Question

复数（2k2－3k－2）+(k2＋k－6)i 胃纯虚数,且虚步小玉0,求实数K

Answer 1

z=(2k²-3k-2)+(k²+k-6)i

=(2k+1)(k-2)+(k-2)(k+3)i

由题设可得

(2k+1)(k-2)=0

(k-2)(k+3)＜0

解得:k=-1/2.