Express conjugate z in terms of z I need to solve for z in equation below.
${{z}^{2} = \bar{z}}$
If I can substitute ${\bar{z}}$ in terms of $ {z}$, then I can go about solving this equation.
Question
What would be this substitute expression?
UPDATE
There are four roots for this equation given at back of the book and they are ${0,1, -\frac{1}{2} + \frac{\sqrt3}{2}i,-\frac{1}{2} - \frac{\sqrt3}{2}i }$.
I can see how to get ${0,1}$, but confused about the last two answers in above list. I can assume ${z^{2}}$ is a pure real number since only for real numbers the  complex number equals it's conjugate, and this will lead me to the first two roots.
 A: We don't need that 
Let $z=a+ib$ where $a,b$ are real.
$\implies a^2-b^2+i(2ab)=a-ib$
Now equate the real & the imaginary parts.
A: Multiply both sides by $z$ to obtain
$$z^3=\bar z z=|z|^2 $$
In particular, $z^3\in\Bbb R$. Now apply modulus to the above to find
$|z|^3=|z|^2$, i.e., $|z|=0$ or $|z|=1$.  Hence you need only check $z=0$ and the three solutions of $z^3=1$.
A: As $|z|=|\overline z|$
Taking modulus we have $|z|^2=|\overline z|\iff|z|(|z|-1)=0$
If $|z|\ne0, |z|=1$ let $z=e^{it}$ where $t$ is real.
So, we have $(e^{it})^2=e^{-it}\iff e^{3it}=1=e^{2m\pi i}$ where $m$ is any integer
$t=2m\pi/3$
Alternatively, start with $z=re^{it}$
A: It seems to me that you are trying to solve the following
$$z^2=\bar{z}$$
and not the following
$$|z|^2=\bar{z}\,.$$
In that case, replacing $\bar{z}$ with a function of $z$ (such as, $\Re(z)+i \Im(z)$, where $\Re$ and $\Im$ denotes the real and imaginary part of $z$) seems overly complicated to me.
Instead, I would suggest to replace $z$ with $x+i\,y$, where $x=\Re(z)$ and $y=\Im(z)$. These replacements will yield the following
$$x^2-y^2+i \,2\, x\,y = x - i\,y\, .$$
Solving for the imaginary part yields the following
$$ 2\,x\,y = -y \implies x = -\frac12 \text{ or } y=0$$
Replacing $x=-\frac12$ this into the real part yields the following
$$ \frac14-y^2 = -\frac12 \implies y=\pm \sqrt{\frac34}$$
Replacing $y=0$ this into the real part yields the following
$$ x^2 = x \implies x\in\{0,1\}$$
Putting all together yields the following four roots for the equation $z^2=\bar{z}$ $$z\in\Big\{0,1,-\frac12\pm\sqrt{\frac34}\Big\}.$$
Hope this helps. 
A: Given $z^{2}=\bar {z} $ , this  implies  $ |z|^{2}=|\bar{z}|$ or,  $ z \bar {z}=|\bar{z}| $  or, $ |z|^{2}=|z| $. Then , $ (|z|-1)|z|=0$ , implies $ either \ |z|-1=0 , \ or \ |z|=0 $.$ \begin {align}Then \ either \ z=1,or, z=i, \ or, -i\ or,  \ or \ z=0 \end{align} $
