Why are the roots of $y^2 - iy + 2 = 0$ not complex conjugates? So I have the equation $y^2 - iy + 2 = 0$. I use the quadratic formula to solve it, and the solutions I got are $2i$ and $-i$. Why aren't the solutions here complex conjugates? Why didn't the quadratic formula produce complex conjugates as solutions in this case?
 A: The 'conjugates rule' only applies when the coefficients of the quadratic are real. Say the roots of a quadratic $f(z)$ are $\alpha=a+bi$ and $\beta=c+di$. Then, by the factor theorem, the factorisation of $f(z)$ must be
$$
(z-\alpha)(z-\beta)=(z^2-(\alpha+\beta)z+\alpha\beta) \, .
$$
This means that in order for $f(z)$ to have real coefficients,
$$
\alpha + \beta \in \mathbb{R} \text{ and } \alpha\beta \in \mathbb{R} \, .
$$
The only time this is true when $\alpha$ and $\beta$ are both real, or when $\alpha$ and $\beta$ form a complex conjugate pair; it is a good exercise to verify this. Otherwise, $f(z)$ would have non-real coefficients, and there would be no requirement that its roots form a complex conjugate pair. I can easily think up a quadratic where the roots don't form a complex conjugate pair. For example, $\alpha = 5+7i$ and $\beta=2+3i$ gives the equation
$$
z^2-(7+10i)z+(-11+29i) = 0 \, .
$$

Throughout this answer, I assume that the leading coefficient of the quadratic is $1$. However, this is okay, because any quadratic equation of the form
$$
az^2+bz+c=0
$$
can be turned into
$$
z^2+Bz+C = 0
$$
if we divide through by $a$.
A: For a degree-$n$ polynomial function $p(z)=\sum_{i=0}^np_iz^i$ with $p_n\ne0$, define $\bar{p}(z):=\sum_ip_i^\ast z^i$, which is the same polynomial iff each $p_i$ is real. Then $\bar{p}(z^\ast)=\sum_ip_i^\ast z^{\ast i}=(p(z))^\ast$, and $p(z)=0\iff\bar{p}(z^\ast)=0$. The complex-conjugates rule for polynomials with real coefficients is a special case of this.
A: You can also see the problem by "completing-the-square":
$$ y^2 \ - \ iy \ + \ 2 \ \ = \ \ \left(y^2 \ - \ 2·\frac{i}{2} · y \ + \ \frac{i^2}{2^2} \right) \ + \ \left(2 \ -  \ \frac{i^2}{2^2} \right) \ \ = \ \ \left(y \ - \ \frac{i}{2}  \right)^2  \ + \ \frac{9}{4} \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \left(y \ - \ \frac{i}{2}  \right)^2  \  = \  -\frac{9}{4} \ \ \Rightarrow \ \ y \ \ = \ \ \frac{i}{2} \ \pm \ \frac{3i}{2} \ \ .  $$
This hopefully makes the issue a bit easier to see than by applying the quadratic formula.
Making reference to the formula, the zeroes of a quadratic polynomial are $ \ -\frac{b}{2a} \ \pm \ \frac{\sqrt{\Delta}}{2a} \ \ , $ with $ \ \Delta \ $ being the discriminant.  With all of the coefficients as real numbers, the only way to obtain a complex number is to have $ \ \Delta < 0 \ \ , $ which produces   zeroes $ \ -\frac{b}{2a} \ \pm \ \ s·i \ \  ; $ these represent a "conjugate-pair" since the second term will be a pure imaginary number.  When we permit complex coefficients, the two zeroes of the quadratic polynomial can become $ \ (\alpha \ + \  \beta·i) \ \pm \ (\gamma \ + \ \delta·i) \ \ , $ which will not generally be conjugate. (With $ \ a \ $ no longer real, the second term is not necessarily purely imaginary.)
