finding all complex roots of equation let $z = 1 +i$ 
Find all complex solutions such that $z^2 + \bar z^2 = 0$. 
My working out:
$z^2 = -\bar z^2 = -(1-i)^2 = 2i$
so $z^2 = 2i$
hence $r^2 = 2 \implies r = \sqrt 2$
mod: $2\theta = \frac{\pi}{2} + 2k\pi \implies \theta = \frac{\pi}{4} + k(\pi)$ where $k = 0, 1$
overall roots are $z = \sqrt 2 \operatorname{cis} \left(\frac\pi4 +k\pi\right)$
Is my working out and solution correct?

You're working seems a bit confused. It's like you're verifying a solution and trying to solve for a solution at the same time. Do one or the other. Either substitute in $z=1+i$ and show that this satisfies the equation (no solving involved!) or substitute in $z=a+bi$ and solve to find the conditions on $z$ to satisfy the equation.
