Question

3．實(shí)數(shù)m為何值時(shí)，復(fù)數(shù)z=$\frac{{m}^{2}+m-6}{m+5}$+（m²+8m+15）i
（Ⅰ）為實(shí)數(shù)；
（Ⅱ）為純虛數(shù)；
（Ⅲ）對(duì)應(yīng)點(diǎn)在第二象限．

Answer 1

分析 （Ⅰ）z為實(shí)數(shù)?m²+8m+15=0且m+5≠0，解得mj即可．
（Ⅱ）z為純虛數(shù)?$\left\{\begin{array}{l}{\frac{{m}^{2}+m-6}{m+5}=0}\\{{m}^{2}+8m+15≠0}\end{array}\right.$，解出即可．
（III）z對(duì)應(yīng)的點(diǎn)在第二象限?$\left\{\begin{array}{l}{\frac{{m}^{2}+m-6}{m+5}＜0}\\{{m}^{2}+8m+15＞0}\end{array}\right.$，解出即可．

解答 解：（Ⅰ）z為實(shí)數(shù)?m²+8m+15=0且m+5≠0，解得m=-3．
（Ⅱ）z為純虛數(shù)?$\left\{\begin{array}{l}{\frac{{m}^{2}+m-6}{m+5}=0}\\{{m}^{2}+8m+15≠0}\end{array}\right.$，
解得m=2；
（III）z對(duì)應(yīng)的點(diǎn)在第二象限?$\left\{\begin{array}{l}{\frac{{m}^{2}+m-6}{m+5}＜0}\\{{m}^{2}+8m+15＞0}\end{array}\right.$，
解得m＜-5或-3＜m＜2．

點(diǎn)評(píng) 本題考查了復(fù)數(shù)為實(shí)數(shù)及純虛數(shù)的充要條件、幾何意義，考查了計(jì)算能力，屬于基礎(chǔ)題．