JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(\left(\frac{1+i}{1-i}\right)^{\frac{m}{2}}=\left(\frac{1+i}{i-1}\right)^{\frac{n}{3}}=1,(m, n \in N)\) then the greatest common divisor of the least values of \(m\) and \(n\) is
- A \(4\)
- B \(8\)
- C \(12\)
- D \(2\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(\left(\frac{1+i}{1-i}\right)^{m / 2}=\left(\frac{1+i}{i-1}\right)^{n / 3}=1\) \(\Rightarrow\left(\frac{(1+i)^{2}}{2}\right)^{m / 2}=\left(\frac{(1+i)^{2}}{-2}\right)^{n / 3}=1\) \(\Rightarrow \quad( i )^{ m / 2}=(- i )^{ n / 3}=1\) \(\Rightarrow \frac{ m }{2}=4 k _{1}\) and…
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