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4-1.Complex numbers
hard
$\mid 1$ - $\left.\mathrm{i}\right|^x=2^x$ ના ઉકેલોની સંખ્યા $\alpha$ અને $\beta=\left(\frac{|z|}{\arg (\mathrm{z})}\right)$, જ્યાં $\mathrm{z}=\frac{\pi}{4}(1+\mathrm{i})^4\left(\frac{1-\sqrt{\pi} \mathrm{i}}{\sqrt{\pi}+\mathrm{i}}+\frac{\sqrt{\pi}-\mathrm{i}}{1+\sqrt{\pi} \mathrm{i}}\right), \mathrm{i}=\sqrt{-1}$ તો $(\alpha, \beta)$ નું $4 x-3 y=7$ થી અંતર મેળવો.
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
