If α denotes the number of solutions of 1 − i x = 2 x and β = z arg z , where z = π 4 1 + i 4 1 − π · i π +…

Given,

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

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

$\Rightarrow x=0$

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

Now, solving

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

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

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

$\Rightarrow z=\frac{\pi i}{2}{\left(1+i\right)}^4$

$\Rightarrow z=-\frac{\pi i}{2}\left(1+4i+6i^2+4i^3+1\right)$

$\Rightarrow z=2\pi i$

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

$\Rightarrow \beta =\frac{\left|z\right|}{{\displaystyle arg\left(z\right)}}=\frac{2\pi }{{\displaystyle \frac{\pi }{2}}}=4$

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