If $f(1)$ and $f(i)$ real, then find minimum value of $|a|+|b|$ A function $f$ is defined by $$f(z)=(4+i)z^2+a  z+b$$ $(i=\sqrt{-1})$for all complex number $z$ where $a$ and $b$ are complex numbers. If $f(1)$ and $f(i)$ are both purely real, then find minimum value of $|a|+|b|$
Now
$f(1)=4+i+a+b$ which means imaginary part of $a+b=-1$ and $f(i)=-4-i+a \cdot i+b$ which gives imaginary part of $a \cdot i+b=1$  but how to proceed further to find desired value?
 A: If $a=a_1+a_2i$ and $b=b_1+b_2i$ then you have that $b_2+a_2=-1, b_2+a_1=1$. So you can pick any $b_1$, which, since you are seeking a minimum, means you can choose $b_1=0$. You get $b=b_2i$ and $a_2=-(b_2+1)$ and $a_1=1-b_2$. So you are trying to minimize:
$$\sqrt{a_1^2+a_2^2}+\sqrt{b_1^2+b_2^2}=\sqrt{(-1-b_2)^2+(1-b_2)^2}+|b_2|=\sqrt{2+2b_2^2}+|b_2|$$
Apply usual calculus tricks to minimize $\sqrt{2+2x^2}+x$ with $x\geq 0$. You should get that the minimum is when $x=0$.
A: There is nothing to force the real part of $b$ to be other than zero and having it nonzero would make $|b|$ greater, so $b$ is purely imaginary.  Let $b=ci, a=a_r+a_ii$.  We know $c+a_i=-1, a_r+c=1$, so $|a|+|b|=\sqrt{(c+1)^2+(c-1)^2}+c$  Minimize this over $c$.
A: You have $\text{Im}(a+b)=-1$ and $\text{Im}(a\text{i}+b)=1$.  That is, $$\text{Im}\big(a(1-\text{i})\big)=\text{Im}(a+b)-\text{Im}(a\text{i}+b)=-2\,.$$
This means 
$$\sqrt{2}|a|=\big|a(1-i)\big|\geq \Big|\text{Im}\big(a(1-\text{i})\big)\Big|=\big|-2\big|=2\,.$$
Hence, $|a|\geq \sqrt{2}$.  The equality holds if and only if $a(1-\text{i})=-2\text{i}$, or $a=1-\text{i}$.  
From the result above, $$|a|+|b|\geq|a|\geq\sqrt{2}\,.$$  The equality holds iff $a=1-\text{i}$ and $b=0$.  Note that $(a,b)=(1-\text{i},0)$ satisfies the required conditions, so the minimum value of $|a|+|b|$ is indeed $\sqrt{2}$.
A: Setting $a = a_1 + i a_2, b = b_1 + i b_2$ for four real unknowns $a_k, b_k$.
Real $f(1) = 4 + i + a + b$ and real $f(i) = -4 - i + a i + b$ give the equations
$$
\text{Im}(f(1)) = 1 + a_2 + b_2 = 0 \\
\text{Im}(f(i)) = -1 + a_1 + b_2 = 0
$$
Then we have to minimize
$$
g(a_1, a_2, b_1, b_2) 
= \lvert a \rvert + \lvert b \rvert 
= \sqrt{a_1^2 + a_2^2} + \sqrt{b_1^2 + b_2^2}
$$
which reduces to a function of two unknowns if we apply the two equations above:
$$
b_2 - 1 = - a_1 \\
b_2 + 1 = - a_2 \\
h(b_1, b_2) 
= \sqrt{(b_2 - 1)^2 + (b_2 + 1)^2} + \sqrt{b_1^2 + b_2^2} \\
= \sqrt{2b_2^2 + 2} + \sqrt{b_1^2 + b_2^2} \\
\ge \sqrt{2}
$$
Here is a graph:

$h$ is shown in red colour, $\partial_{b_1} h$ in blue and $\partial_{b_2} h$ in green.
So we have a minimum at $b = 0$ which implies $a = 1-i$ via the constraints.
The minimal value for $\lvert a \rvert + \lvert b \rvert $ is $\sqrt{2}$.
