令 $z=re^{i\theta}$, 写出 $\frac{e^{it}+z}{e^{it}-z}$ 的实部和虚部
\[
\mathrm{Re}(\frac{e^{it}+z}{e^{it}-z})=\frac{1-r^2}{1-2r\cos(\theta-t)+r^2},
\]
\[
\mathrm{Im}(\frac{e^{it}+z}{e^{it}-z})=\frac{2r\sin(\theta-t)}{1-2r\cos(\theta-t)+r^2}.
\]
若令 $r\in[0,1)$, 则 $\mathrm{Re}(\frac{e^{it}+z}{e^{it}-z}) > 0$. 我们记 $P_r(\theta-t)=\mathrm{Re}(\frac{e^{it}+z}{e^{it}-z})$.
References:
W. Rudin, 《实分析和复分析》
1
\[
\begin{split}
\frac{e^{it}+z}{e^{it}-z}&=\frac{e^{it}+re^{i\theta}}{e^{it}-re^{i\theta}}\\
&=\frac{(\cos t+i\sin t)+r(\cos\theta+i\sin\theta)}{(\cos t+i\sin t)-r(\cos\theta+i\sin\theta)}\\
&=\frac{(\cos t+r\cos\theta)+i(\sin t+r\sin\theta)}{(\cos t-r\cos\theta)+i(\sin t-r\sin\theta)}\\
&=\frac{\bigl[(\cos t+r\cos\theta)+i(\sin t+r\sin\theta)\bigr]\cdot\bigl[(\cos t-r\cos\theta)-i(\sin t-r\sin\theta)\bigr]}{(\cos t-r\cos\theta)^2+(\sin t-r\sin\theta)^2}\\
&=\frac{(\cos t+r\cos\theta)(\cos t-r\cos\theta)+(\sin t+r\sin\theta)(\sin t-r\sin\theta)+i\Bigl[(\sin t+r\sin\theta)(\cos t-r\cos\theta)-(\cos t+r\cos\theta)(\sin t-r\sin\theta)\Bigr]}{\cos^2 t-2r\cos t\cos\theta+r^2\cos^2\theta+\sin^2 t-2r\sin t\sin\theta+r^2\sin^2\theta}\\
&=\frac{\cos^2 t-r^2\cos^2\theta+\sin^2 t-r^2\sin^2\theta+i\Bigl[\sin t\cos t-r\sin t\cos\theta+r\sin\theta\cos t-r^2\sin\theta\cos\theta-(\cos t\sin t-r\sin\theta\cos t+r\cos\theta\sin t-r^2\sin\theta\cos\theta)\Bigr]}{1+r^2-2r(\cos t\cos\theta+\sin t\sin\theta)}\\
&=\frac{1-r^2+i\bigl[2r(\sin\theta\cos t-\sin t\cos\theta)\bigr]}{1+r^2-2r\cos(t-\theta)}\\
&=\frac{1-r^2+i2r\sin(\theta-t)}{1-2r\cos(t-\theta)+r^2}\\
\end{split}
\]
因此
\[
\mathrm{Re}(\frac{e^{it}+z}{e^{it}-z})=\frac{1-r^2}{1-2r\cos(t-\theta)+r^2},
\]
\[
\mathrm{Im}(\frac{e^{it}+z}{e^{it}-z})=\frac{2r\sin(\theta-t)}{1-2r\cos(t-\theta)+r^2}.
\]