How do you multiply \[(3 + 2i)(1 - 3i)\]?

Hint: Real numbers and imaginary numbers are the two types of numbers, real numbers are the numbers that can be plotted on a number line while imaginary numbers, as the name suggests, cannot be represented on the number line. Sometimes while solving equations under the square root, we get a negative answer but we know that the square root of a negative number doesn’t exist so we had to think of a way to represent them that’s why we take \[\sqrt { - 1} = i\]. In the given problem it is actually equal to\[(3 + \sqrt { - 2} )(1 - \sqrt { - 3} )\] and these are called complex numbers. To solve this we need to know the value of \[{i^2}\] that is \[{i^2} = - 1\].

Complete step-by-step solution:
We know that
\[i = \sqrt { - 1} \]
Squaring on both side we have,
\[{i^2} = - 1\].
Now we have,
\[\Rightarrow (3 + 2i)(1 - 3i) = 3(1 - 3i) + 2i(1 - 3i)\]
\[\Rightarrow 3(1 - 3i) + 2i(1 - 3i)\]
\[\Rightarrow 3 - 9i + 2i - 6{i^2}\]
Adding like terms we have
\[\Rightarrow 3 - 8i - 6{i^2}\]
We know \[{i^2} = - 1\],
\[\Rightarrow 3 - 8i - 6( - 1)\]
We know that the product of two negative numbers gives us the positive number. Then we have
\[\Rightarrow 3 - 8i + 6\]
\[\Rightarrow 9 - 8i\]
Thus we have \[(3 + 2i)(1 - 3i) = 9 - 8i\]

Note: For multiplying the terms written in the parentheses like\[(a + b)(c + d)\] we first multiply the first term of the first bracket with the whole second bracket and then multiply the second term of the first bracket with the whole second bracket that is
\[(a + b)(c + d) = a((c + d) + b(c + d)\]
\[(a + b)(c + d) = ac + ad + bc + dc\]
But there are various identities to make the calculations easier. Suppose if we have \[(3 + 2i)(3 - 2i)\], we know the identity \[{a^2} - {b^2} = (a + b)(a - b)\]. Using this we can solve it easily. We also have \[{(a + b)^2} = {a^2} + 2ab + {b^2}\] also we know \[{(a - b)^2} = {a^2} - 2ab + {b^2}\]. Depending on the given problem we apply these identities.