# How do you simplify  (-3i) * (4i) ?

Dec 6, 2015

The simplified form of $\left(- 3 i\right) \cdot \left(4 i\right)$ is $12$.

#### Explanation:

A lot of people are intimidated by imaginary numbers, but you don't have to be!

Let's think of $\left(i\right)$ as a variable. We know that $i = \sqrt{- 1}$. Therefore, $i \cdot i$, or ${i}^{2}$, = $\sqrt{- 1} \cdot \sqrt{- 1} , \mathmr{and} {\left(\sqrt{- 1}\right)}^{2}$. A common saying is "the square undoes the square root." This means that when you square a quantity under a square root, the act of taking the square root cancels with the act of multiplying it by itself. This makes sense because if we were to take the square root of $25$ and then square it, it would look like this:

$\sqrt{25} = 5 , - 5$ ---> ${\left(5\right)}^{2} = 25 , \mathmr{and} {\left(- 5\right)}^{2} = 25$.

You can see that we end up where we started, with $25$, thus proving that taking the square root of something cancels with squaring it. So, this means that ${\left(\sqrt{- 1}\right)}^{2} = \left(- 1\right)$.

The first thing you can do when multiplying terms with imaginary numbers is to multiply the real numbers first, in this case that would be $\left(- 3\right) \cdot 4$.

$\left(- 3\right) \cdot 4 = \left(- 12\right)$, so now we have $\left(- 12\right) \cdot i \cdot i$. The $\left(- 12\right)$ is the product from the real numbers, and the two $i$'s are the "left over" imaginary parts. We know that when we multiply a quantity by itself it's the same thing as squaring it, so this can further be simplified to $\left(- 12\right) \cdot \left({i}^{2}\right)$.

As we found earlier, $\left({i}^{2}\right) = \left(- 1\right)$. So our expression is now $\left(- 12\right) \cdot \left(- 1\right)$, which we can easily conclude is equal to $12$.