Ta có $wzi + 3w = z + 1 - i$
$\iff z(wi - 1) = 1 - i - 3w$
Lấy mô-đun hai vế có $2\sqrt{2} |wi - 1| = |1 - i - 3w|$
Đặt $w = a + bi$
Suy ra $2\sqrt{2} |-b - 1 + ai| = |1 - 3a - (3b+1)i|$
$\iff 8[(b+1)^2 + a^2] = (1-3a)^2 + (3b+1)^2$
$\iff (a-3)^2 + (b-5)^2 = 40$
...