Roots, polynomials and equality breaking

Question

Suppose I have

$$ z=(a+ib)\implies z^2=(a+ib)^2\implies z=\pm(a+ib) $$

But if I do this instead:

$$ \begin{align} z&=a+bi\\ \implies z^2&=a^2-b^2+2iab\\ &=a(2a+2ib-a)-b^2\\ &=a(2z-a)-b^2\\ z^2-2az+a^2+b^2&=0\\ &\implies z=a\pm ib \end{align} $$

Why the contradiction?

Answer 1

There is no contradiction. In the first derivation, you have shown that if $z=a+ib$, then either $z=a+ib$ or $z=-a-ib$. That is true, because $z=a+ib$.

In the second, you have shown that if $z=a+ib$, then either $z=a+ib$ or $z=a-ib$. That is true, because $z=a+ib$.

Answer 2

There is no contradiction: $z=a+ib$ implies both $z=\pm (a+ib)$ and $z=a\pm ib$. To be clear, a statement like $z=\pm(a+ib)$ in this context means "$z=a+ib$ or $z=-(a+ib)$". If $z=a+ib$, then certainly any "or" statement where one of the options is $z=a+ib$ is also true, and that applies to both $z=\pm(a+ib)$ and $z=a\pm ib$.