9.22

\(\bm{设z_1,z_2,z_3为复数，满足：\dfrac{z_2-z_1}{z_3-z_1} = \dfrac{z_1-z_3}{z_2-z_3}}\)
\(\bm{求证：|z_1-z_2| = |z_2-z_3| = |z_1-z_3|}\)

\(\bm{解法一：}设 k = \dfrac{z_2-z_1}{z_3-z_1} = \dfrac{z_1-z_3}{z_2-z_3}\)
\(即z_2-z_1 = k(z_3-z_1) = -k^2(z_2-z_3)①\)
\(只需证:|k| = 1\)
\(设z_2-z_1 = x,z_3-z_1 = y,则z_2-z_3 = x-y\)
\(①式化为：x = ky = -k^2(x-y)\)
\(代入x = ky\Rightarrow ky = -k^2(k-1)y\)
\(若y为0,z_1 = z_2 = z_3，会出现分母为0的情况，舍弃.\)
\(\therefore y\neq0,此时k\neq0，k^2-k+1 = 0\Rightarrow k = \dfrac{1±\sqrt{3}i}{2}\)
\(\therefore |k| = 1,结论成立\)

\(\bm{解法二：}\dfrac{z_2-z_1}{z_3-z_1} = \dfrac{z_1-z_3}{z_2-z_3}\Rightarrow |\dfrac{z_2-z_1}{z_3-z_1}| = |\dfrac{z_1-z_3}{z_2-z_3}|①\)
\(\dfrac{z_2-z_1}{z_3-z_1} = \dfrac{z_1-z_3}{z_2-z_3}\Rightarrow \dfrac{z_2-z_1}{z_3-z_1} - 1 = \dfrac{z_1-z_3}{z_2-z_3} - 1\)
\(即\dfrac{z_2-z_3}{z_3-z_1} = \dfrac{z_1-z_2}{z_2-z_3}\Rightarrow|\dfrac{z_2-z_3}{z_3-z_1}| = |\dfrac{z_1-z_2}{z_2-z_3}|②\)
\(由①②变形即证\)

\(\bm{解法三：}由解法二①②可知\angle z_1z_2z_3 = \angle z_2z_1z_3 = \angle z_1z_3z_2知三角形为等边三角形.\)