Question

If ${\left(1-i\right)}^n=2^n$, then $n$ is equal to

• A. $1$
• B. $0$
• C. $-1$
• D. None of these

Answer 1

The correct option is B

$0$

The explanation for the correct option:

Find the value of $n$:

Given, ${\left(1-i\right)}^n=2^n$

$\Rightarrow$ ${\left(1-i\right)}^n=2^n+0i$

We know that if two complex numbers are equal, their moduli must also be equal,

$\Rightarrow$ $\begin{array}{rcl}\left|{\left(1-i\right)}^n\right|& =& \left|2^n\right|\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\left|\left(1-i\right)\right|}^n& =& \left|2^n\right|\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\left(\sqrt{{\left(1{)}^2+(-1\right)}^2}\right)}^n& =& 2^n\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\sqrt{2^n}& =& 2^n\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{2}^{\frac{n}{2}}& =& 2^n\end{array}$

$\because 2^{\frac{n}{2}}=2^n$

$\Rightarrow$ $\begin{array}{rcl}\frac{n}{2}& =& n\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{n}{2}& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\mathit{n}& \mathbf{=}& \mathbf{0}\end{array}$

Hence option (B) is correct.