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

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

Explanation

Solution

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 (5+2i)(5−2i)\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.
⇒(5+2i)(5−2i)=5(5)+5(−2i)+2i(5)+2i(−2i)\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,
⇒(5+2i)(5−2i)=25−10i+10i−4i2\Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 25 - 10i + 10i - 4{i^2}, here the −10i - 10iand+10i + 10igets cancel each other and we get,
⇒(5+2i)(5−2i)=25−4i2\Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 25 - 4{i^2}
Since in complex analysis, i2=−1{i^2} = - 1, the expression simplifies to,
⇒(5+2i)(5−2i)=25−4(−1)\Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 25 - 4\left( { - 1} \right), product of two negative terms becomes positive.
⇒(5+2i)(5−2i)=25+4\Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 25 + 4
On adding the term and we get,
⇒(5+2i)(5−2i)=29\Rightarrow \left( {5 + 2i} \right)\left( {5 - 2i} \right) = 29
So the answer for the product two given complex terms is 2929.
There is no complex term in the answer which we got, so we add 0i0i to the answer.

Therefore, the required answer is 29+0i29 + 0i

Note: Here we the term −4i2 - 4{i^2} by multiplying 2i2i and −2i - 2i then we raised ii to the power of 11 and we get −4(i1i1) - 4({i^1}{i^1}). Then we applied the power rule axay=ax+y{a^x}{a^y} = {a^{x + y}} to combine exponents.
By this way we got −4i2 - 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 0i0i to the real number.
That will be in the form of a+bia + bi.