I'm trying to solve the equation $$S^2 = \begin{bmatrix}1&0\\0&-1\end{bmatrix}$$ to conclude at least one coefficient of $S$ will have to be a complex number. I let $$S = \begin{bmatrix} a & b \\ c & d\end{bmatrix},$$ from which I obtain
\begin{align*}
a^2 + bc &= 1\\
ab + bd &= 0\\
ac + dc &= 0\\
bc + d^2 &= – 1.
\end{align*}
Question. How should I do it? It seems a found one solution in my investigation $4$ below. Perhaps you could optimize your reading here by going straight to it and then seeing my question in investigation $5$. (Thank you!)
Investigation 1. I noticed $ab$ and $bd$ either cannot both have the same sign (because they must sum up to zero) or then must both be zero. But if they're both zero, then either $b = 0$ or $a = d = 0$. (But I don't know where else to go. Too many either-or here.)
Investigation 2. I know $bc = 1 – a^2$ and then substituting $bc$ in $bc + d^2 = -1$, I get $d^2 – a^2 = -2$, which looks interesting and implies $(d – a)(d + a) = -2$, but I didn't find a way to deduce a complex solution from there either.
Investigation 3. I know $b$ and $a + d$ cannot both be nonzero because one of the equations in the middle implies $b (a + d) = 0$ and the complex field is an integral domain. (Similarly for $c$ and $a + d$.)
Investigation 4. From investigation $1$ and $3$, I can conclude that if $b$ is not zero zero then $ab$ and $bd$ would both be zero meaning $a = d = 0$. So maybe there's a solution here. Why don't I let $b = 0$ first? That implies $a = \pm 1$ and $bc = 0$, from which we can deduce that $d^2 = -1$, so $d$ will have to be $\pm \sqrt{-1}$, as desired.
Investigation 5. Because I found one solution in investigation $4$ and it is a complex solution, that doesn't mean that all possible solutions would all be complex. So I would somehow need to prove that this system has a unique solution, which is the one I found in investigation $4$.
Best Answer
Perhaps simpler:
$$Eq. 2:\;ab+bd=0\iff(a+d)b=0\implies b=0\;\;or\;\;a=-d$$
$$(1)\;\;\text{If}\;\;b=0\implies \;Eq. \;4\;\;d^2=-1\implies d=\pm i\in\Bbb C\setminus\Bbb R$$
$$(2)\;\;\;\;\;\;\;\;\;\;\text{If}\;\;a=-d\implies\;Eqs.\; 2,3\;\;\text{are always true and}\;\; Eq. 4\;\;a^2+bc=-1$$
But as remarked by the comment below, this last equation equation is then identical with equation 1, so they contradict each other and thus this case cannot be.