

$⇒ \beta = \dfrac{\left|z\right|}{arg\left(z\right)} = \dfrac{2\pi}{\dfrac{\pi}{2}} = 4$

Hence, the distance from \left(\alpha, \beta\right) ≡ \left(1, 4\right)$to$4x - 3y = 7$will be \left|\dfrac{4 - 3 \times 4 - 7}{\sqrt{{4}^{2} + {3}^{2}}}\right| = \dfrac{15}{5} = 3$