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Find the product of each expression using properties of complex numbers.

1. (x + 3i)(x − 3i)
2. (x − 4i)(x + 4i)
3. (x + 8i)(x − 8i)

User Kema
by
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1 Answer

2 votes

Answer:


(x+3\mathbf{i})(x-3\mathbf{i})=x^2+9


(x-4\mathbf{i})(x+4\mathbf{i})=x^2+16


(x+8\mathbf{i})(x-8\mathbf{i})=x^2+64

Explanation:

Complex Numbers

The complex numbers are known by having an 'imaginary' part along with the real part. They can be expressed like a + bi, where


\mathbf{i}=√(-1)

Or, equivalently:


\mathbf{i}^2=-1

Let's find the result of the following products. On each one of them, there is a sum of a binomial multiplied by the subtraction of a binomial with the very same terms. It leads to a well-known polynomial identity:


(a+b)(a-b)=a^2-b^2

1.
(x+3\mathbf{i})(x-3\mathbf{i})

Applying the above-mentioned identity:


(x+3\mathbf{i})(x-3\mathbf{i})=(x^2-(3\mathbf{i})^2)


=x^2-9\mathbf{i}^2

Since
\mathbf{i}^2=-1


(x+3\mathbf{i})(x-3\mathbf{i})=x^2-9(-1)


=x^2+9

2.
(x-4\mathbf{i})(x+4\mathbf{i})

Proceed in the same way as before:


(x-4\mathbf{i})(x+4\mathbf{i})=(x^2-16\mathbf{i}^2)


=(x^2-16(-1))


=x^2+16

3.
(x+8\mathbf{i})(x-8\mathbf{i})


(x+8\mathbf{i})(x-8\mathbf{i})=(x^2-64\mathbf{i}^2)


=(x^2-64(-1))


=x^2+64

User Nephewtom
by
5.3k points