Question

已知|z|=r，求2z+3-4i對(duì)應(yīng)的點(diǎn)的軌跡方程.

Answer 1

思路點(diǎn)撥：設(shè)出2z+3-4i所對(duì)應(yīng)的復(fù)數(shù)ω，則2z=ω-3+4i，利用減法的幾何意義可得點(diǎn)的軌跡.

解：設(shè)ω=2z+3-4i，則2z=ω-3+4i，|ω-(3-4i)|=2r.∴ω對(duì)應(yīng)點(diǎn)的軌跡是以(3,-4)為圓心2r為半徑的圓.

[一通百通]由減法的幾何意義知|z-z₁|表示復(fù)平面上兩點(diǎn)Z,Z₁間的距離.當(dāng)|z-z₁|=r時(shí)，表示復(fù)數(shù)z對(duì)應(yīng)的點(diǎn)的軌跡是以z₁對(duì)應(yīng)的點(diǎn)為圓心，半徑為r的圓；當(dāng)|z-z₁|=|z-z₂|時(shí)，表示以復(fù)數(shù)z₁,z₂的對(duì)應(yīng)點(diǎn)為端點(diǎn)的線段的垂直平分線；當(dāng)|z-z₁|+|z-z₂|=2a(2a＞||＞0)時(shí)，表示以復(fù)數(shù)z₁,z₂的對(duì)應(yīng)點(diǎn)為焦點(diǎn)的橢圓；當(dāng)|z-z₁|-|z-z₂|=2a(0＜2a＜||)時(shí)，表示以復(fù)數(shù)z₁,z₂的對(duì)應(yīng)點(diǎn)為焦點(diǎn)的雙曲線.