Find the smallest possible whole numbers $a$ and $b$.

Question

Find the smallest possible whole numbers $a$ and $b$ such that the product $(a+bi)(3-6i)$ simplifies to a real number.

Multiplying by these two conjugates gives me a real number, but I don't know where to begin. Do I make the expression equal to $a$ or $b$ and just multiply the numbers out?

My teacher wrote the answer as $a=1$ and $b=2$, if that helps.

Answer 1

$$...=a+6b +i(3b-6a)\implies 3b-6a = 0\implies b= 2a$$

So if $a=1$ we get $b=2$ and this are smallest choice when both are positive integers. Else there is no solution.