Question

How do you multiply ${{(-8i)}^{2}}$ ?

Answer 1

Hint: In this question we will split the expression using the property of exponents which is ${{(a\times b)}^{n}}={{a}^{n}}\times {{b}^{n}}$ and then we will use the property of complex numbers that the value of ${{i}^{2}}=-1$ to simplify the equation to get the final answer.

Complete step-by-step answer:
We have the expression as:
\[\Rightarrow {{(-8i)}^{2}}\]
In this expression we have two elements which have to be square to get the required solution.
we have two terms in multiplication therefore, the element can be split up and written as:
$\Rightarrow {{(-8\times i)}^{2}}$
Now we know that ${{(a\times b)}^{n}}={{a}^{n}}\times {{b}^{n}}$ therefore on using this property of exponents we can write the expression as:
$\Rightarrow -{{8}^{2}}\times {{i}^{2}}$
We will now solve each element separately and then multiply them both to get the required solution.
We know the value of $i=\sqrt{-1}$ therefore, on squaring the term, we get ${{i}^{2}}=-1$.
On substituting it in the expression, we get:
$\Rightarrow -{{8}^{2}}\times -1$
Now we know $-{{8}^{2}}$ can be written as $-8\times -8$ therefore on multiplying, we get $64$.
On substituting it in the expression, we get:
$\Rightarrow 64\times -1$
Which can be simplified as:
$\Rightarrow -64$, which is the required solution therefore, ${{(-8i)}^{2}}=-64$.

Note: It is to be remembered that when two negative numbers are multiplied their product is positive therefore on squaring a negative number, the solution will always be positive.
The same is not the case when finding the cube of a negative number, the cube of a negative number will be negative. It is a general rule when the power of a negative number is odd, the solution will be negative and when the power is even, the solution will be a positive number.
In this question we have the expression in the complex form, which is also called as the imaginary form since it has the term $i$ which is used when we have to simplify a negative number which is in the square root.