
Evaluate the following, expressing your answer in cartesian form $\left( a+bi \right):{{\left( 1-3i \right)}^{3}}$.
Answer
613.5k+ views
Hint: We know that the cartesian form of complex numbers is nothing but the way of representing a complex number in the form of $\left( a+bi \right)$, where $a,b$ are real numbers. We also know that, we can write, ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$. So, we have to use this in the question to get the cartesian form of the complex number.
Complete step-by-step answer:
In this question, we have been asked to find the cartesian form of complex number ${{\left( 1-3i \right)}^{3}}$, that is we have to express ${{\left( 1-3i \right)}^{3}}$ in the form of $\left( a+bi \right)$. Now, we know that ${{\left( a-b \right)}^{3}}$ can be written as, ${{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$, or as ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$. Now we will use this formula to simplify ${{\left( 1-3i \right)}^{3}}$. By simplifying, we get,
$\begin{align}
& {{\left( 1-3i \right)}^{3}}={{\left( 1 \right)}^{3}}-{{\left( 3i \right)}^{3}}-3{{\left( 1 \right)}^{2}}\left( 3i \right)+3\left( 1 \right){{\left( 3i \right)}^{2}} \\
& \Rightarrow {{\left( 1-3i \right)}^{3}}=1-27{{i}^{3}}-9i+27{{i}^{2}} \\
\end{align}$
Now, we know that ${{i}^{2}}=-1$ and that, ${{i}^{3}}=-i$. By substituting them in the above equation, we get,
$\begin{align}
& {{\left( 1-3i \right)}^{3}}=1-27\left( -i \right)-9i+27\left( -1 \right) \\
& \Rightarrow {{\left( 1-3i \right)}^{3}}=1+27i-9i-27 \\
\end{align}$
By adding all the like terms in the above equation, we get,
$\begin{align}
& {{\left( 1-3i \right)}^{3}}=\left( 1-27 \right)+\left( 27-9 \right)i \\
& \Rightarrow {{\left( 1-3i \right)}^{3}}=-26+18i \\
\end{align}$
After simplification, we get the value of ${{\left( 1-3i \right)}^{3}}$as $-26+18i$. Hence the expression ${{\left( 1-3i \right)}^{3}}$ can be written in the cartesian form as $-26+18i$, where $\left( -26 \right)$ is the real part of the cartesian form and $\left( 18i \right)$ is the imaginary part of the cartesian form.
Note: The possible mistakes the students can do while solving this type of questions are by getting confused with the terms like, ${{i}^{2}}$ or terms like ${{i}^{3}}$. The students can also make a mistake of writing, ${{i}^{2}}=1$ and ${{i}^{3}}=i$ while trying to solve the question in a hurry, which will give them a wrong answer.
Complete step-by-step answer:
In this question, we have been asked to find the cartesian form of complex number ${{\left( 1-3i \right)}^{3}}$, that is we have to express ${{\left( 1-3i \right)}^{3}}$ in the form of $\left( a+bi \right)$. Now, we know that ${{\left( a-b \right)}^{3}}$ can be written as, ${{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$, or as ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$. Now we will use this formula to simplify ${{\left( 1-3i \right)}^{3}}$. By simplifying, we get,
$\begin{align}
& {{\left( 1-3i \right)}^{3}}={{\left( 1 \right)}^{3}}-{{\left( 3i \right)}^{3}}-3{{\left( 1 \right)}^{2}}\left( 3i \right)+3\left( 1 \right){{\left( 3i \right)}^{2}} \\
& \Rightarrow {{\left( 1-3i \right)}^{3}}=1-27{{i}^{3}}-9i+27{{i}^{2}} \\
\end{align}$
Now, we know that ${{i}^{2}}=-1$ and that, ${{i}^{3}}=-i$. By substituting them in the above equation, we get,
$\begin{align}
& {{\left( 1-3i \right)}^{3}}=1-27\left( -i \right)-9i+27\left( -1 \right) \\
& \Rightarrow {{\left( 1-3i \right)}^{3}}=1+27i-9i-27 \\
\end{align}$
By adding all the like terms in the above equation, we get,
$\begin{align}
& {{\left( 1-3i \right)}^{3}}=\left( 1-27 \right)+\left( 27-9 \right)i \\
& \Rightarrow {{\left( 1-3i \right)}^{3}}=-26+18i \\
\end{align}$
After simplification, we get the value of ${{\left( 1-3i \right)}^{3}}$as $-26+18i$. Hence the expression ${{\left( 1-3i \right)}^{3}}$ can be written in the cartesian form as $-26+18i$, where $\left( -26 \right)$ is the real part of the cartesian form and $\left( 18i \right)$ is the imaginary part of the cartesian form.
Note: The possible mistakes the students can do while solving this type of questions are by getting confused with the terms like, ${{i}^{2}}$ or terms like ${{i}^{3}}$. The students can also make a mistake of writing, ${{i}^{2}}=1$ and ${{i}^{3}}=i$ while trying to solve the question in a hurry, which will give them a wrong answer.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

India is a sovereign socialist secular democratic republic class 12 social science CBSE

The correct structure of ethylenediaminetetraacetic class 12 chemistry CBSE

How many states of matter are there in total class 12 chemistry CBSE

