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JEE Mains · Maths · STD 11 - 4.1 complex nubers

ન્યૂનતમ ધન પૂર્ણાંક \(\mathrm{n}\) મેળવો કે જેથી \(\frac{(2 \mathrm{i})^{\mathrm{n}}}{(1-\mathrm{i})^{\mathrm{n}-2}}, \mathrm{i}=\sqrt{-1}\) એ ધન પૃણાંક બને.

  1. A \(2\)
  2. B \(4\)
  3. C \(6\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(6\)

Step-by-step Solution

Detailed explanation

\(\frac{(2 \mathrm{i})^{\mathrm{n}}}{(1-\mathrm{i})^{\mathrm{n}-2}}=\frac{(2 \mathrm{i})^{\mathrm{n}}}{(-2 \mathrm{i})^{\frac{\mathrm{n}-2}{2}}}\)…
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