Question

Express ${{\left( 1-2i \right)}^{-3}}$ in the form of $a+ib$.

Answer 1

Hint : We first find the simplification of the given polynomial ${{\left( 1-2i \right)}^{3}}$ according to the identity ${{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}$. We need to simplify the cubic polynomial of difference of two terms. We replace it with $x=1;y=i$. We also use ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$. Then we take the inverse and the rationalisation to find the form of $a+ib$.

Complete step-by-step answer :
We need to find the simplified form of ${{\left( 1-2i \right)}^{-3}}$. We know ${{\left( 1-2i \right)}^{-3}}=\dfrac{1}{{{\left( 1-2i \right)}^{3}}}$.
We are going to use the identity ${{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}$.
We express ${{\left( 1-2i \right)}^{3}}$ as the cube of difference of two numbers. We take $x=1;y=2i$ for the identity of ${{\left( x-y \right)}^{3}}={{x}^{3}}-3{{x}^{2}}y+3x{{y}^{2}}-{{y}^{3}}$.
${{\left( 1-2i \right)}^{3}}={{1}^{3}}-3\times {{1}^{2}}\times 2i+3\times 1\times {{\left( 2i \right)}^{2}}-{{\left( 2i \right)}^{3}}$
We have the relations for imaginary $i$ where ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$.
Therefore, the simplified form of ${{\left( 1-2i \right)}^{3}}$ is

& {{\left( 1-2i \right)}^{3}} \\\ & =1-6i+12{{i}^{2}}-8{{i}^{3}} \\\ & =1-6i-12+8i \\\ & =-11+2i \\\ \end{aligned}$$ So, ${{\left( 1-2i \right)}^{-3}}=\dfrac{1}{{{\left( 1-2i \right)}^{3}}}=\dfrac{1}{-11+2i}$. We take the rationalisation. We multiply both numerator and denominator with $-11-2i$. $\begin{aligned} & \dfrac{1}{-11+2i} \\\ & =\dfrac{\left( -11-2i \right)}{\left( -11+2i \right)\left( -11-2i \right)} \\\ & =\dfrac{\left( -11-2i \right)}{{{\left( -11 \right)}^{2}}-{{\left( 2i \right)}^{2}}} \\\ & =\dfrac{\left( -11-2i \right)}{121+4} \\\ & =-\dfrac{11}{125}-\dfrac{2}{125}i \\\ \end{aligned}$ The $a+ib$ form is for ${{\left( 1-2i \right)}^{-3}}=-\dfrac{11}{125}-\dfrac{2}{125}i$ where $a=-\dfrac{11}{125};b=-\dfrac{2}{125}$. Therefore, expressing ${{\left( 1-2i \right)}^{3}}$ in $a+ib$ form, we get $-\dfrac{11}{125}-\dfrac{2}{125}i$. **So, the correct answer is “$-\dfrac{11}{125}-\dfrac{2}{125}i$”.** **Note** : We also can use the binomial theorem to find the general form and then put the value of 3. We have ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+....+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$. We need to find the cube of the sum of two numbers. So, we put $n=3$. ${{\left( a+b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}+{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}+{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$. In this way we also simplify the term of ${{\left( a-b \right)}^{3}}$.