复数Z=[(a+2i)/(1+i)]+(3-i)为纯虚数,则实数a=______.

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复数Z=[(a+2i)/(1+i)]+(3-i)为纯虚数,则实数a=______.

复数Z=[(a+2i)/(1+i)]+(3-i)为纯虚数,则实数a=______.
复数Z=[(a+2i)/(1+i)]+(3-i)为纯虚数,则实数a=______.

复数Z=[(a+2i)/(1+i)]+(3-i)为纯虚数,则实数a=______.
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=0
a=-8

Z=(a+2i)(1-i)/2+3-i
=a/2+3-ai/2
复数Z为纯虚数,a/2+3=0
a=-8

Z=[(a+2i)/(1+i)]+(3-i)
=a+(a+2)i-2+3-i
=a+1+(a+1)i
Z不可能是纯虚数

Z=(a+2i)(1-i)/((1+i)(1-i))+(3-i)
=(a/2+4)-a/2i
a/2+4=0
a=-8

Z=[(a+2i)(1-i)/(1+i)(1-i)]+(3-i)
=[(a+2)+(2-a)i]/2+(3-i)
=[(a+2)/2+3]+[(2-a)/2-1]i
Z为纯虚数,则
(a+2)/2+3=0, (2-a)/2-1≠0
解得a=-8
代入(2-a)/2-1=4≠0
∴a=-8