How do you simplify $\left( -3i \right)\left( 2i \right)$?

Hint: We will use the concept of complex numbers to solve the above question. We know that ‘i’ represents iota and the value of $i=\sqrt{-1}$ and so ${{i}^{2}}=-1$. Now, to simplify $\left( -3i \right)\left( 2i \right)$, we will multiply the constant terms and $i$ separately.
$\Rightarrow \left( -3\times 2 \right)\times {{i}^{2}}$
$\Rightarrow -6{{i}^{2}}$
Now, we will substitute the value of ${{i}^{2}}$in the above expression and then simplify it to get the required answer.

Complete step-by-step solution:
We will use the concept of the complex number to solve the above question. We know that ‘i’ represents the iota where $i=\sqrt{-1}$.
Since, we have to simplify $\left( -3i \right)\left( 2i \right)$.
$\Rightarrow \left( -3i \right)\left( 2i \right)$
Now, after arranging the constant and variable terms we will get:
$\Rightarrow \left( -3 \right)\left( 2 \right)\left( i\times i \right)$
$\Rightarrow -6{{i}^{2}}$
Now, we know that $i=\sqrt{-1}$, so squaring both sides we will get:
${{i}^{2}}=-1$
Now, after substituting the value of ${{i}^{2}}$ in the above we will get:
$\Rightarrow -6\left( -1 \right)$
Now, we know that when we multiply two negative numbers then we get positive numbers.
$\Rightarrow 6$
Hence, the simplified value of $\left( -3i \right)\left( 2i \right)$ is 6.
This is our required solution.

Note: Students are required to note that when we have to simplify expressions which contain iota (i) then we simplify the constant and iota terms separately and then we put the value of iota if it is even power. Also, note that the value of iota repeats after every ${{4}^{th}}$ power. We have value of $i=\sqrt{-1}$, ${{i}^{2}}=-1$, ${{i}^{3}}= -i$ and ${{i}^{4}}=1$. So, we will have ${{i}^{5}}={{i}^{4}}\times i=i$, ${{i}^{6}}={{i}^{4}}\times {{i}^{2}}=-1$, ${{i}^{7}}={{i}^{4}}\times {{i}^{3}}=i$ and ${{i}^{8}}={{i}^{4}}\times {{i}^{4}}=1$. So, we can see that the value of iota repeats after every ${{4}^{th}}$power.