问题
填空题
若z1=a+3i，z2=3+4i，且
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答案
=z1 z2
=a+3i 3+4i
=(a+3i)(3-4i) (3+4i)(3-4i)
，∴(3a+12)+(9-4a)i 25
解得a=-43a+12=0 9-4a≠0
故答案为：-4
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若z1=a+3i，z2=3+4i，且
|
=z1 z2
=a+3i 3+4i
=(a+3i)(3-4i) (3+4i)(3-4i)
，∴(3a+12)+(9-4a)i 25
解得a=-43a+12=0 9-4a≠0
故答案为：-4