How do you simplify $(3 + 5i)(2 + 4i)$ ?

Hint: We know that when two brackets are multiplied, then their product is equal to the first term of the first bracket multiplied with the whole second bracket and the next term multiplied with the whole second and so on, that is $(a + b)(c + d) = a(c + d) + b(c + d)$ . Then we will apply the distributive property which states that the product of a number with the sum of two other numbers is equal to the product of that number with the first number plus the product of that number with the second number and vice versa, that is the product of a number with the sum of two other numbers is equal to the product of that number with the first number plus the product of that number with the second number and vice versa, that is, $a(b + c) = ab + ac$.

Complete step-by-step solution:
We have to simplify $(3 + 5i)(2 + 4i)$
We know that $(a + b)(c + d) = a(c + d) + b(c + d)$ , so we get –
$\Rightarrow (3 + 5i)(2 + 4i) = 3(2 + 4i) + 5i(2 + 4i)$
On applying the distributive property, we get –
$
 \Rightarrow (3 + 5i)(2 + 4i) = 6 + 12i + 10i + 20{i^2} \\
   \Rightarrow (3 + 5i)(2 + 4i) = 6 + 22i + 20{i^2} \\
 $
We know that $i = \sqrt { - 1} $ , so we get ${i^2} = {(\sqrt { - 1} )^2} = - 1$
$
   \Rightarrow (3 + 5i)(2 + 4i) = 6 + 22i + 20( - 1) \\
   \Rightarrow (3 + 5i)(2 + 4i) = 6 - 20 + 22i \\
   \Rightarrow (3 + 5i)(2 + 4i) = - 14 + 22i \\
 $
Hence the simplified form of $(3 + 5i)(2 + 4i)$ is $ - 14 + 22i$.

Note: In this question, we are given the product of two complex numbers. Numbers are mainly of two types – Real numbers and complex numbers. The numbers that can be represented on a number line are real numbers and the numbers that cannot be represented on a number line are known as complex numbers. The complex numbers are of the form $a + ib$ where $a$ is the real part as it can be represented on a number line and $ib$ is the imaginary part as it contains the imaginary number $i$ .