complex conjugate variables in a system of complex linear equations I was wondering how to solve a linear equation of the below type. 
In the below equation, z=x+yi, a complex number. 
$$(2+2i)z^*+(2-2i)z=10+(2-2i)(2+2i)$$
So, in the above equation, there are two unknown variables and only one equation. I have learned about normal con-Sylvester matrix equation, but they don't seem to fit this problem. 
I understand that it could be like:
$$2*Re((2+2i)z^*)=10+(2-2i)(2+2i)$$
But the problem is how to find the value of x and y individually. 
 A: Let $z=x+yi.$  The equation $(2+2i)z^*+(2-2i)z=10+(2-2i)(2+2i)=10+4+4=18$
then becomes $(2+2i)(x-yi)+(2-2i)(x+yi)$
$=(2x+2y+(2x-2y)i)+(2x+2y+(2y-2x)i)=4x+4y=18,$ 
so it is solved when $x+y=\dfrac{18}4=\dfrac9 2$.  
There are infinitely many solutions. 
A: Hint:
I suggest that you write $z$ as $x + i y$ and then expand all products.
Afterwards, split the equations in two parts: the real part and the imaginary part. This will leave you with a system of two equations with two unknowns.

Edit: Just for the sake of completeness, and after NIT_GUP solved the problem, here is the solution:
$$
(2+2i)z^∗ + (2−2i)z = 10 + (2−2i)(2+2i) \\
(2+2i)(x - i y) + (2−2i)(x + i y) = 10 + (2^2 + 2^2) \\
(2x \color{red}{+ 2ix} \color{blue}{- 2iy} + 2y) + (2x \color{red}{- 2ix} \color{blue}{+ 2iy} + 2y) = 18 \\
4x + 4y = 18 \\
x + y = \frac{9}{2}
$$
So, in this case we have a degenerate solution because the complex terms cancelled out in the equations above. In this case, the solution is a straight line in the complex plane with the equation:
$$
y = - x + \frac{9}{2} \ .
$$

Only after writing this I've noticed that J. W. Tanner had already written a solution.
