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?
Why are the roots of $y^2 – iy + 2 = 0$ not complex conjugates
algebra-precalculuscomplex numbersquadratics
Best Answer
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 $a$.