Question

How do you write the expression in standard form $a+ib$ given ${{\left( 1-i \right)}^{5}}$?

Answer 1

Hint: The above given question is the mixture of the concept of the complex number and the binomial expansion. From the complex number we will use the concept that $i=\sqrt{-1}$ , hence ${{i}^{2}}=-1$, ${{i}^{3}}=-\sqrt{-1}=-i$, and ${{i}^{4}}=1$. Hence, after ${{4}^{th}}$ power of $i$ value of $i$ goes on repeating it as in the previous 4 sequences. And, from binomial expansion we will use the concept of expansion of ${{\left( a+b \right)}^{n}}$ to expand ${{\left( 1-i \right)}^{5}}$ to that it can be written in standard form $a+ib$. The expansion of ${{\left( a+b \right)}^{n}}$ is given as \[{}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{b}^{3}}+........+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}\].

Complete step by step answer:
We will use the concept of binomial expansion and complex numbers both to solve the above question.
We can see from the question that we have to express ${{\left( 1-i \right)}^{5}}$ in standard form $a+ib$, and this is possible only if we write the expansion of ${{\left( 1-i \right)}^{5}}$.
Since, we know that the binomial expansion of ${{\left( a+b \right)}^{n}}$is given as:
${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{b}^{3}}+........+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$
So, we can similarly expand ${{\left( 1-i \right)}^{5}}$, and we can see here a = 1, b = $-i$ and n = 5.
Thus, we can write ${{\left( 1-i \right)}^{5}}$ as:${{\left( 1-i \right)}^{5}}={}^{5}{{C}_{0}}\times {{1}^{5}}\times {{\left( -i \right)}^{0}}+{}^{5}{{C}_{1}}\times {{1}^{4}}\times \left( -i \right)+{}^{5}{{C}_{2}}\times {{1}^{3}}\times {{\left( -i \right)}^{2}}+{}^{5}{{C}_{3}}\times {{1}^{2}}\times {{\left( -i \right)}^{3}}+{}^{5}{{C}_{4}}\times 1\times {{\left( -i \right)}^{4}}+{}^{5}{{C}_{5}}\times {{1}^{0}}\times {{\left( -i \right)}^{5}}$
$\Rightarrow {{\left( 1-i \right)}^{5}}={}^{5}{{C}_{0}}-{}^{5}{{C}_{1}}\times i+{}^{5}{{C}_{2}}\times {{\left( i \right)}^{2}}-{}^{5}{{C}_{3}}\times {{\left( i \right)}^{3}}+{}^{5}{{C}_{4}}\times {{\left( i \right)}^{4}}-{}^{5}{{C}_{5}}\times {{\left( i \right)}^{5}}$
Now, from complex number we know that $i=\sqrt{-1}$, so we can write ${{i}^{2}}=-1$, ${{i}^{3}}=-\sqrt{-1}=-i$, ${{i}^{4}}=1$, also ${{i}^{5}}=1\times i=i$.
So, after substituting value of $i,{{i}^{2}},{{i}^{3}},{{i}^{4}},{{i}^{5}}$ in the above equation we will get:
$\Rightarrow {{\left( 1-i \right)}^{5}}={}^{5}{{C}_{0}}-{}^{5}{{C}_{1}}\times i+{}^{5}{{C}_{2}}\times \left( -1 \right)-{}^{5}{{C}_{3}}\times \left( -i \right)+{}^{5}{{C}_{4}}\times 1-{}^{5}{{C}_{5}}\times i$
We also know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\times \left( n-r \right)!}$.
So, ${}^{5}{{C}_{0}}=\dfrac{5!}{0!\times \left( 5-0 \right)!}=1$, ${}^{5}{{C}_{1}}=\dfrac{5!}{1!\times \left( 5-1 \right)!}=\dfrac{5!}{4!}=5$, ${}^{5}{{C}_{2}}=\dfrac{5!}{2!\times \left( 5-2 \right)!}=\dfrac{5!}{2!\times 3!}=\dfrac{5\times 4}{2}=10$,
 Similarly, ${}^{5}{{C}_{3}}=\dfrac{5!}{3!\times \left( 5-3 \right)!}=10$, ${}^{5}{{C}_{4}}=5$, and ${}^{5}{{C}_{5}}=1$.
So, after putting all of these we will get:
$\Rightarrow {{\left( 1-i \right)}^{5}}=1-5\times i+10\times \left( -1 \right)-10\times \left( -i \right)+5\times 1-1\times i$
$\Rightarrow {{\left( 1-i \right)}^{5}}=1-5i-10+10i+5-i$
$\Rightarrow {{\left( 1-i \right)}^{5}}=-4+4i$

Hence, ${{\left( 1-i \right)}^{5}}$ in standard form is given as $-4+4i$.

Note: Students are required to memorize the binomial expression formula. Also, note that we can write ${}^{n}{{C}_{r}}$ as ${}^{n}{{C}_{n-r}}$ because they both are equal. ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$ will always helps us to easily simply ${}^{n}{{C}_{r}}$.