KCET · Maths · Complex Number
If \(\left(\frac{1+i}{1-i}\right)^{x}=1\), then
- A \(x=4 n+1, n \in N\)
- B \(x=2 n+1, n \in N\)
- C \(x=2 n, n \in N\)
- D \(x=4 n, n \in N\)
Answer & Solution
Correct Answer
(D) \(x=4 n, n \in N\)
Step-by-step Solution
Detailed explanation
Given, \(\left(\frac{1+i}{1-i}\right)^{x}=1\)
\(\begin{aligned}
&\Rightarrow \quad\left[\frac{(1+i)}{(1-i)} \times \frac{(1+i)}{(1+i)}\right]^{x}=1 \\
&\Rightarrow \quad\left[\frac{1+i^{2}+2 i}{1^{2}-i^{2}}\right]^{x}=1 \\
&\Rightarrow \quad\left[\frac{1-1+2 i}{1+1}\right]^{x}=1 \\
&\Rightarrow \quad i^{x}=1 \\
&\therefore \quad 4 n, n \in N .
\end{aligned}\)
\(\begin{aligned}
&\Rightarrow \quad\left[\frac{(1+i)}{(1-i)} \times \frac{(1+i)}{(1+i)}\right]^{x}=1 \\
&\Rightarrow \quad\left[\frac{1+i^{2}+2 i}{1^{2}-i^{2}}\right]^{x}=1 \\
&\Rightarrow \quad\left[\frac{1-1+2 i}{1+1}\right]^{x}=1 \\
&\Rightarrow \quad i^{x}=1 \\
&\therefore \quad 4 n, n \in N .
\end{aligned}\)
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