Question

If the equation ${{\left( 1-i \right)}^{n}}={{2}^{n}}$ holds for a particular n, then the value of n will be given by
a)1
b)0
c)-1
d)None of these

Answer 1

Hint: In this question, we have to find the value of n such that the nth power of $(1-i)$ and 2 has the same value. Now, we know that anything to the power zero should be equal to one. Therefore, we should prove that no non zero value of n satisfies the equation and this equation is valid only for n=0.

Complete step-by-step answer:
In this question we need to find the value of n such that ${{\left( 1-i \right)}^{n}}={{2}^{n}}$. Now, if $n\ne 0$, we take the nth root on both sides, we obtain
$\begin{aligned} & {{\left( 1-i \right)}^{\dfrac{n}{n}}}={{2}^{\dfrac{n}{n}}}\Rightarrow 1-i=2 \\\ & \Rightarrow i=1-2=-1........................(1.1) \\\ \end{aligned}$
However, we know that the value of i in complex numbers is equal to $\sqrt{-1}.........................(1.2)$
Therefore, we find that if $n\ne 0$, then equation (1.1) and (1.2) contradict each other and thus, the given equation in the question cannot be true.
However, for n=0, we get
$\begin{aligned} & {{\left( 1-i \right)}^{n}}={{\left( 1-i \right)}^{0}}=1={{2}^{0}} \\\ & \Rightarrow {{\left( 1-i \right)}^{n}}={{2}^{n}}\text{ (for n=0)} \\\ \end{aligned}$
As the zeroth power of any number is equal to 1. This answer matches with option (b) given in the question. Hence, option (b) is the correct answer to this question.

Note: Note that the equation derived in 1.1 is true only for $n\ne 0$, as otherwise the expression $\dfrac{n}{n}$ in the power of $(1-i)$ and 2 would be undefined and thus we cannot equate $(1-i)$ to 2 to get the contradiction.