Complex number quadratic Is the following equation

$$z^2 + z^* + \frac14 = 0$$

where $z$ is a complex number and $z^*$ is its conjugate
completely separate from ordinary quadratic equations? i.e. can I use the discriminant, quadratic formula etc. If not what, what type of equation is this? Can z* be treated independently from z? How is the degree related to the number of roots which is 4 (2 real, 2 complex) I believe.
p.s. which specific topics could I look at to help me understand this further?
 A: Well, notice that all complex numbers can be written as:
$$\text{s}=\Re\left(\text{s}\right)+\Im\left(\text{s}\right)i\tag1$$
So, we get (when $\text{z}\in\mathbb{C}$ and $\text{n}\in\mathbb{R}$):


*

*$$\Re\left(\text{z}^2+\overline{\text{z}}+\text{n}\right)=\Re^2\left(\text{z}\right)-\Im^2\left(\text{z}\right)+\Re\left(\text{z}\right)+\text{n}\tag2$$

*$$\Im\left(\text{z}^2+\overline{\text{z}}+\text{n}\right)=2\cdot\Re\left(\text{z}\right)\cdot\Im\left(\text{z}\right)-\Im\left(\text{z}\right)\tag3$$


We end up with the following system of equations:
$$
\begin{cases}
\Re^2\left(\text{z}\right)-\Im^2\left(\text{z}\right)+\Re\left(\text{z}\right)+\text{n}=0\\
\\
2\cdot\Re\left(\text{z}\right)\cdot\Im\left(\text{z}\right)-\Im\left(\text{z}\right)=0
\end{cases}\space\Longleftrightarrow\space\begin{cases}\Re^2\left(\text{z}\right)-\Im^2\left(\text{z}\right)+\Re\left(\text{z}\right)+\text{n}=0\\
\\
\Im\left(\text{z}\right)\cdot\left(2\cdot\Re\left(\text{z}\right)-1\right)=0
\end{cases}\tag4
$$
So, we get:
$$\Im\left(\text{z}\right)\cdot\left(2\cdot\Re\left(\text{z}\right)-1\right)=0\space\Longleftrightarrow\Im\left(\text{z}\right)=0\space\vee\space\Re\left(\text{z}\right)=\frac{1}{2}$$
A: As $z^*=-z^2-\frac{1}{4}$
$$ z^2+z^*+ \frac{1}{4} =0\Leftrightarrow (z^*)^2+z+\frac{1}{4} =0\Leftrightarrow (-z^2-\frac{1}{4})^2+z+\frac{1}{4}=0 $$
$$\Leftrightarrow z^4 +\frac{z^2}{2}+\frac{1}{16}+z+\frac{1}{4}=0$$ 
$$\Leftrightarrow 16z^4+8z^2+16z+5=0$$
$$\Leftrightarrow (2z+1)^2(4z^2-4z+5) = 0$$
A: Here is a more geometric approach. We have 
$$z^{2}+z^{*} = -\frac{1}{4} \in \mathbb{R}$$
so $z^{2}+z^{*}$ is real. Write this in polar form 
$$r^{2}e^{2i\theta}+re^{-i\theta}=-\frac{1}{4}$$
to obtain the constraint
$$r^{2}\sin(2\theta)-r\sin(\theta)=0$$
so that either $r=0$ (so $z=0$), or $\sin(\theta)=0$ (so $z \in \mathbb{R}$), or 
$$2r\cos(\theta)-1=0 \implies r\cos(\theta)=\frac{1}{2}$$
which we recognise as the condition $\Re(z)=\frac{1}{2}$.
$z=0$ is not a solution; for real $z$ we get the repeated solution $z=-1/2$, and if we say $z=1/2+iy$, the equation becomes:
$$\frac{1}{4}+iy-y^{2}+\frac{1}{2}-iy+\frac{1}{4}=0$$
which simplifies to $y^{2}=1$. We easily verify that $1/2 \pm i$ are both solutions. As you note, this is four roots (with multiplicity) - contrast this with an ordinary quadratic equation, which always has $2$ roots over $\mathbb{C}$.
A: $$z=a+ib\\z^*=a-ib\\\text{while $a,b\in\mathbb{R}$}\\z^2+z^*+\frac{1}{4}=\\(a+ib)^2+(a-ib)+\frac{1}{4}=\\a^2+2iba-b^2+a-ib=-\frac{1}{4}\in\mathbb{R}\\\therefore a^2+2iba-b^2+a-ib\in\mathbb{R}$$
so we can create 2  equations:
$$\begin{cases}
2iba-ib=0
\\[2ex]
a^2-b^2+a=-\frac{1}{4}
\end{cases}$$
we can solve this:$$ib(2a-1)=0\implies b=0\text{ or }a=\frac{1}{2}$$
let's solve for $b=0$ first: $a^2+a=-\frac{1}{4}\implies a=-\frac{1}{2}\\\therefore z=a+ib=-\frac{1}{2}+0i=-\frac{1}{2}$
now let's solve for a=$\frac{1}{2}$: $\frac{1}{2}^2+\frac{1}{2}-b^2=-\frac{1}{4}\implies1=b^2\implies b=\pm1\\\therefore z=a+ib=\frac{1}{2}\pm1i=\frac{1}{2}\pm i$
∎
A: Answering your question in order:


*

*Is it completely separate from ordinary quadratic equations? Yes, it is.

*can I use the discriminant, quadratic formula etc.? Not at all.

*If not what, what type of equation is this? It is a kind of non-$\Bbb{C}$-linear non-polynomial equation, in particular it is a system of $2$ real equations in $2$ unknowns, one quadratic and the other of first degree in both variables.

*Can z* be treated independently from z? No, they are related by a non-$\Bbb{C}$-linear operator.

*How is the degree related to the number of roots which is 4 (2 real, 2 complex)? No relation you can envisage generally speaking, that is, the "fundamental theorem of Algebra" cannot be applied here.

*which specific topics could I look at to help me understand this further? Complex field fundamentals.
Your confusion, maybe, comes from the fact that you want to intepret this $z^2+z^*+\frac{1}{4}=0$ as a quadratic equation over $\Bbb{C}$. But on the left hand side there is not a polynomial in one indeterminate over $\Bbb{C}$ because the operator $^*$ complex conjugation is not representable as a multiplication by a complex scalar (there does not exists an element $a\in\Bbb{C}$ such that $az=z^*$). (So you cannot apply here the "fundamental theorem of Algebra": that is, you can't immediatly say that such non-$\Bbb{C}$-linear non-polynomial equation has a root over $\Bbb{C}$). So you need to revert to interpreting it as a system of $2$ real equations over $\Bbb{R}$ in two indeterminates. As a tutorial you can easily approach it by using vector notation over $\Bbb{R}$ and translating complex numbers into $\Bbb{R}$-bidimensional vectors and operator on $\Bbb{C}$ into operator on $\Bbb{R}^2$. The only operator here is the complex conjugate operator which is a $\Bbb{R}$-linear operator on $\Bbb{R}^2$ representable as a multiplication by the following matrix:
$$
J=
\begin{bmatrix}
1&\phantom-0\\
0&-1
\end{bmatrix}
$$
So you have
$
\begin{bmatrix}
x^2-y^2\\
2xy
\end{bmatrix}
+
J\begin{bmatrix}
x\\
y
\end{bmatrix}
+\frac{1}{4}
\begin{bmatrix}
1\\0
\end{bmatrix}
=
\begin{bmatrix}
0\\0
\end{bmatrix}
\tag{1}$
From here the real vector solutions can also be interpreted as complex scalar solutions, while complex vector solutions have no interpretation in terms of complex scalar roots (try with a constant term of $1$ instead of $1/4$ and you'll see). You can equally have obtained all real vector solutions with the $y$ component not null that means that no real scalar solution would exist.
The fact that you set deliberately $z$ to be a real scalar in this non-linear equations does not mean that you get a real solution. 
Concluding. This is a scenario of non-$\Bbb{C}$-linear non-polynomial equation:
if you search for solutions over $\Bbb{C}$, you can get zero solutions or one or more real and/or  complex solutions, if you search for solutions over $\Bbb{R}^2$, you can get zero solutions or one or more real and/or complex vector solutions, among which real vector solutions can be interpreted as complex scalar solutions over $\Bbb{C}$.
