#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)#