For $az^2 + bz + c = 0$ with $a, b, c \in \mathbb{R}$:
$$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
| $\Delta = b^2 - 4ac$ | Roots |
|---|---|
| $\Delta > 0$ | Two distinct real roots |
| $\Delta = 0$ | One repeated real root |
| $\Delta < 0$ | Two complex conjugate roots |
When $\Delta < 0$: $\sqrt{\Delta} = i\sqrt{|\Delta|}$.
Theorem: If $P(z)$ has real coefficients and $P(\alpha + \beta i) = 0$ (with $\beta \neq 0$), then $P(\alpha - \beta i) = 0$.
Proof: Since $P$ has real coefficients, $\overline{P(z)} = P(\bar{z})$. Taking conjugates of $P(\alpha+\beta i) = 0$ gives $P(\alpha-\beta i) = \overline{P(\alpha+\beta i)} = 0$. $\square$
Implication: Non-real roots of polynomials with real coefficients always come in conjugate pairs. A polynomial of odd degree over $\mathbb{R}$ must have at least one real root.
Example 1: Solve $z^2 + 4z + 13 = 0$.
$$\Delta = 16 - 52 = -36$$
$$z = \frac{-4 \pm 6i}{2} = -2 \pm 3i$$
Verification (Vieta): sum $= -4 = -4/1$ \checkmark; product $= (-2+3i)(-2-3i) = 4+9 = 13$ \checkmark
Example 2: A cubic $P(z) = z^3 - 3z^2 + 7z - 5$ has $z = 1$ as a root. Find all roots.
Divide: $P(z) = (z-1)(z^2 - 2z + 5)$.
Solve $z^2 - 2z + 5 = 0$: $\Delta = 4-20 = -16$, so $z = 1 \pm 2i$.
All roots: $1,\ 1+2i,\ 1-2i$.
Example 3: Find a quadratic (real coefficients) with $z = 3-2i$ as a root.
Other root: $3+2i$. Sum $= 6$, product $= 9+4 = 13$.
$$z^2 - 6z + 13 = 0$$
Example 4: $z^2 + pz + q = 0$ ($p, q \in \mathbb{R}$) has root $z = 2+i$. Find $p$ and $q$.
Other root: $2-i$. By Vieta: $p = -(2+i+2-i) = -4$; $q = (2+i)(2-i) = 5$.
For $az^2 + bz + c = 0$ with roots $z_1, z_2$:
$$z_1 + z_2 = -\frac{b}{a}, \qquad z_1 z_2 = \frac{c}{a}$$
These hold over $\mathbb{C}$ and provide quick checks without fully solving.
KEY TAKEAWAY: The conjugate root theorem halves the work for polynomials with real coefficients: knowing one complex root immediately gives its pair.
EXAM TIP: When asked to find unknowns given a complex root, use Vieta’s formulas directly rather than substituting and equating real and imaginary parts.
COMMON MISTAKE: Forgetting to divide by $2a$ in the quadratic formula, or dropping the conjugate when asked to list all roots.
