[a (3-2a)i 6] 5为纯虚数怎么判断
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/10 10:36:04
为纯小数则0
z=a+2i/1+i+(3-i)=(a+2i)(1-i)/(1+1)+3-i=(a+2-ai+2i)/2+3-i=a/2+4-a/2i因为纯虚数所以实部a/2+4=0a/2≠0a=-8再问:=(a+2
(a-i)(3-i)=3a-1-(3+a)i因为(a-i)(3-i)为纯虚数所以3a-1=0a=1/3
(根号5)-2
由复数a+3i1+2i=(a+3i)(1−2i)(1+2i)(1−2i)=(a+6)+(3−2a)i5=a+65+3−2a5i是纯虚数,则a+65=03−2a5≠0,解得a=-6.故选A.
a+3i/1+2i=(a+3i)(1-2i)/[(1+2i)(1-2i)]=[(a+6)+(3-2a)i]/5=(a+6)/5+(3-2a)i/5要是纯虚数,则实部等于0,即(a+6)/5=0,即a=
(a+3i)/(1+2i)=(a+3i)(1-2i)/(1-4i²)=(a-2ai+3i-6)/(1+4)=(a-6)/5+(3-2a)i/5;因为是纯虚数;所以a-6=0;a=6;很高兴为
/>(a+3i)/(1+2i)=(a+3i)(1-2i)/[(1+2i)(1-2i)]=(a-2ai+3i+6)/5=(a+6)/5+(3-2a)i/5复数为纯虚数,实部=0,虚部≠0令(a+6)/5
(a+3i)/(1+2i)=(a+3i)(1-2i)/[(1+2i)(1-2i)]=(a-2ai+3i+6)/(1+4)=[a+6+(3-2a)i]/5因为复数a+3i/1+2i(a属于R,i为虚数单
再问:为什么就等于0了
z=(a²-3a+2)+(a-1)i是纯虚数,则:1、a²-3a+2=0,得:a=1或a=22、a-1≠0,得:a≠1则:a=2
(a+i)/(2-i)=[(a+i)(2+i)]/[(2+i)(2-i)]=(2a+ai+2i+i²)/(4-i²)=[(2a-1)+(a+2)i]/5=(2a-1)/5+(a+2
复数(a+3i)(1-2i)=a+6+(3-2a)i,复数是纯虚数,所以a+6=0,可得a=-6.故选C.
由z=(a2+2a-3)+(a+3)i为纯虚数可得(a2+2a-3)=0,且(a+3)≠0解方程可a=1故选D
复数a+2i与1+3i的积为(a+2i)(1+3i)=a-6+(3a+2)i为纯虚数,∴a-6=0,3a+2≠0,∴a=6,故选C.
(a^2+a-6)/(a+2)=0,a^2-5a+6≠0所以选B
Z=(a+2i)(1-i)/2+3-i=a/2+4+(1-a)i复数Z=[(a+2i)/(1+i)]+(3-i)为纯虚数a/2+4=0a=-8
因为复数z=(a*2-3a+2)+(a-1)i为纯虚数,所以(a*2-3a+2)=0,(a-1)≠0(a-1)(a-2)=0所以a=2
1.(a-i)^2=a²-1-2ai,所以a²-1=0,a=±12.sin[(pi/2)-x]=cosx=3/5cos2x=2cos²x-1=2*9/25-1=-7/25
若复数z=(2-i)(a-i),(i为虚数单位)为纯虚数,则实数a的值为______. 答案: