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?

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 $x=0$.
