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已知复数Z=3+bi，其中b∈R，i为虚数单位且（1+3i）z为纯虚数。⑴求复数z⑵若ω=z/2+i，求复数的ω模
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Answer 1

z=3+bi
(1+3i)*z = 3+bi+9i-3b = 3-3b+(9+b)i 为纯虚数
因此 3-3b=0，b=1
得 z=3+i
w=z/(2+i)
|w| = |z| / |2+i| = √10 / √5 = √2

Answer 2

解：
（1）z=3+bi
(1+3i)z
=(1+3i)(3+bi)
= 3+bi+9i-3b
= 3-3b+(9+b)i 为纯虚数
因此 3-3b=0，b=1
得 z=3+i
（2）
w=z/(2+i)
|w| = |z| / |2+i| = √10 / √5 = √2

Answer 3

（1+3i）z=3-3b+（9+b）i
又因为上式是纯虚数
故3-3b=0
b=1
z=3+i
w=3/2+3/2i
w的摸等于1.5x根号2

Answer 4

(1)
(1+3i)(3+bi)
=3+bi+9i+3bi^2
=(3-3b)+(9+b)i
∵是纯虚数
∴3-3b=0
b=1
∴z=3+i

(2)
w=z/(2+i)
=(3+i)(2-i)/(2+i)(2-i)
=(6-3i+2i-i^2)/(4-i^2)
=(7-i)/5
=7/5-i/5
|w|=√[(7/5)^2+(-1/5)^2]=√2

Answer 5

①
Z=3+bi
(1+3i)(3+bi)=3+bi+9i+3bi²=(3-3b)+(9+b)i
若(1+3i)(3+bi)为虚数
则9+b=0则b=-9则z=3-9i

②
∵ω=z/(2+i)=(3+i)(2-i)/(2+i)(2-i)=(6-3i+2i-i²)/(4-i²)=(7-i)/5=7/5-i/5
∴|ω|=√[(7/5)²+(-1/5)²]=√2