KCET · Maths · Application of Derivatives
If \( \left(\frac{1+i}{1-i}\right)^{\mathrm{m}} \), then the least positive integralvalue of \( \mathrm{m} \) is
- A \( 02 \)
- B \( 13 \)
- C \( 04 \)
- D \( 11 \)
Answer & Solution
Correct Answer
(C) \( 04 \)
Step-by-step Solution
Detailed explanation
Given that, \(\left(\frac{1+i}{1-i}\right)^{m}=1\)
\(\Rightarrow\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1\)
\(\Rightarrow\left(\frac{(1+i)^{2}}{1-(i)^{2}}\right)^{m}=1\)
\(\Rightarrow\left(\frac{1-1+2 i}{1+1}\right)^{m}=1\)
\(\Rightarrow\left(\frac{2 i}{2}\right)^{m}=1\)
\(\Rightarrow i^{m}=1\)
For \(m=1\), we have \(i=1\), but \(i \neq 1\)
For \(m=2\), we have \(i^{2}=-1=1\), but \(-1 \neq 1\)
For \(m=3\), we have \(i^{3}=-i=1\), but \(-i \neq 1\)
For \(m=4\), we have \(i^{4}=1\)
Therefore, \(m=4\)
\(\Rightarrow\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1\)
\(\Rightarrow\left(\frac{(1+i)^{2}}{1-(i)^{2}}\right)^{m}=1\)
\(\Rightarrow\left(\frac{1-1+2 i}{1+1}\right)^{m}=1\)
\(\Rightarrow\left(\frac{2 i}{2}\right)^{m}=1\)
\(\Rightarrow i^{m}=1\)
For \(m=1\), we have \(i=1\), but \(i \neq 1\)
For \(m=2\), we have \(i^{2}=-1=1\), but \(-1 \neq 1\)
For \(m=3\), we have \(i^{3}=-i=1\), but \(-i \neq 1\)
For \(m=4\), we have \(i^{4}=1\)
Therefore, \(m=4\)
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