If $\left|{\mathrm{z}}_1\right|=2$ and $(1-\mathrm{i}){\mathrm{z}}_2+(1+\mathrm{i}){\overline{\mathrm{z}}}_2=8\sqrt{2}$ then the  minimum value of $\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|$ is

We have,$\left|{\mathrm{z}}_1\right|=2$

z₁ lies on the circle having center at origin and radius 2

and $(1-\mathrm{i}){\mathrm{z}}_2+(1+\mathrm{i}){\overline{\mathrm{z}}}_2=8\sqrt{2}$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(1-\mathrm{i})\left({\mathrm{x}}_2+{\mathrm{iy}}_2\right)+(1+\mathrm{i})\left({\mathrm{x}}_2-{\mathrm{iy}}_2\right)=8\sqrt{2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{x}}_2+{\mathrm{y}}_2=4\sqrt{2}\end{array}$

So, z₂ lies on the straight line $\mathrm{x}+\mathrm{y}=4\sqrt{2}$

${\left|{\mathrm{z}}_1-{\mathrm{z}}_2\right|}_{min}=$ shortest distance between circle and straight line = AB = OB- OA =4-2=2