Question

完成下列复数运算！请高手帮忙看看!

(1) (2 + j3) + (5 - j2)
(2) 5 - (3 + j2)
(3) (2 + j) × (-1 + j4)
(4) (1 + j8) ÷ (1 - j2)
(5) (3 + j6 ) ÷ 3j

Answer 1

1）原式=7+j
2) 原式=2-j2
3)原式=-2-j+j8+j²4=-2+j7-4=-6+j7
4)设原式=u+jw
(u+jw)(1-j2)=(u+2w)+j(w-2u)=1+j8
∴u+2w=1 w-2u=8 => 5w=10 => w=2 => u=-3
∴原式=-3+j2
5)原式=1/j+2=2-j 【=j/j²+2=-j+2】

Answer 2

(2 + j3) + (5 - j2)=7+i
5 - (3 + j2)=2-2i
(2 + j) × (-1 + j4)=-2-i+8i-4=-2+7i
(1 + j8) ÷ (1 - j2)=(1 + j8) *(1 +j2)/((1-2i)(1+2i))=(1+2i+8i-16)/(1+4)=-3+2i
(3 + j6 ) ÷ 3j=(1 + 2i ) ÷ j=2-i