請教高手解第8題複數方程式?
回答 (1)
2017-11-15 1:24 pm
8
x^2－(1+3i)x+(－2+i)=0
D=(1+3i)^2－4*1*(－2+i)
=1+6i－9+8－4i
=2i
=2[Cos(2nπ+π)+iSin(2nπ+π)]
D^(1/2)=√2*[Cos(nπ+π/2)+iSin(nπ+π/2)]
(1) n=0
D^(1/2)=√2*[Cos(π/4)+iSin(π/4)]=√2*(√2/2+i√2/2)=1+i
x=[(1+3i)+(1+i)]/2=1+2i
(2) n=1
D^(1/2)=√2*[Cos(3π/4)+iSin(3π/4)]=√2*(－√2/2+i√2/2)=－1+i
x=[(1+3i)－1+i]/2=i
x^2－(1+3i)x+(－2+i)=0
D=(1+3i)^2－4*1*(－2+i)
=1+6i－9+8－4i
=2i
=2[Cos(2nπ+π)+iSin(2nπ+π)]
D^(1/2)=√2*[Cos(nπ+π/2)+iSin(nπ+π/2)]
(1) n=0
D^(1/2)=√2*[Cos(π/4)+iSin(π/4)]=√2*(√2/2+i√2/2)=1+i
x=[(1+3i)+(1+i)]/2=1+2i
(2) n=1
D^(1/2)=√2*[Cos(3π/4)+iSin(3π/4)]=√2*(－√2/2+i√2/2)=－1+i
x=[(1+3i)－1+i]/2=i
收錄日期: 2021-04-30 22:31:49
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171115042828AANsXpX