is $\ c= \big( \overline{x+yi} \big)^2 \ $ equivalent to $ \ \overline{c^{1/2}}=x+yi $? is $\ c= \bigg( \overline{x+yi} \bigg)^2 \ $ equivalent to $ \ \overline{c^{1/2}}=x+yi $?
EDIT
$c$ is complex, but not real. 
 A: Take $x=0, y=-1$. Then statement 1 becomes $c=-1$.
However, statement 2 becomes $ \ \overline{c^{1/2}}=i$. This is definitely not the same as the first statement, since $c=-1$. implies $ \ \overline{c^{1/2}}=-i $ or $i$. So it is a more general statement than statement 2.
A: You have $c=a+ib$, then $c^{\frac 12}=[r(\cos x+i \sin x)]^{\frac 12}$, where $r=\sqrt{a^2+b^2}$ and $x=arctan(\frac{b}{a})$
Now by Moivre's formula 
$c^{\frac 12}=
r^{\frac 12} \left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin 
\left( \dfrac {x+2\pi k}{2}\right )\right ],k=0,1$.
Finally when calculating the conjugate you will have to  
$\overline{c^{\frac 12}}=
\overline{r^{\frac 12} \left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin 
\left( \dfrac {x+2\pi k}{2}\right )\right ]} ,k=0,1$.
$\overline{c^{\frac 12}}=
\overline{r^{\frac 12}} \overline{\left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin 
\left( \dfrac {x+2\pi k}{2}\right )\right ]} ,k=0,1$.
$\overline{c^{\frac 12}}=
r^{\frac 12} \overline{\left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )+i \sin 
\left( \dfrac {x+2\pi k}{2}\right )\right ]} ,k=0,1$.
$\overline{c^{\frac 12}}=
r^{\frac 12} \left[ \cos \left(
\dfrac {x+2\pi k}{2}\right )-i \sin 
\left( \dfrac {x+2\pi k}{2}\right )\right ] ,k=0,1$.
On the other hand
$c=\overline{(x+iy)}^{2}=(x-iy)^2=x^2 -2ixy+i^2 y=x^2-y^2-2xyi$
