we will proceed to verify each case to determine the solution
remember that
\(i^{2} =-1\)
case A) \((1 + 3i)(6i)\)
applying distributive property
\((1 + 3i)(6i)=1*6i+3i*6i \\=6i+18i^{2}\\=6i+18*(-1)\\=6i-18\)
\(-18+6i\) ------> is not a real number
therefore
the case A) is not a real number product
case B) \((1 + 3i)(2-3i)\)
applying distributive property
\((1 + 3i)(2-3i)=1*2+1*(-3i)+3i*(2)+3i*(-3i)\\=2-3i+6i-9i^{2}\\=2+3i-9*(-1) \\=11+3i\)
\(11+3i\) ------> is not a real number
therefore
the case B) is not a real number product
case C) \((1 + 3i)(1-3i)\)
applying difference of square
\((1 + 3i)(1-3i)=(1)^{2}-(3i)^{2}\\=1-9i^{2}\\=1-9*(-1) \\=10\)
\(10\) ------> is a real number
therefore
the case C) is a real number product
case D) \((1 + 3i)(3i)\)
applying distributive property
\((1 + 3i)(3i)=1*3i+3i*3i \\=3i+9i^{2}\\=3i+9*(-1) \\=3i-9\)
\(-9+3i\) ------> is not a real number
therefore
the case D) is not a real number product
the answer is
\((1 + 3i)(1-3i)\)
