Question

設(shè)關(guān)于x的方程是x²-(tanθ+i)x-(2+i)=0,

(1)若方程有實(shí)根,求銳角θ和實(shí)數(shù)根;

(2)求證：對(duì)任意θ≠kπ+(k∈Z)方程無純虛數(shù)根.

Answer 1

（1）解：設(shè)方程的實(shí)根為a，則a²-（tanθ+i）a-（2+i）=0，

即a²-tanθ·a-2-(a+1)i=0.

∵a·tanθ∈R，∴

∴a=-1且tanθ=1.

又0＜θ＜,∴θ=,a=-1.

(2)證明：假設(shè)方程有純虛數(shù)根bi(b∈R,b≠0),則

(bi)²-(tanθ+i)·bi-(2+i)=0.

∴-b²+b-2=0,                              ①

且-tanθ·b-1=0.                                   ②

由②得b=-cotθ,代入①得cot²θ+cotθ+2=0,此方程Δ=1-8＜0,∴cotθ為虛數(shù),與cotθ∈R矛盾,假設(shè)不成立.∴原方程對(duì)于任意實(shí)數(shù)θ不可能有純虛數(shù)根.