Multiply $2i\left( 3-8i \right)$ . Write the resulting number in the form of $a+bi$ .

- Hint: The given problem is related to multiplication of two complex numbers. In this question, we will use the concept of multiplication of complex numbers using distributive law. According to the distributive law, a(b + c) = ab + ac. Also, we will use the fact that the value of ${{i}^{2}}=-1$ to further simplify the expression. Then, we will express the result in the form of a + ib.

Complete step-by-step solution -

In this question, we are given two complex numbers $2i$ and $3-8i$ which are to be multiplied.
Complex numbers are multiplied in the same way as multiplying two linear equations of one variable, where i is treated as a variable. Value of imaginary number I is $\sqrt{-1}$ , therefore, its square is,
${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$
Now, multiplying $2i$ and $3-8i$, we get,
$2i\left( 3-8i \right)$
Applying distributive law, we get,
$\begin{align}
  & 2i\left( 3-8i \right)=2i\times 3-2i\times 8i \\
 & =6i-16{{i}^{2}} \\
\end{align}$
Putting the value of ${{i}^{2}}$ , we get,
$\begin{align}
  & 2i\left( 3-8i \right)=6i-16\left( -1 \right) \\
 & =6i+16 \\
 & =16+6i \\
\end{align}$
Therefore, value of $2i\left( 3-8i \right)$is $16+6i$ which is written in the form of $a+bi$,
Where, a is 16 and b is 6.

Note: While solving such problems, make sure to use correct signs. Students often write ${{i}^{2}}=1$ , which is wrong. Such mistakes should be avoided as they make the final answer incorrect.