<s>[INST] 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?
 [/INST] 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.:
  $$x+yi=
\left(
  \sqrt{\frac{\sqrt{x^2+y^2}+x}{2}}+
  i\frac{y}{|y|} \sqrt{\frac{\sqrt{x^2+y^2}-x}{2}} \,
\right)^2$$

</s>