Question

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

Answer 1

设z=x+yi，
(1+3i)(x+yi)=(x-3y)+(3x+y)i
为纯虚数，所以x-3y=0，x=3y，
w=(x+yi)/(2+i)
|w|^2=(x^2+y^2)/(2^2+1^2)=10y^2/5=2y^2
5√2=|w|=√2|y|,
|y|=5,
y=±5
x=±15
w=±5(3+i)/(1+2i)
=±5(3+i)(1-2i)/[(1+2i)(1-2i)]
=±5(5-5i)/5
=±(5-5i)
w=±(5-5i).

Answer 2

设(1＋3i)z=bi，则z=(bi)/(1＋3i)，w=z/(2＋i)=(bi)/[(1＋3i)(2＋i)]，|w|=|(bi)/[(1＋3i)(2＋i)]|=|b|/[√10×√5]=5√2，解得|b|=50，则：w=(bi)/[(1＋3i)(2＋i)]=±(1－i)

Answer 3

设(1+3i）z=mi,z=(-3m+mi)/10,w=(-3m+mi)/20+i=(-3m+(m+20)i)/20,|w|=5倍根号2;求出m,可得w

Answer 4

解:设W=a+bi，∵（１+３i)Z为纯虚数，∴(1+3i)(a+bi)=(a-3b)+(3a+3b)i中a-3b=0,∴a=3b
W=3b+bi,且|w|=5倍根号2,(3b)平方＋（b)平方＝５０，即b的平方＝５，∴Ｗ＝３倍根号５＋３倍根号５i
或w＝－３倍根号５－３倍根号５i