Question

How do you solve \[4(-2i)(-2i)\]?

Answer 1

Hint: We are given a complex expression (one that involves the use of iota \[i\]). We will be using the properties of arithmetic multiplication to simplify the given expression. We will also use the fact that the product of iotas is iota raised to the power 2, that is, \[i.i={{i}^{2}}=-1\]. We will multiply the terms in succession, the two negative signs will get positive and hence we will get the answer.

Complete step by step solution:
According to the given question, we have been asked to solve the given expression. The expression involves the use of iota used in representing imaginary numbers. In order to solve the expression, we will be using the properties related to iota as well as arithmetic multiplication.
We will start by writing the expression we have,
\[4(-2i)(-2i)\]
We will first multiply the two \[(-2i)\]’s with each other. The number -2 is multiplied by -2 and ‘I’ is multiplied by ‘I’. We get,
\[\Rightarrow 4((-2)(-2).(i.i))\]
We know that, -2 when multiplied to -2 we get 4 and \[i\] multiplied by \[i\] gives \[{{i}^{2}}\]. We get it as,
\[\Rightarrow 4((4).({{i}^{2}}))\]
We know that, \[{{i}^{2}}=-1\], applying this in the above expression, we have
\[\Rightarrow 4((4).(-1))\]
Now, we will multiply the terms we have, we then get,
\[\Rightarrow 4(-4)\]
On solving further, we get the value of the expression as,
\[\Rightarrow -16\]
Therefore, the answer of the given expression is \[-16\].

Note:
The iota (\[i\]) used in the above question is a complex number representation. When we solve the quadratic equations, when the discriminant is less than 0, the imaginary roots (not real roots) are found using iota (\[i\]). The value of \[i=\sqrt{(-1)}\] then \[{{i}^{2}}=-1\].