ANSWER
A. True
EXPLANATION
Let the complex number be z = a+ ib, so its complex conjugate is z* = a - ib, where a and b are real numbers. Let's find the product,
The product is,
[tex](a+ib)(a-ib)=a\cdot a-a\operatorname{\cdot}ib+ib\operatorname{\cdot}a-ib\operatorname{\cdot}ib[/tex]
Solve the products,
[tex](a+ib)(a-ib)=a^2-iab+iab-i^2b^2[/tex]
Simplify: note that the second and third terms are opposites, so they cancel out. Remember that i² is equal to -1,
[tex](a+ib)(a-ib)=a^2+0-(-1)b^2=a^2+b^2[/tex]
Since a and b were real numbers, then the sum of their squares is also a real number.
Hence, this statement is true.
