Thank you for reporting, we will resolve it shortly
Q.
The centre of a regular hexagon is at the point $ z=i $ . If one of its vertices is at $ 2+i, $ then the adjacent vertices of $ 2+i $ are at the points
Given, $ z=i $ Let $ {{z}_{1}}+1+i(1\pm \sqrt{3}) $ and $ {{z}_{2}}=2+i $ Now, $ |{{z}_{2}}-z|=|2+i-i| $
$=2 $
As we know that the distance from the centre to every vertices is equal.
Now, $ |{{z}_{1}}-z|=|1+i(1\pm \sqrt{3})-i| $
$=|1\pm i\sqrt{3}|=\sqrt{{{1}^{2}}+{{(\sqrt{3})}^{2}}} $
$=2 $
Hence, option (d) is correct.