Question

已知复数zi+z/(1-i)=3+4i(i是虚数单位),则z=

Answer 1

设z=a+bi
则原式为(a+bi)i+z/(1-i)=3+4i
(a-3b)/2+(3a+b)/2=3+4i
从而(a-3b)/2=3,(3a+b)=4(虚部和实部对应相等)
解得：a=3,b=-1
∴就有z=3-i
祝你好运！~~

Answer 2

zi+z/(1-i)=3+4i
z[i+1/(1-i)]=3+4i
z(i-i²+1)/(1-i)=3+4i
z(2+i)/(1-i)=3+4i
z=(3+4i)(1-i)/(2+i)
z=(3+4i-3i-4i²)/(2+i)
z=(7+i)/(2+i)
z=(7+i)(2-i)/[(2+i)(2-i)]
z=(14+2i-7i-i²)/5
z=(15-5i)/5
z=3-i

Answer 3

好难，不会。