Complex Number ( 高手請進 )

(1) 簡單來說,每一個complex number 都可以用Argand Diagram 上的一點代表,這是定義, ,就像座標一樣,你會比較座標上兩點的大小嗎?

(2) If Arg x = Arg y+ 2nπ, 從Argand Diagram 上來說, 它們是在同一位置的,而且x是y的某一個倍數, 例如 x=3+4i, y=9+12i, 得出來的argument 會一樣, y=3x, 所以其實x=ky for any non negative real k.
| x + y |=| ky + y |=(k+1)|y|

| x | + | y | = | ky | + | y |= (k+1)|y|= | x + y |

If | x + y | = | x | + | y | , then squaring both side (x bar is x conjugate)

因為 xxbar=| x |^2, so

| x + y |^2 = (| x | + | y |)^2

(x+y)(x+y)bar= | x |^2+| y |^2+2| x || y |

(x+y)(x+y)bar= | x |^2+| y |^2+xybar+yxbar = | x |^2+| y |^2+2| x || y |

because xybar+yxbar= xybar+bar(xybar) =2 Re(xybar)= 2| x || y |

so Re(xybar)= | x || y |= | x || y bar| = | xybar| (因為| y |= | y bar |)

一個複數的modulus會等於real part, 除非Imaginary part=0, 所以 xybar is a real number

A real number has a argument of 0

so Arg(xybar)=0

but Arg(xybar)= Arg(x)-Arg(y)+2kπ=0

所以 Arg(x) = Arg(y)-2kπ = Arg(y)+2nπ where n=-k of some integer

2007-08-26 02:55:37 補充：
應該話係 A real number has a argument of kπ where k is a integer

1. Suppose ordering is meaningful, WLOG, assume i > 0, then i^2 = -1 > 0 is impossible. SImilarly, assume i < 0, then i^2 = -1 > 0 (note: multiply a negative number will change the inequality sign)

In both cases, contradiction exists, so ordering is not possible.


2. | x + y | = | x | + | y |
iff O, x, y are collinear
iff arg x - arg y = 2nπ for some integer n (by definition of colinearity)
iff arg x = arg y + 2nπ for some integer n

1)
咁梗係啦, 對 real number 黎講,
佢地係條 number line 上面有左右之分,
左邊梗係大過右邊果 d
而 complex number 係個平面上面,
咁你話點至為之大同細
唔係好難明者
2)
let y = r(cos@ + i sin@)
so, x = R [ cos(@+ 2nπ) + i sin(@+ 2nπ) ]
x = R(cos@ + i sin@)
| x + y | = | (r+R)(cos@ + i sin@) | = r+R = | x | + | y |
so, equality sign holds
if part 搞掂, 下面做埋 only if 果 part
if | x + y | = | x | + | y |
呢度慢一步比埋你...