Adding Absolute value to a complex number: $ z+| z|=2+8i$ I would like to know my error in this problem.
Find the complex number such that:
$$ z+|z|=2+8i$$
So far, I have:
$$
\begin{split}
a+bi+\sqrt{a^2+b^2} &= 2 + 8i\\
a^2-b^2+a^2+b^2&=4-64\\
2a^2 -b^2 + b^2&=-60\\
a^2&=-30
\end{split}
$$
But I should end up with
$$a^2=-15$$
No matter how hard I try, I can't seem to find what I did wrong. Any suggestions?

$a+bi+\sqrt{a^2+b^2} = 2 + 8i$ so
$a + \sqrt{a^2 + b^2} = 2$ and $b = 8$.
So $a + \sqrt{a^2 + 64} = 2$
So $\sqrt{a^2 + 64} = 2- a$
$a^2 + 64 = 4 -4a + a^2$
$4a = -60$
$a = -15$.
$z = -15 + 8i$.
....
To do what you were attempting
You have to realize that the $Re(z) = a + \sqrt{a^2 + b^2}$ and $Im(z) = b$.  I think somehow you were thinking there were threee parts $Re(z)=a$ and $Im(z) = b$ and some $Weird(z) = \sqrt{a^2 + b^2}$ and that $z\overline z = Re^2(z) - Im^2(z) + Weird^2(z)$.  That simply isn't true....
$(a+bi+\sqrt{a^2+b^2})(a - bi +\sqrt{a^2 + b^2}) = (2 + 8i)(2-8i)$
$(a + \sqrt{a^2 + b^2})^2 - b^2 = 4 - 64$
$2a^2 + b^2 + 2a\sqrt{a^2 + b^2} -b^2 = -60$
$a^2 + a \sqrt{a^2 + b^2} = -30$
which is a pain to solve but can be done.
