Question

The number \[\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}\] is equal to:

Answer 1

Hint:To find the value of the term we use the complex number multiplication expressed in form of the complex number as \[a+ib\] where the variable a and b are real numbers with the value of \[i\] as imaginary unit, when multiplying the value of \[i\times i\] we get the result of product as \[-1\] which can be used in the above question.

Complete step by step solution:
Now as given in the question, the term \[\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}\] first need to simplify it in terms of \[i\] and constant numbers. Expanding the numerator and denominator in terms of simpler complex numbers, we get the numerator as:
\[\Rightarrow {{\left( 1-i \right)}^{3}}=1-i-3i+3{{i}^{2}}\]
And the denominator is written as:
\[\Rightarrow 1-{{i}^{3}}=1+i\]
Now placing the expanding part of the numerator and the denominator we get the term as:
\[\Rightarrow \dfrac{1-\left( -i \right)-3i+3{{i}^{2}}}{1+i}\]
\[\Rightarrow \dfrac{-2-2i}{1+i}\]
\[\Rightarrow \dfrac{-2\left( 1+i \right)}{1+i}\]
\[\Rightarrow -2\]
Therefore, the value of the term \[\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}\] is \[-2\].

Note: Another method to solve the question is by:
\[\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=\dfrac{\left( 1-i \right)\left( 1-i \right)\left( 1-i \right)}{1-1\times -i}\]
\[\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=\dfrac{\left( 1-i \right)\left( 1-i \right)\left( 1-i \right)}{1+i}\]
\[\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=\dfrac{-2\left( 1+i \right)}{1+i}\]
\[\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=-2\]