Solving $\overline {(z+1+5i)}= 3z+9−9i$ 
Solve for $z$, and give your answer in the form $a+bi$.
$$\overline {(z+1+5i)}= 3z+9−9i$$

I have gotten to where I have:
$$z=3\bar z+8+4i$$
But I am stuck here and could use some guidance.
 A: Let $z=a+ib$ with $a$ and $b$ real. Then you get $a+ib=3(a-ib)+8+4i$. Compare real and imaginary parts and you will see that $a=-4, b=1$. So the answer is $z=-4+i$.
A: Given
$z=3\overline z+8+4i,$
next take the conjugate equation:
$\overline z=3z+8-4i.$
Eliminate $\overline z$ between these equations to get
$z=3(3z+8-4i)+8+4i$
and solve by ordinary algebraic methods.

 $-8z=32-8i, z=-4+i.$

A: We wish to solve the equation:
$$\overline {(z+1+5i)}= 3z+9−9i$$
(Here, $z \in \mathbf{C}$)
We simplify the given equation as
$$
\bar{z} + 1 - 5 i = 3 z + 9 - 9 i
$$
i.e.
$$
\bar{z} - 3 z + 4 i - 8 = 0 \tag{1}
$$
Let $z = x + i y$. Then $\bar{z} = x - i y$.
Substituting above into (1), we get
$$
x - i y - 3 (x + i y) + 4 i - 8  = 0
$$
or
$$
- 2 x - 4 i y + 4 i - 8 = 0
$$
or
$$
x + 2 i y - 2 i + 4 = 0
$$
or
$$
(x + 4) + 2 i (y - 1) = 0 \tag{2}
$$
Equating the real and imaginary parts on both sides of (2), we get
$$
x = -4 \ \ \mbox{and} \ \ y = 1
$$
Thus, the solution of the given equation is:
$$
z = x + i y = -4 +  i 
$$
i.e.
$$
\boxed{z = -4 + i}
$$
