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JEE Mains · Maths · STD 11 - 4.1 complex nubers

\(|1\) - \(\left.\mathrm{i}\right|^x=2^x\) 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.

  1. A \(2\)
  2. B \(3\)
  3. C \(4\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

\((\sqrt{2})^x=2^x \Rightarrow x=0 \Rightarrow \alpha=1\) \(z=\frac{\pi}{4}(1+i)^4\left[\frac{\sqrt{\pi}-\pi i-i-\sqrt{\pi}}{\pi+1}+\frac{\sqrt{\pi}-i-\pi i-\sqrt{\pi}}{1+\pi}\right]\) \(=-\frac{\pi i}{2}\left(1+4 i+6 i^2+4 i^3+1\right)\) \(=2 \pi i\)…
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