If ${\mathrm{x}}_{\mathrm{n}}+{\mathrm{iy}}_{\mathrm{n}}={(1+\mathrm{i})}^{\mathrm{n}}$ then ${\mathrm{x}}_{\mathrm{n}-1}{\mathrm{y}}_{\mathrm{n}}-{\mathrm{x}}_{\mathrm{n}}{\mathrm{y}}_{\mathrm{n}-1}=$

$$\begin{gathered} {\mathrm{x}}_{\mathrm{n}}+{\mathrm{iy}}_{\mathrm{n}}={(1+\mathrm{i})}^{\mathrm{n}}={\sqrt{2}}^{\mathrm{n}} {\left(\frac{1+\mathrm{i}}{\sqrt{2}}\right)}^{\mathrm{n}} \\ \Rightarrow {\mathrm{x}}_{\mathrm{n}}+{\mathrm{iy}}_{\mathrm{n}}={\sqrt{2}}^{\mathrm{n}}{\left(\cos \frac{\mathrm{\pi }}{4}+\mathrm{isin}\frac{\mathrm{\pi }}{4}\right)}^{\mathrm{n}} \\ = {\sqrt{2}}^{\mathrm{n}}\left(\cos \frac{\mathrm{n\pi }}{4}+\mathrm{isin}\frac{\mathrm{n\pi }}{4}\right) \\ \Rightarrow {\mathrm{x}}_{\mathrm{n}}={\sqrt{2}}^{\mathrm{n}} \cos \frac{\mathrm{n\pi }}{4}, {\mathrm{y}}_{\mathrm{n}}={\sqrt{2}}^{\mathrm{n}} \sin \frac{\mathrm{n\pi }}{4} \\ \mathrm{and} {\mathrm{x}}_{\mathrm{n}-1}={\sqrt{2}}^{\mathrm{n}-1} \cos (\mathrm{n}-1)\frac{\mathrm{\pi }}{4} {\mathrm{y}}_{\mathrm{n}-1}={\sqrt{2}}^{\mathrm{n}-1} \sin (\mathrm{n}-1)\frac{\mathrm{\pi }}{4} \\ \Rightarrow {\mathrm{x}}_{\mathrm{n}-1} {\mathrm{y}}_{\mathrm{n}}-{\mathrm{x}}_{\mathrm{n}} {\mathrm{y}}_{\mathrm{n}-1}={\sqrt{2}}^{\mathrm{n}}.{\sqrt{2}}^{\mathrm{n}-1}\left(\sin \frac{\mathrm{n\pi }}{4} \cos (\mathrm{n}-1)\frac{\mathrm{\pi }}{4}-\cos \frac{\mathrm{\pi }}{4}\mathrm{Tan}(\mathrm{n}-1)\frac{\mathrm{\pi }}{4}\right) \\ =\frac{2^{\mathrm{n}}}{\sqrt{2}} \sin \left(\frac{\mathrm{n\pi }}{4}-(\mathrm{n}-1)\frac{\mathrm{\pi }}{4}\right)=\frac{2^{\mathrm{n}}}{\sqrt{2}}.\sin \frac{\mathrm{\pi }}{4} \\ =\frac{2^{\mathrm{n}}}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=\frac{2^{\mathrm{n}}}{2} =2^{\mathrm{n}-1} \end{gathered}$$