Question

How do you simplify ${{\left( 7-3i \right)}^{2}}$ ?

Answer 1

Hint: We are asked to simplify ${{\left( 7-3i \right)}^{2}}$.
To simplify the given term we will learn how complex numbers are written, how they are arranged, and what ‘I’ stand for.
What are the various powers of iota, after that we will learn how to multiply term we will use ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ .To expand our term we use ${{i}^{2}}=-1$ to simplify our given problem.

Complete step by step answer:
We are given ${{\left( 7-3i \right)}^{2}}$ , we are asked to simplify.
Before we simplify, we will learn about complex numbers.
We know the term $a+ib$ is known as a complex number. Where ‘I’ is known as iota , and ‘b’ are any real numbers.
Iota (I) is defined as $i=\sqrt{-1}$ .
Iota has properly that $i'=i,{{i}^{2}}=-1,{{i}^{3}}=-i,\text{ and }{{i}^{4}}=1$
We should know that for any complex number $a+ib$ , conjugate of it is given as $a-ib$ ,
We are asked to simplify ${{\left( 7-3i \right)}^{2}}$.
We firstly learnt that ${{\left( 7-3i \right)}^{2}}$mean as we know that ${{a}^{2}}$ mean $a\times a$ so,
${{\left( 7-3i \right)}^{2}}=\left( 7-3i \right)\left( 7-3i \right)$
Now to expand this we will multiply.
We will multiply each element of the first bracket to the other brackets. So,
$\begin{align}
  & {{\left( 7-3i \right)}^{2}}=\left( 7-3i \right)\left( 7-3i \right) \\
 & =7\left( 7-3i \right)-3i\left( 7-3i \right) \\
\end{align}$
By opening brackets, we get –
$7\times 7-7\times 3i-3i\times 7-3i\times \left( -3i \right)$
By simplifying, we get –
$=49-21i-21i+9{{i}^{2}}$
As ${{i}^{2}}=-1$ so,
$\begin{align}
  & 49-42i+9\left( -1 \right) \\
 & =49-42i-9 \\
\end{align}$
By simplifying further, we get –
$=40-42i$
So, we get –
${{\left( 7-3i \right)}^{2}}=40-42i$
We can also expand our whole square using the formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ .
So, using this formula, applying on ${{\left( 7-3i \right)}^{2}}$, we get –
${{\left( 7-3i \right)}^{2}}={{7}^{2}}+{{\left( 3i \right)}^{2}}-2\times 7\times 3i$
By simplifying, we get –
$=49+9{{i}^{2}}-42i$ .
As ${{i}^{2}}=-1$ we get –
$\begin{align}
  & 49-9-42i \\
 & 40-42i \\
\end{align}$

So,${{\left( 7-3i \right)}^{2}}=40-42i$

Note: While simplifying, always remember that in product we multiply each term of the first bracket by each term of the other bracket. Do not make errors like $\left( a+b \right)\left( c+d \right)=ac+bd$ This is wrong.
Also while simplifying things in addition or subtraction we have to apply the sign that is outside the bracket to all terms of the bracket.
Do not make mistake like –
$\left( a+b \right)-\left( c+a \right)=a+b-c+a$ this is wrong.
It should be like –
$\left( a+b \right)-\left( c+d \right)=a+b-c-d$ .