For certain real values of $a, b, c,$ and $d_{},$  the equation $x^4+ax^3+bx^2+cx+d=0$ has four non-real roots.  The product of two of these roots is $13+i$ and the sum of the other two roots is $3+4i,$ where $i=\sqrt{-1}.$  Find $b.$
Since the [coefficients](https://artofproblemsolving.com/wiki/index.php/Coefficient) of the [polynomial](https://artofproblemsolving.com/wiki/index.php/Polynomial) are real, it follows that the non-real roots must come in [complex conjugate](https://artofproblemsolving.com/wiki/index.php/Complex_conjugate) pairs. Let the first two roots be $m,n$. Since $m+n$ is not real, $m,n$ are not conjugates, so the other pair of roots must be the conjugates of $m,n$. Let $m'$ be the conjugate of $m$, and $n'$ be the conjugate of $n$. Then,
\[m\cdot n = 13 + i,m' + n' = 3 + 4i\Longrightarrow m'\cdot n' = 13 - i,m + n = 3 - 4i.\]
By [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas), we have that $b = mm' + nn' + mn' + nm' + mn + m'n' = (m + n)(m' + n') + mn + m'n' = \boxed{051}$.