Question

How do you simplify $\left( {5 + 2i} \right)\left( {5 - 2i} \right)$ and write in $a + bi$ form?

Answer 1

Hint: In this problem, we have given a product two complex numbers and we asked to simplify the given product of two complex numbers. And also we are asked to write the answer after simplification in the form of a complex number. That is our answer will be an addition of two terms in which one term will be multiplied by a complex number.

Complete step-by-step solution:
Given term is $\left( {5 + 2i} \right)\left( {5 - 2i} \right)$
Now we are going to use the distributive property to expand the brackets.
That is, we are going to multiply the first term of the first bracket with the every term of second bracket and second term of first bracket with every term of second bracket.
$ \Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 5\left( 5 \right) + 5\left( { - 2i} \right) + 2i\left( 5 \right) + 2i\left( { - 2i} \right)$
Next we simplify the above term,
$ \Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 25 - 10i + 10i - 4{i^2}$, here the $ - 10i$and$ + 10i$gets cancel each other and we get,
$ \Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 25 - 4{i^2}$
Since in complex analysis, ${i^2} = - 1$, the expression simplifies to,
$ \Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 25 - 4\left( { - 1} \right)$, product of two negative terms becomes positive.
$ \Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 25 + 4$
On adding the term and we get,
$ \Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 29$
So the answer for the product two given complex terms is $29$.
There is no complex term in the answer which we got, so we add $0i$ to the answer.

Therefore, the required answer is $29 + 0i$

Note: Here we the term $ - 4{i^2}$ by multiplying $2i$ and $ - 2i$ then we raised $i$ to the power of $1$ and we get $ - 4({i^1}{i^1})$. Then we applied the power rule ${a^x}{a^y} = {a^{x + y}}$ to combine exponents.
By this way we got $ - 4{i^2}$. And the important thing about this problem is every real number can be written in the form of a complex number by adding $0i$ to the real number.
That will be in the form of $a + bi$.