Real Divides A Complex(Finding the Unknown)

Multiplying both sides by $2+ni$ ,

$13=(2-ni)(2+ni)$

$13=4-n^{2}i^{2}$

$n^{2}=9 \implies n=\pm 3$

We can't enter $-3$ as our answer in Brilliant (lol).

Therefore, $\boxed {n=3}$

In doing this kind of problem, lets find the two numbers whose sum is 13 and form it as a complex number in form a+bi. We all know that a is given and that is 2, we cannot assume that the unknown is 7 because your logic that 2+7=9 is wrong. But how do i get 3 in easiest way. As we all know, the product of a complex and its conjugate yields into a solution of a real number but there's a twist. Why there's a twist? Its because there are cases that yields into a factors of a complex radical expression. In order to obtain 3, first, square the a which is 2 and the answer is 4.Then, find the other addend that if we add it to its 4, its sum would be 13. So, I get 9 because 9+4=13. Then find the square root of 9 which is +3 to make the real number or the dividend be positive so the final answer for what is the value of N is 3.