已知z,w为复数,(1+3i)乘z为纯虚数,w=2+i/5z,且lwl=10倍根号2,求w

Question

已知z,w为复数,(1+3i)乘z为纯虚数,w=2+i/5z,且lwl=10倍根号2,求w

Answer 1

设z=a+bi(a,b为实数),
(1+3i)(a+bi)=a+bi+3ai-3b=(a-3b)+(b+3a)i,
则a-3b=0,a=3b,
z=3b+bi=b(3+i)
w=2+i／5z
=2+i／5b(3+i)
=2+i(3-i)／5b(9+1)
=2+1／50b+3i／50b,
|w|=✓[(2+1／50b)^2+(3／50b)^2]=10✓2,
(100b+1)^2+3^2=200×2500b^2
10000b^2+200b+10=500000b^2
49000b^2-20b-1=0
△<0,
b不为实数,无解.