PURE MATHS

Let z' is the conjuate of z

(a) |(z - 1)/(z - 4)| = 2

|z - 1| = 2|z - 4|

(z - 1)(z - 1)' = 4(z - 4)(z - 4)'

z^2 - (z + z') + 1 = 4[z^2 - 4(z + z') + 16]

3z^2 - 15(z + z') + 63 = 0

z^2 - 5(z + z') + 21 = 0

(z - 5)(z - 5)' + 21 - 25 = 0

|z - 5| = 2

(b) Method 1:

|(z - 1)/(z - 4)| < 2

z^2 - (z + z') + 1 < 4[z^2 - 4(z + z') + 16]

3z^2 - 15(z + z') + 63 > 0

z^2 - 5(z + z') + 21 > 0

(z - 5)(z - 5)' + 21 - 25 > 0

|z - 5| > 2

So, the shaded area is the outside region of the circle |z - 5| = 2

Method 2 :

Sub. 6 into |(z - 1)/(z - 4)| = 5/2 = 2.5 > 2.

Since 6 is in |z - 5| = 2, we conclude that the shaded area is the outside region of the circle |z - 5| = 2.

因為 |z| = r < a
因為 a 大於其 半徑 r 所以就響出面=.=

我唔想下載個檔案
但係我相信都係complex number既past paper
用一個set話某一個modulus of a complex number < a real number果種題目
做呢種題目，首先要畫左個啱既圓圈，仲要畫啱佢d intercepts
之後你代(0+0i)入個complex number入面
睇下佢符唔符合個condition
如果符合，就係包含(0+0i)果個係個shaded area

如果我解錯題你再講啦。

2012-03-08 20:43:45 補充：
BTW, 以我做左n年complex number pp既經驗，通常|z| a係圓既入面

未代高考生共勉之

2012-03-08 20:45:11 補充：
|z| < a 通常係圓既外面
|z| > a 通常係圓既入面