If ${z}_{1},{z}_{2},..{z}_{n}$ lie on the circle $|z|=2$ then the value of $|{z}_{1}+{z}_{2}+..{z}_{n}|-4|\dfrac {1}{{z}_{1}}+\dfrac {1}{{z}_{2}}++\dfrac {1}{{z}_{n}}|=$

$\begin{array}{l}|z|=2 \\ \Rightarrow|z \cdot|^{2}=4 \\ \Rightarrow z \cdot \bar{z}=4 \\ \therefore \frac{1}{z}=\frac{\bar{z}}{4}\end{array}$

As $z_{1}, z_{2}, z_{3} \ldots, z_{n}$ lie on $|z|=2$ circle

$\begin{array}{l} \therefore\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=\cdots \cdot=\left|z_{n}\right|=2 \\ \therefore \frac{1}{z_{1}}=\frac{\bar{z}_{1}}{4}, 1 / z_{2}=\bar{z}_{2} / 4,1 / z_{3}=\bar{z}_{3} / 4 \cdots, \frac{1}{z_{n}}=\bar{z}_{n} / 4 \\ \end{array}$

Given expression,

$\begin{array}{l} \left|z_{1}+z_{2}+z_{3}+\cdots+z_{n}\right|-4\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}+\cdots+\frac{1}{z_{n}}\right| \\ =\left|z_{1}+z_{2}+z_{3}+\cdots+z_{n}\right|-4\left|\frac{\bar{z}_{1}}{4}+\frac{\bar{z}_{2}}{4}+\frac{\bar{z}_{3}}{4}+\cdots+\frac{\bar{z}_{n}}{4}\right| \end{array}$

$\begin{array}{l}=\left|z_{1}+z_{2}+z_{3}+\cdots \cdot+z_{n}\right|-4 \times \frac{1}{4}\left|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}+\cdots+\bar{z}_{n}\right| \\ =\left|z_{1}+z_{2}+z_{3}+\cdots \cdot+z_{n}\right|-\left|\overline {\overline{z_{1}}+\bar{z}_{2}+\bar{z}_{3}+\cdots+\cdots+\bar{z}_{n}}\right| \\ {[\because \bar{a}+\bar{b}=\overline{a+b}]}\end{array}$

$\begin{array}{l}=\left|z_{1}+z_{2}+z_{3}+\cdots \cdot+z_{n}\right|-\left|z_{1}+z_{2}+z_{3} \cdots+z_{n}\right| \\ {[\because|z|=|\bar{z}|]} \\ =0 \text { (Ans) } \\\end{array}$