我有兩條數想問equality of complex num

1.)if the result of(4a＋６i)-(5+8i) is a pure imaginary number,find the value of the real number a.

(4a＋6i)-(5+8i) = (4a – 5) + (6i -8i)
= (4a – 5) + (6i -8i)
=(4a – 5) - 2i

For a pure number – 2i, 4a -5 = 0
4a = 5
a = 5/4 = 1.25
For a pure number, a = 1.25


2.)if (1-2i)除(2-i)+a+bi+b-ai,find the values of values of real number a and b.
(1- 2i)    (1-2i) ( 2+ i)      2 -4i +i -2i^2          2 – 3i -2(-1)         4 – 3i
------- = ----------------- = --------------------- = ------------------ = -------------- 
(2 – i)    (2-i)(2 - i)           4 – i^2                    4 –(-1)                  5

0.8 – 0.6 i

a + b + bi – ai
(a + b) + (b-a)i

a + b = 0.8 -----------(1)
b-a = 0.6 ----------- (2)
From equation (2)
b = 0.6 +a ------- (3)
Substitute equation (3) into equation (1)
a + (0.6 + a) = 0.8
a + a = 0.8 – 0.6
2a = 0.2
a = 0.2/2 = 0.1
Substitute a = 0.1 into equation (3)
b = 0.6 + 0.1
b = 0.7

a = 0.1 and b = 0.7

2010-09-12 01:32:13 補充：
你在哪裡不明? 你匿名不收任何E-MAIL,沒法詳細解釋.
你要識 complex no. 除complex no. 才可計算
Both分子 and 分母要乘 (2 +i). 分母 (2-i) 要乘 (2 +i) 才得 4 – i^2

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