Question

若z1=a+3i,z2=3+4i,且z1/z2为纯虚数，则实数a=？详解 急

Answer 1

z1/z2
=(a+3i)/(3+4i)
=[(a+3i)(3-4i)]/[(3+4i)(3-4i)]
=(3a+12+9i-4ai)/(9+16)
=(3a+12+9i-4ai)/25
=(3a+12)/25 + [(9-4a)i]/25
z1/z2为纯虚数
所以(3a+12)/25=0
3a+12=0
a=-4

Answer 2

z1/z2为纯虚数,设z1/z2=bi(b∈R,且b≠0),
则z1=z2×bi,a+3i=(3+4i)bi=3bi+4b(i平方)=-4b+3bi,
两复数相等，实部虚部分别相等，得a=-4b且3=3b,解得b=1,a=-4,

Answer 3

z1/z2
=(a+3i)/(3+4i)
=(a+3i)(3-4i)/(3+4i)(3-4i)
=(3a+9i-4ai-12i²)/（9-16i²）
=(3a+12+9i-4ai)/25
=(3a+12)/25+(9-4a)i/25
实部(3a+12)/25，虚部(9-4a)i/25
为纯虚数则实部为0，即(3a+12)/25=0， 所以a=-4

Answer 4

解：由题意，可设z1/z2=ti(t∈R,t≠0).则z1=ti×z2.∴由题设条件可得：a+3i=(3+4i)ti=-4t+3ti.∴结合虚数相等定义，对比可得：a=-4t且3=3t.解得t=1,a=-4.∴a=-4.