回復 7# Christina 的帖子
第9題
湊一湊
\(\begin{align}
& z=\frac{32}{x-yi}=\left( \frac{32x}{{{x}^{2}}+{{y}^{2}}} \right)+\left( \frac{32y}{{{x}^{2}}+{{y}^{2}}} \right)i \\
& z=\left( x',y' \right)=\left( \frac{32x}{{{x}^{2}}+{{y}^{2}}},\frac{32y}{{{x}^{2}}+{{y}^{2}}} \right) \\
& \\
& {{\left( x+2 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=8 \\
& x=2\sqrt{2}\cos \theta -2,y=2\sqrt{2}\sin \theta +2 \\
& {{x}^{2}}+{{y}^{2}}={{\left( 2\sqrt{2}\cos \theta -2 \right)}^{2}}+{{\left( 2\sqrt{2}\sin \theta +2 \right)}^{2}}=16+8\sqrt{2}\left( \sin \theta -\cos \theta \right) \\
& x-y=-4-2\sqrt{2}\left( \sin \theta -\cos \theta \right) \\
& \\
& x'-y'=\frac{32\left( x-y \right)}{{{x}^{2}}+{{y}^{2}}}=\frac{32\left[ -4-2\sqrt{2}\left( \sin \theta -\cos \theta \right) \right]}{16+8\sqrt{2}\left( \sin \theta -\cos \theta \right)}=-8 \\
& x-y+8=0 \\
\end{align}\)
