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kuing
Posted 2021-10-11 19:22
记 `z=x+iy`,易得
\begin{align*}
x^2-2ixy+x-y(y-i)&=z+\bar z^2,\\
x^2+2ixy+x-y(y+i)&=\bar z+z^2,
\end{align*}由 `z` 在单位圆上知 `z\bar z=\abs z^2=1`,故
\begin{align*}
\LHS&=\frac {\bar z(z+\bar z^2)}{z(\bar z+z^2)}=\frac {1+\bar z^3}{1+z^3}=\frac {z^3+z^3\bar z^3}{z^3(1+z^3)}=\frac 1{z^3},\\
\RHS&=\frac {(\bar z+1)^3}{(z+1)^3}=\frac {(z\bar z+z)^3}{z^3(z+1)^3}=\frac 1{z^3}.
\end{align*} |
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