Asked by Lacey
                What is the product of a complex conjugated?
The product of complex conjugates is a difference of two squares it is always a real number.
The product a complex conjugates may be written in standard form as a + bi where neither a nor b is zero.
The product of complex conjugates is a sum of two squares and is always a real number.
The product of complex conjugates is the same as the product of opposites.
            
            
        The product of complex conjugates is a difference of two squares it is always a real number.
The product a complex conjugates may be written in standard form as a + bi where neither a nor b is zero.
The product of complex conjugates is a sum of two squares and is always a real number.
The product of complex conjugates is the same as the product of opposites.
Answers
                    Answered by
            oobleck
            
    (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 - b^2 i^2
But i^2 = -1, so that makes it a^2 + b^2
    
But i^2 = -1, so that makes it a^2 + b^2
                    Answered by
            swear on your fricking yeezys
            
    how does this help im still confused
    
                    Answered by
            Sherlock411
            
    The answer is A: the product is the difference of 2 squares and always a real number. 
***Do NOT let the words "Complex Number" freak you out.***
It's still the same as any real numbers used before with just the addition of the imaginary unit of (i). When you have (i)^2 then it becomes a -1 in place of that square. So an example would be: (2+3i)(2-3i)=2(2+3i) + -3i(2+3i) with the "distributive property" of algebra/mathematics. <JUST TAKE THE 2nd PART AND PUT IT THROUGH THE 1st>
Then do the math operation like you've always done to get: (4+6i)+(-6i-9i^2) = (4+6i)+(-6i-9(-1))
combine like terms, simplify if possible, then write in standard form: the +6i & -6i cancel each other out, the i^2 becomes a -1 to be multiplied into the -9 to = +9 then write correctly after simplifying: 4+9=13 so 13 is the answer
***hope that helps***
    
***Do NOT let the words "Complex Number" freak you out.***
It's still the same as any real numbers used before with just the addition of the imaginary unit of (i). When you have (i)^2 then it becomes a -1 in place of that square. So an example would be: (2+3i)(2-3i)=2(2+3i) + -3i(2+3i) with the "distributive property" of algebra/mathematics. <JUST TAKE THE 2nd PART AND PUT IT THROUGH THE 1st>
Then do the math operation like you've always done to get: (4+6i)+(-6i-9i^2) = (4+6i)+(-6i-9(-1))
combine like terms, simplify if possible, then write in standard form: the +6i & -6i cancel each other out, the i^2 becomes a -1 to be multiplied into the -9 to = +9 then write correctly after simplifying: 4+9=13 so 13 is the answer
***hope that helps***
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