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Question: How do you find the product of complex numbers: \(\left( 5-2i \right)\left( 5+2i \right)\)?...

How do you find the product of complex numbers: (5−2i)(5+2i)\left( 5-2i \right)\left( 5+2i \right)?

Explanation

Solution

We first explain the meaning of the process ‘FOIL’. We multiply the terms according to their positions. There are four multiplications to be done. We complete all four multiplications according to the previously mentioned process.

Complete step by step answer:
We have been given multiplication of two linear equations. We have to do the breakings of the polynomials in order of FOIL. The word FOIL stands for First-Outside-Inside-Last. It is a technique to distribute the multiplication of polynomials.
There are two imaginary terms in each polynomial.
We start by multiplying the first terms of (5−2i)\left( 5-2i \right) and (5+2i)\left( 5+2i \right). The terms are 5 and 5.
The multiplication gives a result of 5×5=255\times 5=25.
We now multiply the outside terms of (5−2i)\left( 5-2i \right) and (5+2i)\left( 5+2i \right). The terms are 5 and 2i2i.
The multiplication gives a result of 5×2i=10i5\times 2i=10i.
Then we multiply the inside terms of (5−2i)\left( 5-2i \right) and (5+2i)\left( 5+2i \right). The terms are −2i-2i and 5.
The multiplication gives the result of (−2i)×5=−10i\left( -2i \right)\times 5=-10i.
We end by multiplying the last terms of (5−2i)\left( 5-2i \right) and (5+2i)\left( 5+2i \right). The terms are −2i-2i and 2i2i.
The multiplication gives the result of (−2i)×2i=−4i2\left( -2i \right)\times 2i=-4{{i}^{2}}.
Now we add all the terms and get the final solution as
(5−2i)(5+2i)=25−10i+10i−4i2=25−4i2\left( 5-2i \right)\left( 5+2i \right)=25-10i+10i-4{{i}^{2}}=25-4{{i}^{2}}.
We have the relations for imaginary ii where i2=−1,i3=−i,i4=1{{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1. We place the values and get (5−2i)(5+2i)=25−4i2=25+4=29\left( 5-2i \right)\left( 5+2i \right)=25-4{{i}^{2}}=25+4=29

Therefore, the multiplied value of (5−2i)(5+2i)\left( 5-2i \right)\left( 5+2i \right) is 29.

Note: We can also solve the equation (5−2i)(5+2i)\left( 5-2i \right)\left( 5+2i \right) using formula of a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right).
For our given problem we take a=5;b=2ia=5;b=2i.
So, (5−2i)(5+2i)=52−(2i)2=25−4i2=25+4=29\left( 5-2i \right)\left( 5+2i \right)={{5}^{2}}-{{\left( 2i \right)}^{2}}=25-4{{i}^{2}}=25+4=29.
Thus, we verify our result of simple multiplication.