Answer

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**Hint:**To solve these questions, we have to simplify the given expression. The expression can be simplified by simply multiplying the terms given in the parenthesis with one another. On further simplification of the expression obtained after multiplying, we will get the required solution.

**Complete step by step answer:**

It is given that, $(5 + 3i)(3 - i)$

To simplify the given expression, we need to multiply the terms given in the parenthesis with each other.

Let us multiply the second expression with the first and second term of the first expression, to get

$(5 + 3i)(3 - i) = 5(3 - i) + 3i(3 - i)$

On multiplying the terms we get

$\Rightarrow 5 \times 3 - 5 \times i + 3i \times 3 + 3i \times ( - i)$

On simplifying the above expression we get

$\Rightarrow 15 - 5i + 9i - 3{i^2}$

Adding the like terms that contain $i$ we get

$\Rightarrow 15 - 4i - 3{i^2}$

In imaginary numbers, we know that the value of ${i^2} = - 1$ , where $i$ is an imaginary number. Therefore, by substituting the value ${i^2}$ in the above expression we get

$\Rightarrow 15 - 4i - 3( - 1)$

$\Rightarrow 15 - 4i + 3$

Adding the terms we get

$\Rightarrow 18 - 4i$

Therefore, $(5 + 3i)(3 - i) = 18 - 4i$

**Hence on simplifying $(5 + 3i)(3 - i)$ we get $18 - 4i$**

**Additional information:**

A complex number can be defined as a number that can be expressed in the form $a + ib$ where $a$ and $b$ are real numbers and $i$ represents the imaginary number and satisfies the equation ${i^2} = - 1$ . It also means that the value of $i$ is $i = \sqrt { - 1}$ . Since no real number satisfies the two given equations $i$ is called an imaginary number. Complex numbers cannot be marked on the number line.

**Note:**While solving these questions it is important to note down that complex numbers are represented in the form of $a + ib$ and that the imaginary part in this representation is $b$ and not $\;ib$ . Also, keep in mind to substitute the value of ${i^2}$ in the expression to completely simplify the expression and then add the like terms.

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