If $(a+bi{)}^{11}=x+iy,$ where $a,b,x,y\in R,$ then ${(b+ai)}^{11}$ equals

$\overline{(a+bi{)}^{11}}=\overline{x+iy}$

$\begin{array}{l}\Rightarrow (a-bi{)}^{11}=x-iy\\ \Rightarrow \left[\right(-i\left)\right(b+ia){]}^{11}=-i(y+ix)\\ \Rightarrow (-i{)}^{11}(b+ia{)}^{11}=-i(y+ix)\end{array}$

As $(-i{)}^{11}=(-i{)}^8(-1{)}^3{i}^3=i,$
We get $(b+ia{)}^{11}=-(y+ix)=-y-ix$