Why are the roots of y2−iy+2=0 not complex conjugates?

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champe547 · Accepted Answer

The &#039;conjugates rule&#039; only applies when the coefficients of the quadratic are real. Say the roots of a quadratic f(z) are α=a+bi and β=c+di. Then, by the factor theorem, the factorisation of f(z) must be(z−α)(z−β)=(z2−(α+β)z+αβ),.This means that in order for f(z) to have real coefficients,α+β∈R  and  αβ∈R,.The only time this is true when α and β are both real, or when α and β 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&#039;t form a complex conjugate pair. For example, α=5+7i and β=2+3i gives the equationz2−(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 formaz2+bz+c=0can be turned intoz2+Bz+C=0if we divide through by a.

RI5N6mv3 · Accepted Answer

For a degree-n polynomial function p(z)=∑i=0npizi with pn≠0, define p―(z)=∑ipi∗zi, which is the same polynomial iff each pi is real. Thenp―(z∗)=∑ipi∗z∗i=(p(z))∗, and p(z)=0⇔p―(z∗)=0. The complex-conjugates rule for polynomials with real coefficients is a special case of this.

Why are the roots of \(\displaystyle{y}^{{2}}-{i}{y}+{2}={0}\) not

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