A rectangle of maximum area is inscribed in the circle $|\mathrm{z}-3-4\mathrm{i}|=1$ .lf one vertex of the rectangle is 4 + 4i,then another adjacent vertex of this rectangle can be

Clearly, the inscribed rectangle is a square. Let the adjacent vertex be z.
Then,

$\frac{3+4\mathrm{i}-\left(\mathrm{z}\right)}{(3+4\mathrm{i})-(4+4\mathrm{i})}={\mathrm{e}}^{\pm \mathrm{i\pi }/2}$ (by rotation about center)

$\begin{array}{l}\Rightarrow 3+4\mathrm{i}-\mathrm{z}=\pm \mathrm{i}(-1)\\ \Rightarrow \mathrm{z}=3+4\mathrm{i}\pm \mathrm{i}=3+5\mathrm{i}\text{ or }3+3\mathrm{i}\end{array}$