Complexed

Given fourth degree equation can be written as product of two quadratic polynomial as $(x^2-2x+5)(x^2-4x+13)=0$ .

Now on solving $x^2-2x+5=0$ give its roots as $1+2i$ and $1-2i$

and on solving $x^2-4x+13=0$ give its roots as $2+3i$ and $2-3i$

Hence $\lambda = |-2|+|2|+|3|+|-3| = 10$

and $\mu = 1+1+2+2 = 6$

$\lambda+\mu = 16$

Since the coefficients are all real numbers, the pairs of roots must be complex conjugates of each other: $a_1 ± ib_1, a_2 ± ib_2$ , where there are four roots.

By Vieta, the sum of the roots is $6 \Rightarrow a_1 + a_ 2 = 3$ , and their product is $65 \Rightarrow (a_1^2 + b_1^2)(a_2^2 +b_2^2) = 65$ . Since $a_k, b_k$ are all integers and $65 = 5 \times 13$ , WLOG let $a_1^2 + b_1^2 = 5, a_2^2 + b_2^2 = 13$ . There are only a few cases to check, so $a_1 = 1, b_2 = 2, a_2 = 2, b_2 = 3$ .

If you want, verify that this gives you the correct polynomial. Thus $\lambda + \mu = (2+3+2+3) + 6 = \boxed{16}$ .