Question

For a positive integer n, find the value of ${{\left( 1-i \right)}^{n}}{{\left( 1-\dfrac{1}{i} \right)}^{n}}$.

Answer 1

Hint: Use the fact that we can simplify the expression of the form $\dfrac{1}{c+id}$ by multiplying and dividing it by $c-id$. Simplify the given expression using algebraic identity $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$ and then further use the fact that $i=\sqrt{-1}$ to calculate higher powers of $i$.

Complete step-by-step solution -
We have to calculate the value of ${{\left( 1-i \right)}^{n}}{{\left( 1-\dfrac{1}{i} \right)}^{n}}$ for any positive integer n.
To do so, we will first simplify the expression $1-\dfrac{1}{i}$.
We know that we can simplify the expression of the form $\dfrac{1}{c+id}$ by multiplying and dividing it by $c-id$.
So, we can rewrite the expression $1-\dfrac{1}{i}$ as $1-\dfrac{1}{i}=1-\left( \dfrac{1}{i}\times \dfrac{-i}{-i} \right)$.
Thus, we have $1-\dfrac{1}{i}=1-\left( \dfrac{1}{i}\times \dfrac{-i}{-i} \right)=1-\dfrac{i}{{{i}^{2}}}$.
We know that $i=\sqrt{-1}$. Thus, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$.
So, we can rewrite the expression $1-\dfrac{1}{i}$ as $1-\dfrac{1}{i}=1-\left( \dfrac{1}{i}\times \dfrac{-i}{-i} \right)=1-\dfrac{i}{{{i}^{2}}}=1-\dfrac{i}{\left( -1 \right)}=1+i$.
Thus, we can rewrite the expression ${{\left( 1-i \right)}^{n}}{{\left( 1-\dfrac{1}{i} \right)}^{n}}$ as ${{\left( 1-i \right)}^{n}}{{\left( 1-\dfrac{1}{i} \right)}^{n}}={{\left( 1-i \right)}^{n}}{{\left( 1+i \right)}^{n}}$.
Simplifying the above expression, we have ${{\left( 1-i \right)}^{n}}{{\left( 1-\dfrac{1}{i} \right)}^{n}}={{\left( 1-i \right)}^{n}}{{\left( 1+i \right)}^{n}}={{\left[ \left( 1+i \right)\left( 1-i \right) \right]}^{n}}$.
We know the algebraic identity $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$.
Thus, we have $\left( 1+i \right)\left( 1-i \right)={{1}^{2}}-{{i}^{2}}$.
We know that $i=\sqrt{-1}$. Thus, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$.
So, we have $\left( 1+i \right)\left( 1-i \right)={{1}^{2}}-{{i}^{2}}=1-\left( -1 \right)=1+1=2$.
Thus, we can rewrite the expression ${{\left( 1-i \right)}^{n}}{{\left( 1-\dfrac{1}{i} \right)}^{n}}={{\left( 1-i \right)}^{n}}{{\left( 1+i \right)}^{n}}={{\left[ \left( 1+i \right)\left( 1-i \right) \right]}^{n}}$ as ${{\left( 1-i \right)}^{n}}{{\left( 1-\dfrac{1}{i} \right)}^{n}}={{\left[ \left( 1+i \right)\left( 1-i \right) \right]}^{n}}={{2}^{n}}$.
Hence, the value of the expression ${{\left( 1-i \right)}^{n}}{{\left( 1-\dfrac{1}{i} \right)}^{n}}$ is ${{2}^{n}}$.

Note: We can substitute any integer value of n to calculate the value of the given expression. We can’t solve this question without simplifying the expression $1-\dfrac{1}{i}$ and using the algebraic identities. We can write any complex number in the form $a+ib$, where $ib$ is the imaginary part and $a$ is the real part.