Quadratic equation, thanks?

Let z=a+bi
z+6-i = (a+bi)+6-i = (a+6)+(b-1)i
z+6-i is an imaginary no. => a+6=0 , ∴ a= -6
So, z= -6+bi

(1-4i)z = (1-4i)(-6+bi) => it's imaginary part = bi-6(-4i) = (b+24)i
(1-4i)z is a real no. => b+24=0, ∴ b= -24

Thus, z=-6-24i
_______________________________________________________
要 Summary 嗎 ?
=============
1) To find complex number z --
1st let z=　a　+　bi, 改為 find a=?, b=?
　　　　　||　　　||
　　　real part　 imaginary part
2)z+6-i is an imaginary no. => "real part of z+6-i" = 0 => 可找到 a=?
3)(1-4i)z is a real no. => "imaginary part of (1-4i)z" = 0 =>可找到 b=?