Solving $a^2-b^2-a=0$ ∧ $-2ab+b={1\over 2}$ I want to solve
 $$a^2-b^2-a=0 ∧ -2ab+b={1\over 2}$$
I'm a bit stuck in the first part. I thought of the solutions, $a={1±\sqrt {1+4b^2} \over 2}$. But can't figure out solutions to the second part. Am I approaching it the right way?
 A: Let $z=a-bi$, then $z^2-z=(a^2-b^2-a)+(-2ab+b)i$
\begin{align*}
  z^2-z &= \frac{i}{2} \\
  z &= \frac{1\pm \sqrt{1+2i}{}}{2} \\
  a-bi &=
  \frac{1}{2}
  \left(
    1 \pm \sqrt{\frac{\sqrt{5}+1}{2}} \pm i\sqrt{\frac{\sqrt{5}-1}{2}} \,
  \right) \\
  a &= \frac{1}{2} \pm \sqrt{\frac{\sqrt{5}+1}{8}} \\
  b &= \mp \sqrt{\frac{\sqrt{5}-1}{8}} \\
\end{align*}

N.B.:
  $$x+yi=
\left(
  \sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}+
  i\frac{y}{|y|} \sqrt{\frac{\sqrt{x^2+y^2}-x}{2}} \,
\right)^2$$

A: I would solve it as $b=\pm \sqrt{a^2-a}$ and $b=\frac{1}{2(1-2a)}$.
So $$a^2-a = \frac{1}{4(1-2a)^2}$$
Or:
$$\left(2a-1\right)^2 - 1 = \frac{1}{(1-2a)^2}$$
Letting $u=(2a-1)^2$, you have $u-1=\frac{1}{u}$ or $u^2-u-1=0$ or $u=\frac{1\pm\sqrt{5}}{2}$. Then: 
$$a=\frac{1\pm\sqrt{\frac{1\pm\sqrt{5}}{2}}}{2}$$
Not that $a$ is complex if $u=\frac{1-\sqrt{5}}{2}$.
So for real solutions, we get $a=\frac{1\pm\sqrt{\frac{1+\sqrt{5}}{2}}}{2}$ and $$b=\frac{1}{2(1-2a)}=\frac{1}{\mp 2\sqrt{\frac{1+\sqrt{5}}{2}}}$$
A: $$
\begin{cases}
\text{a}^2-\text{b}^2-\text{a}=0\\
\\
\text{b}-2\text{a}\text{b}=\frac{1}{2}
\end{cases}\space\space\space\Longleftrightarrow\space\space\space
\begin{cases}
\text{a}=\frac{1\pm\sqrt{1+4\text{b}^2}}{2}\\
\\
\text{b}-2\times\frac{1\pm\sqrt{1+4\text{b}^2}}{2}\times\text{b}=\frac{1}{2}
\end{cases}
$$
Now:
$$\text{b}-2\times\frac{1\pm\sqrt{1+4\text{b}^2}}{2}\times\text{b}=\pm\text{b}\sqrt{1+4\text{b}^2}=\frac{1}{2}\space\space\space\Longleftrightarrow\space\space\space\text{b}=\begin{cases}
\pm\frac{1}{2}\sqrt{\frac{\sqrt{5}-1}{2}}\\
\\
\pm\frac{i}{2}\sqrt{\frac{1+\sqrt{5}}{2}}
\end{cases}$$
