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?

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 $\sqrt{2}$.
