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4-1.Complex numbers
hard
If $\alpha$ denotes the number of solutions of $|1-i|^x=2^x$ and $\beta=\left(\frac{|z|}{\arg (z)}\right)$, where $z=\frac{\pi}{4}(1+i)^4\left(\frac{1-\sqrt{\pi i}}{\sqrt{\pi}+i}+\frac{\sqrt{\pi}-i}{1+\sqrt{\pi} \mathrm{i}}\right), i=\sqrt{-1}$, then the distance of the point $(\alpha, \beta)$ from the line $4 x-3 y=7$ is
A
$2$
B
$3$
C
$4$
D
$5$
(JEE MAIN-2024)
Solution
$(\sqrt{2})^x=2^x \Rightarrow x=0 \Rightarrow \alpha=1$
$z=\frac{\pi}{4}(1+i)^4\left[\frac{\sqrt{\pi}-\pi i-i-\sqrt{\pi}}{\pi+1}+\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi}}{1+\pi}\right]$
$=-\frac{\pi i}{2}\left(1+4 i+6 i^2+4 i^3+1\right)$
$=2 \pi i$
$\beta=\frac{2 \pi}{\frac{\pi}{2}}=4$
Distance from $(1,4)$ to $4 x-3 y=7$
Will be $\frac{15}{5}=3$
Standard 11
Mathematics
