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JEE Mains · Maths · STD 11 - 4.1 complex nubers

\(n\) ની ન્યૂનતમ કિમત મેળવો કે જેથી \(\left( \frac{1 + i\sqrt 3 }{1 - i\sqrt 3 }\right)^n = 1,\) થાય.

  1. A \(2\)
  2. B \(6\)
  3. C \(5\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3\)

Step-by-step Solution

Detailed explanation

Let \(l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right)\) \(\therefore l=\left(\frac{1+i \sqrt{3}}{1-i \sqrt{3}}\right) \times\left(\frac{1+i \sqrt{3}}{1+i \sqrt{3}}\right)\) \(=\left(\frac{-2+i 2 \sqrt{3}}{4}\right)=\left(\frac{1-i \sqrt{3}}{-2}\right)\) Also,…
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