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

The number (1−i)31−i3\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}} is equal to:

Explanation

Solution

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

Complete step by step solution:
Now as given in the question, the term (1−i)31−i3\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}} first need to simplify it in terms of ii and constant numbers. Expanding the numerator and denominator in terms of simpler complex numbers, we get the numerator as:
⇒(1−i)3=1−i−3i+3i2\Rightarrow {{\left( 1-i \right)}^{3}}=1-i-3i+3{{i}^{2}}
And the denominator is written as:
⇒1−i3=1+i\Rightarrow 1-{{i}^{3}}=1+i
Now placing the expanding part of the numerator and the denominator we get the term as:
⇒1−(−i)−3i+3i21+i\Rightarrow \dfrac{1-\left( -i \right)-3i+3{{i}^{2}}}{1+i}
⇒−2−2i1+i\Rightarrow \dfrac{-2-2i}{1+i}
⇒−2(1+i)1+i\Rightarrow \dfrac{-2\left( 1+i \right)}{1+i}
⇒−2\Rightarrow -2
Therefore, the value of the term (1−i)31−i3\dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}} is −2-2.

Note: Another method to solve the question is by:
⇒(1−i)31−i3=(1−i)(1−i)(1−i)1−1×−i\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=\dfrac{\left( 1-i \right)\left( 1-i \right)\left( 1-i \right)}{1-1\times -i}
⇒(1−i)31−i3=(1−i)(1−i)(1−i)1+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}
⇒(1−i)31−i3=−2(1+i)1+i\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=\dfrac{-2\left( 1+i \right)}{1+i}
⇒(1−i)31−i3=−2\Rightarrow \dfrac{{{\left( 1-i \right)}^{3}}}{1-{{i}^{3}}}=-2