is $\ c= \big( \overline{x+yi} \big)^2 \ $ equivalent to $ \ \overline{c^{1/2}}=x+yi $?

Question

is $\ c= \bigg( \overline{x+yi} \bigg)^2 \ $ equivalent to $ \ \overline{c^{1/2}}=x+yi $?

EDIT

$c$ is complex, but not real.

Answer 1

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.

Answer 2

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

$c=\overline{(x+iy)}^{2}=(x-iy)^2=x^2 -2ixy+i^2 y=x^2-y^2-2xyi$