we will proceed to solve each case to determine the solution of the problem
we know that
\(i^{2} =-1\)
case A. \(15*(-15i)\)
\(15*(-15i)=-225i\)
The result of case A is not a real number product
case B. \(3i*(1-3i)\)
\(3i*(1-3i)=3i*(1)-(3i)*(3i)\\=3i-9i^{2}\\=3i-9*(-1)\\=3i+9\)
The result of case B is not a real number product
case C. \((8+20i)*(-8-20i)\)
\((8+20i)*(-8-20i)=8*(-8)+8*(-20i)+20i*(-8)-20i*(20i)\\=-64-160i-160i-400i^{2}\\=-64-320i-400*(-1)\\=-64-320i+400\\=336-320i\)
The result of case C is not a real number product
case D. \((4+7i)*(4-7i)\)
\((4+7i)*(4-7i)=4^{2} -(7i)^{2}\\=16-49i^{2}\\=16-49*(-1)\\=65\\\)
The result of case D is a real number product
therefore
the answer is the option
\((4+7i)*(4-7i)\)
