complex number shading

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.

你 a part 都成功搵到 l z - 5 l = 2, 你只是唔知佢代表什麼.
其實, z = x + yi, 所以
l z - 5 l = 2
==> |x + yi - 5| = 2
==> (x - 5)^2 + y^2 = 4
這是一條圓形方程, 圓心是 (5, 0), 半徑是 2.
所以 l z - 5 l < 2 的範圍是一個圓形, 就是以 (5, 0) 為圓心, 半徑為 2. 不包括圓周. (因為題目用 "<")