Question

Let $\alpha$ and $\beta$ be complex numbers satisfying $|\alpha +1+i|=1$ and $|\beta -2-3i|=6$ such that $6|\alpha {|}_{max}-|\beta {|}_{max}=\sqrt{a}-\sqrt{b},$ where $a,b\in R^{+}.$ Then the value of $\sqrt{b^2-2a}$ is

Answer 1

$|\alpha +1+i|=1$
$\Rightarrow |\alpha -(-1-i)|=1$
$\alpha$ represents circle with centre at $C_1(-1,-1)$ and radius = $1$ unit.



$|\alpha |=|\alpha -0|=$ distance from point on circle to origin.
$|\alpha {|}_{max}=O{C}_1+r$ $=\sqrt{2}+1$
Similarly, for $|\beta -(2+3i)|=6$
$|\beta {|}_{max}=\sqrt{13}+6$
So, $6|\alpha {|}_{max}-|\beta {|}_{max}=6\sqrt{2}-\sqrt{13}=\sqrt{72}-\sqrt{13}$
$\therefore \sqrt{b^2-2a}=\sqrt{169-144}=5$