[UPLIFTJEE]

If $\alpha$ denotes the number of solutions of $|1-\mathrm{i}|^{\mathrm{x}}=2^{\mathrm{x}}$ and $\beta=\left(\frac{|\mathrm{z}|}{\arg (\mathrm{z})}\right)$, where $\mathrm{z}=\frac{\pi}{4}(1+\mathrm{i})^{4}\left[\frac{1-\sqrt{\pi \mathrm{i}}}{\sqrt{\pi}+\mathrm{i}}+\frac{\sqrt{\pi}-\mathrm{i}}{1+\sqrt{\pi \mathrm{i}}}\right], \mathrm{i}=\sqrt{-1}$ then the distance of the point $(\alpha, \beta)$ from the line $4 \mathrm{x}-3 \mathrm{y}=7$ is [JEE(Main) 01 Feb 2024 S I]

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[UPLIFTJEE]

Solution

$|1-\mathrm{i}|^{\mathrm{x}}=(\sqrt{2})^{\mathrm{x}}=2^{\mathrm{x}} \Rightarrow \mathrm{x}=0$ one solution, hence $\alpha=1$

\[ \begin{aligned} & z=\frac{\pi}{4}(1+i)^{4}\left[\frac{1-\sqrt{\pi i}}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} i}\right] \Rightarrow z=\frac{\pi}{4}(2 i)^{2}\left[\frac{(1-\sqrt{\pi} i)}{(\sqrt{\pi}+i)} \frac{(\sqrt{\pi}-i)}{(\sqrt{\pi}-i)}+\frac{(\sqrt{\pi}-i)}{(\sqrt{\pi}+i)} \frac{(1-\sqrt{\pi} i)}{(1-\sqrt{\pi} i)}\right] \\ & z=-\pi\left[\frac{\sqrt{\pi}-\pi i-i-\sqrt{\pi}}{\pi+1}+\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi}}{1+\pi}\right] \Rightarrow z=-\pi(-2 i)=2 \pi i \\ & \beta=\left(\frac{|z|}{\arg (z)}\right)=\frac{2 \pi}{\pi / 2}=4 \text { Distance of point }(\alpha, \beta) \equiv(1,4) \text { from line } 4 x-3 y-7=0=\left|\frac{4-12-7}{\sqrt{16+9}}\right|=\frac{15}{5}=3 \end{aligned} \]

[UPLIFTJEE]

Answer

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