Question

If ${\left(\frac{1+\iota }{1-\iota }\right)}^x=1$ then

• A. $x=4n,$where $n$ is any positive integer.
• B. $x=2n,$where $n$ is any positive integer.
• C. $x=4n+1$, where $n$ is any positive integer.
• D. $x=2n+1,$where $n$ is any positive integer.

Answer 1

The correct option is A

$x=4n,$where $n$ is any positive integer.

To find the value of $x$:

The given equation is ${\left(\frac{1+\iota }{1-\iota }\right)}^x=1$

$\begin{array}{ll}\Rightarrow {\left(\frac{1+\iota }{1-\iota }\right)}^x& =1\\ \Rightarrow {\left[\frac{(1+\iota )(1+\iota )}{(1-\iota )(1+\iota )}\right]}^x& =1\\ \Rightarrow {\left[\frac{(1+\iota {)}^2}{2}\right]}^x& =1\\ \Rightarrow {\left(\frac{1+2\iota -1}{2}\right)}^x& =1\\ \Rightarrow {\left(\frac{2\iota }{2}\right)}^x& =1\\ \Rightarrow {\iota }^x& =1\\ \Rightarrow {\iota }^x& ={\iota }^{4n}\\ \Rightarrow x& =4n\end{array}$

where $n$is a positive integer.

Hence option A: $x=4n,$where $n$ is any positive integer.