Question

If ${{\left( \dfrac{1+i}{1-i} \right)}^{x}}=1$ , then
(a) $x=2n$ , where $n$ is any positive integer
(b) $x=4n+1$ , where $n$ is any positive integer
(c) $x=2n+1$ , where $n$ is any positive integer
(d) $x=4n$ , where $n$ is any positive integer

Answer 1

Hint: For solving this question first we will assume $z=\dfrac{1+i}{1-i}$ and then we will rationalize the fraction. After that, we will be using the concept that if ${{i}^{x}}=1$ , then $x$ should be multiple of 4 for writing the value of $x$ and then we will select the correct option.

Complete step-by-step answer:
It is given that if ${{\left( \dfrac{1+i}{1-i} \right)}^{x}}=1$ and we have to find a suitable value of $x$.
Now, let $z=\dfrac{1+i}{1-i}$ and first we will simplify it. Then,
$\begin{align}
  & z=\dfrac{1+i}{1-i} \\
 & \Rightarrow z=\dfrac{\left( 1+i \right)}{\left( 1-i \right)}\times \dfrac{\left( 1+i \right)}{\left( 1+i \right)} \\
 & \Rightarrow z=\dfrac{{{\left( 1+i \right)}^{2}}}{\left( 1-i \right)\left( 1+i \right)} \\
 & \Rightarrow z=\dfrac{1+{{i}^{2}}+2i}{\left( {{1}^{2}}-{{i}^{2}} \right)} \\
\end{align}$
Now, as we know that $i=\sqrt{-1}$ so, we can substitute the value of ${{i}^{2}}=-1$ in the above equation. Then,
$\begin{align}
  & z=\dfrac{1+{{i}^{2}}+2i}{\left( {{1}^{2}}-{{i}^{2}} \right)} \\
 & \Rightarrow z=\dfrac{1-1+2i}{\left( 1+1 \right)} \\
\end{align}$
$\begin{align}
  & \Rightarrow z=\dfrac{2i}{2} \\
 & \Rightarrow z=i \\
\end{align}$
Now, from the above result, we can write that $z=\dfrac{1+i}{1-i}=i$ . And it is given that ${{\left( \dfrac{1+i}{1-i} \right)}^{x}}=1$ . Then,
$\begin{align}
  & {{\left( \dfrac{1+i}{1-i} \right)}^{x}}=1 \\
 & \Rightarrow {{z}^{x}}=1 \\
 & \Rightarrow {{i}^{x}}=1 \\
\end{align}$
Now, we know that if ${{i}^{x}}=1$ , then $x$ should be multiple of 4. Then,
$\begin{align}
  & {{i}^{x}}=1 \\
 & \Rightarrow x=4n \\
\end{align}$
Now, from the above result, we conclude that if ${{\left( \dfrac{1+i}{1-i} \right)}^{x}}=1$ , then the value of $x$ should be multiple of 4 and we can write the value of $x=4n$ where $n$ is any positive integer.
Hence, option (d) is the correct answer.

Note:Here, the student should first try to understand what is asked in the question. Although the problem is very easy, we should solve it properly to get the correct answer. And as we should make use of every information which we know about the imaginary number iota $\left( i \right)$ to write the value of $x$ correctly. Moreover, here we should be careful while selecting the correct option and some students may get confused between option (a) and (d) but we should be able to figure out that option (a) is not correct as it has some extra values of $x$ which are multiple of 2 but not multiple of 4.