答案:6 11
7.计算|(3-i)+(-1+2i)-(-1-3i)|=________.
解析:|(3-i)+(-1+2i)-(-1-3i)|=|(2+i)-(-1-3i)|=|3+4i|= =5.
答案:5
8.已知z1=a+(a+1)i,z2=-3b+(b+2)i(a,b∈R),若z1-z2=4,则a+b=________.
解析:∵z1-z2=a+(a+1)i-[-3b+(b+2)i]=+(a-b-1)i=4,
由复数相等的条件知
解得∴a+b=3.
答案:3
9.计算:
(1)(1+2i)+(3-4i)-(5+6i);
(2)5i-[(3+4i)-(-1+3i)];
(3)(a+bi)-(2a-3bi)-3i(a,b∈R).
解:(1)(1+2i)+(3-4i)-(5+6i)
=(1+3-5)+(2-4-6)i=-1-8i.
(2)5i-[(3+4i)-(-1+3i)]
=5i-(4+i)
=-4+4i.
(3)(a+bi)-(2a-3bi)-3i
=(a-2a)+[b-(-3b)-3]i
=-a+(4b-3)i(a,b∈R).
10.设z1=x+2i,z2=3-yi(x,y∈R),且z1+z2=5-6i,求z1-z2.
解:∵z1=x+2i,z2=3-yi,
∴z1+z2=x+3+(2-y)i=5-6i,