1. 把复数的辐角主值限定在 $$(-\pi, \pi]$$. 因为 $$z_k$$ 在以 $$(1,0)$$ 为心, $$r$$ 为半径的圆中, 从而

$|\arg z_k|\leq\arccos\sqrt{1-r^2},$

$|\arg z_k|\leq\arctan{\frac r{\sqrt{1-r^2}}}.$

$$\theta\in\left(0,\dfrac\pi2\right)$$, 且 $$n$$ 个复数 $$z_1$$, $$z_2$$, $$\dotsc$$, $$z_n$$ 满足 $$|\arg z_k|\leq\theta$$($$k=1$$, $$2$$, $$\dotsc$$, $$n$$), 则

$$\left|\sum_{k=1}^n z_k\right|\left|\sum_{k=1}^n\frac{1}{z_k}\right|\geq n^2\cos^2\theta.$$

$\Re(z_k)\Re\left(\frac{1}{z_k}\right)=\cos^2(\arg z_k)\geq\cos^2\theta.$

\begin{equation*}\begin{split}\left|\sum z_k\right|\left|\sum\frac1{z_k}\right| &\geq\Re\left(\sum z_k\right)\Re\left(\sum\frac1{z_k}\right)\\ &=\left(\sum\Re z_k\right)\left(\sum\Re\left(\frac1{z_k}\right)\right)\\ &\geq\left(\sum\sqrt{\Re z_k\Re\left(\frac1{z_k}\right)}\right)^2\\ &\geq(n\cos\theta)^2.\end{split}\end{equation*}

math.stackexchange 的 davik 有比 (1) 更强的

$$\theta\in\left(0,\dfrac\pi2\right)$$, 且 $$n$$ 个复数 $$z_1$$, $$z_2$$, $$\dotsc$$, $$z_n$$ 满足 $$|\arg z_k|\leq\theta$$($$k=1$$, $$2$$, $$\dotsc$$, $$n$$), 则

$$\left|z_1+z_2+\dotsb+z_n\right|\geq n\sqrt[n]{\left|z_1z_2\dotsm z_n\right|}\cos\theta.$$

\begin{equation*}\begin{split}\big|z_1+z_2+\dotsb+z_n\big| &\geq\Re\left(\sum_{k=1}^n z_k\right)\\&=\sum_{k=1}^n\Re z_k\\ &\geq n\sqrt[n]{\prod_{k=1}^n\Re z_k}\\ &\geq n\sqrt[n]{\prod_{k=1}^n \Big(|z_k|\cos\theta\Big)}\\ &=n\sqrt[n]{\left|\prod_{k=1}^n z_k\right|}\cos\theta.\end{split}\end{equation*}

\begin{equation*} \left|\frac1{z_1}+\frac1{z_2}+\dotsb+\frac1{z_n}\right|\geq \frac{n\cos\theta}{\sqrt[n]{\left|z_1z_2\dotsm z_n\right|}}.\end{equation*}

$$\dfrac{\Re{z_k}}{|z_k|}\geq\sqrt{1-r^2}$$

($$k=1$$, $$2$$, $$\dotsc$$, $$n$$) 为真.

$\left|\frac{\Im{z_k}}{\Re{z_k}}\right|\leq\frac r{\sqrt{1-r^2}}.$

$z_1+z_2+\dotsb+z_n=(x_1+x_2+\dotsb+x_n)+i(y_1+y_2+\dotsb+y_n),$

$\sum_{k=1}^n\frac1{z_k}=\sum_{k=1}^n\frac{\bar{z}_k}{|z_k|^2}=\sum_{k=1}^n\frac{x_k}{x_k^2+y_k^2}-i\sum_{k=1}^n\frac{y_k}{x_k^2+y_k^2}.$

\begin{equation*}\begin{split}\left |\sum_{i=1}^n z_i\right|\left|\sum_{i=1}^n\frac1{z_i}\right | &=\sqrt{\left(\sum_{k=1}^nx_k\right)^2+\left(\sum_{k=1}^ny_k\right)^2}\sqrt{\left(\sum_{k=1}^n\frac{x_k}{x^2_k+y^2_k}\right)^2+\left(\sum_{k=1}^n\frac{y_k}{x^2_k+y^2_k}\right)^2}\\&\geq\left|\left(\sum_{k=1}^n x_k\right)\left(\sum_{k=1}^n\frac{x_k}{x^2_k+y^2_k}\right)\right|+\left |\left(\sum_{k=1}^ny_k\right)\left (\sum_{k=1}^n\frac{y_k}{x^2_k+y^2_k}\right)\right|\\ &\geq \left(\sum_{k=1}^n x_k\right)\left (\sum_{k=1}^n\frac{x_k}{x^2_k+y^2_k}\right)\\&\geq\left(\sum_{k=1}^n\frac{x_k}{\sqrt{x^2_k+y^2_k}}\right)^2\\ &\geq\left(n\sqrt{1-r^2}\right)^2\\&=n^2(1-r^2).\end{split}\end{equation*}

2. Pascal theorem 说明 $$P$$, $$Q$$, $$R$$ 三点共线.

$\frac{SK}{AC}=\frac{SQ}{AR}.$

$\frac{TK}{AB}=\frac{TQ}{AP}.$

$\frac{SK}{TK}=\frac{SQ}{TQ}\cdot\frac{AP}{AR}=\frac{SQ}{TQ}\cdot\frac{\sin\angle ARP}{\sin\angle APR}=\frac{\frac{SQ}{\sin\angle SPQ}}{\frac{TQ}{\sin\angle TRQ}}=\frac{\frac{PQ}{\sin\angle QSP}}{\frac{RQ}{\sin\angle QTR}}=\frac{PQ}{RQ}.$

3. 首先, 我们有 $$0\notin B$$.

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