Question

已知復(fù)平面內(nèi)正方形的三個(gè)頂點(diǎn)所對(duì)應(yīng)的復(fù)數(shù)分別是1+2i,-2+i,-1-2i,求第四個(gè)頂點(diǎn)所對(duì)應(yīng)的復(fù)數(shù).

Answer 1

解：設(shè)正方形的三個(gè)頂點(diǎn)Z₁、Z₂、Z₃對(duì)應(yīng)的復(fù)數(shù)分別為1+2i、-2+i、-1-2i，點(diǎn)Z₄為正方形的第四個(gè)頂點(diǎn),它對(duì)應(yīng)的復(fù)數(shù)為x+yi,則

∴（-2+i）-(1+2i)=(-1-2i)-(x+yi),

即-3-i=(-1-x)+(-2-y)i.

∴

∴第四個(gè)頂點(diǎn)對(duì)應(yīng)的復(fù)數(shù)為2-i.