Question

The minimum value of the expression $E={\left|z\right|}^2+{|z-3|}^2+{|z-6i|}^2$ is m, then the value of $\frac{m}{5}$ is _____.

Answer 1

$E={\left|z\right|}^2+{|z-3|}^2+{|z-6i|}^2$
Substitute $z=r(\cos A+\sin A)$
$E=r^2+{(r\cos A-3)}^2+r^2{\sin }^{2}A+r^2+{(r\sin A-6)}^2=3r^2+44-6r(\cos A+2\sin A)$
Since, maximum value of $a\cos A+b\sin A$ is $\sqrt{a^2+b^2}$
Therefore, $E\ge 3r^2+44-6\sqrt{5}r=3{(r-\sqrt{5})}^2+30\ge 30$
Thus, $m=30$
$\Rightarrow \frac{m}{5}=6$
Ans: 6