~~簡易不等式(2) <--> 絕對值不等式上篇~~

Let a,b be the complex number , and the conj. of a,b are A,B respectively.
Firstly, we consider
Re(bA) ≤ |bA| (by triangle law)
(即一個直角三角形的斜邊長過其他兩個邊)
2 Re(bA) ≤ 2|bA| (注1)
( Re(bA) means real part of bA)
bA+conj.(bA) ≤ 2|bA|
bA+conj.(bA) ≤ 2|ab| (Since |a|=|A|)
aA+bB+bA+aB ≤ aA + bB +2|ab| (Since conj.(bA)=aB)
(a+b)(A+B) ≤ |a|^2+|b|^2+2|ab| (since aA=|a|^2)
|a+b|^2 ≤ (|a|+|b|)^2
|a + b| ≤ |a| + |b| (since |a+b| >0)------(1)
for all a,b belong to complex


Now, Since
|aB| >= Re(aB) (by triangle Law)
-|aB| ≤ Re(aB) (注2)
-2|aB| ≤ 2 Re(aB)
-2|aB| ≤ conj.(aB)+aB (注1)
-2|ab| ≤ bA+aB (since |B|=|b|)
|a|^2 + |b|^2 -2|ab| ≤ |a|^2 + |b|^2 +bA+aB
|a|^2 + |b|^2 - 2|ab| ≤ aA + bB + bA+aB
(since aA=|a|^2,bB=|b|^2 )
|a|^2 + |b|^2 - 2|ab| ≤ (a+b)(A+B)
| |a| - |b| |^2 ≤ |a+b|^2 ( |a|,|b| are real )
| |a| - |b| | ≤ |a + b| (since | |a|-|b| |>0)----(2)
Then, - |a + b| ≤ |a| - |b| ≤ |a + b| (注3)
By (1), (2) , hence , the result follow.

(注1) For complex number a and the conj. a , where a=c+di ,so conj.(a)=c-di
=> a+conj.(a)= 2c = 2 Re(a)
(注2) if a>b (a,b are real), then -a<b
for example, 5>3 , hence -5<3
(注3) by definition, for real number c, -c ≤ |c| ≤c

2011-06-12 11:17:24 補充：
有關直角三角形的斜邊長過其他兩個邊...
Let直角三角形的三條邊係a,b,c,where c 係斜邊 and a,b,c>0
=> a^2+b^2=c^2
=> c^2=a^2+b^2 >= a^2 or c^2>=b^2 (since a,b >0)
有關conj. 即是 conjugate(共軛)

2011-06-13 14:02:43 補充：
To lop****** :
題目無指出a,b係實數 ,即題目沒有限制a,b係咩野...
你只證明兩題係成立當a,b係實數...
To 發問者 :
你只係想問兩題係成立當a,b係實數...??

2011-06-13 14:21:45 補充：
如果只須證明當a,b係實數..
by definition, a ≤ |a| and b ≤ |b| for any real number
| a+b | ≤ | |a| + |b| | = |a| + |b| (since |a|, |b|>0 )
'' |a + b| ≤ |a| + |b| '' ---(1) is true for all a,b are real number

2011-06-13 14:23:35 補充：
Using (1) with a|-->a+b and b|--> -b

2011-06-13 14:25:50 補充：
如果只須證明當a,b係實數..
by definition, a ≤ |a| and b ≤ |b| for any real number
| a+b | ≤ | |a| + |b| | = |a| + |b| (since |a|, |b|>0 )
'' |a + b| ≤ |a| + |b| '' ---(1) is true for all a,b are real number

2011-06-13 14:26:43 補充：
Using (1) with a|-->a+b and b|--> -b
|(a + b)+(-b)| ≤ |a+b| + |-b|
|a| ≤ |a+b| +|b| (since |b|=|-b|)
=>|a| - |b| ≤ |a + b| ------(2)
Since a,b are real, by (注2) and (2)
=> |a| - |b| >= -|a + b| (since |a+b|>0)-----(3)

2011-06-13 14:27:05 補充：
by (2),(3), - |a + b| ≤ |a| - |b| ≤ |a + b|
i.e. | |a| - |b| | ≤ |a + b| ------(4)
by (1),(4) , |a| - |b| ≤ |a + b| ≤ |a| + |b|
and | |a| - |b| | ≤ |a + b| ≤ |a| + |b| are true for all a,b are real number.

2011-06-15 07:17:10 補充：
指長度..
| 2 - 3i | = [2^2+(-3)^2]^(1/2)

回答者 :

請問複數有絕對值麼?
如果 2 - 3i 的絕對值 = ??