f(1) = (4+i)(1)^2 + a(1) + b
= 4+i + a + b
For this value to be real, the imaginary part of a+b must cancel the i, that is, "Im"(a+b) = -1
f(i) = (4+i)(i)^2 + a(i) + b
=-4-i+ai+b
For this value to be real, we must have the imaginary part of ai+b cancel the -i, that is, "Im"(ai+b) = 1
Let a = a_r+a_ii and b = b_r+b_ii. Then ai = -a_i+a_ri, so the above two observations give us the system
{(a_i+b_i = -1), (a_r+b_i = 1):}
Putting a_i and a_r in terms of b_i, this gives
{(a_i = -b_i-1), (a_r = -b_i+1):}
Finally, before proceeding, notice that nowhere did we place any restrictions on b_r. If the above system holds, then we will have f(1) in RR and f(i) in RR regardless of the value of b_r. As such, we will assume b_r = 0 for the purpose of minimizing |b|.
Now, putting everything in terms of b_i, we have
|a| + |b| = sqrt(a_r^2+a_i^2)+sqrt(b_r^2+b_i^2)
=sqrt((-b_i+1)^2+(-b_i-1)^2) + sqrt(0^2+b_i^2)
=sqrt(2b_i^2+2) + |b_i|
By inspection, this value has a minimum at b_i = 0, as any change in b_i in either the positive or negative direction will result in an increase in both terms. Thus, setting b_i = 0, we get our final answer as
a = a_r+a_ii = (-b_i+1)+(-b_i-1)i = 1-i
b = b_r+b_ii = 0+0i = 0
and
|a|+|b| = sqrt(2(0)^2+2)+|0| = sqrt(2)
