Z1=2-3i Z2=15-5i/(2+i)^2 求Z1*Z2和Z1除以Z2

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Z1=2-3i Z2=15-5i/(2+i)^2 求Z1*Z2和Z1除以Z2
Z1=2-3i Z2=15-5i/(2+i)^2
求Z1*Z2和Z1除以Z2

Z1=2-3i Z2=15-5i/(2+i)^2 求Z1*Z2和Z1除以Z2
z2=(15-5i)/(2+i)² 【(2+i)²=3+4i】
=(15-5i)/(3+4i)
=[5(3-i)(3-4i)]/[(3+4i)(3-4i)]
=[5(5-15i)]/(25)
=1-3i
1、z1z2=(2-3i)(1-3i)=-7-9i
2、z1/z2=(2-3i)/(1-3i)=[(2-3i)(1+3i)]/[(1-3i)(1+3i)]=[11+3i]/(10)=(11/10)+(3/10)i

z2=(15-5i)/(2+i)^2 其中(2+i)^2=3+4i
=(15-5i)/(3+4i)
=[5(3-i)(3-4i)]/[(3+4i)(3-4i)]
=[5(5-15i)]/(25)
=1-3i
z1*z2=(2-3i)(1-3i)=-7-9i
z1/z2=2-3i/(1-3i)=(2-3i)(1+3i)/(1-3i)(1+3i)=(11+3i)/10