find the complex number that satisfies the following conditions Find all values of $z \in \Bbb C$ such that: $z + \bar{z} = 18$ and $z.\bar{z} = 84$.
I don't know how to get that values, someone can help me to solve this?
 A: The only complex number that satisfies $z + z = 2z = 18$ is $z = 9$, but $9^2 = 81$, so there are no values that satisfy both conditions.
A: If $z = a+bi$ then the first equation results in $2a = 18$ so $a = 9$.
The second equation results in $a² + b² = 84$ with $a = 9$ gives $b = \pm \sqrt{3}$.
So there are 2 solutions: $9 + \sqrt{3}i$ and its conjugate  $9 - \sqrt{3}i$. No heavy algebra needed!
A: Hint: (Please check first that you have written the question correctly, because $z+z$ looks odd) 
Any complex number can be written as $z=a+i b$ where $a,b$ are real numbers. Substitute into the given conditions,...., and solve for $a,b$.

Edit:
Now your question makes sense. Try this: $$z.\bar{z} =84$$
$$\Rightarrow z.(18-z) =84$$ $$z^2-18z+84=0$$ This is a quadratic equation. So....
A: $z$ and $\bar{z}$ are roots to the quadratic equation $X^2 - 18X + 84 = 0 $. By the quadratic formula, the roots are
$$ X = \frac{ 18 \pm \sqrt{ 18^2 - 4 \times 84 } } { 2 } = 9 \pm \sqrt{3} i$$

Note, you could do it slowly by setting $ z= a+bi$ and then solving for real values of $a$ and $b$. This gives you a direct immediate approach. 
A: $z + \bar{z} = 18$ and $z \bar{z} = 84$
.$z + \bar{z} = 2Re(z)$ 
so(...) 
$z + \bar{z} = 18 \rightarrow 2a = 18 \rightarrow a = 9$
. $z  \bar{z} = |z|^2$ 
so(...) 
$\left(\sqrt{9^2+b^2}\right)^2 = 84 \rightarrow 81 + b^2 = 84 \rightarrow b^2 = 3 \rightarrow b = \pm\sqrt{3}$
So, the solutions:
$9+\sqrt3i$ & $9-\sqrt3i$
