Complex Roots of Polynomials and De Moivre's Theorem

Complex Conjugate Roots

When working with polynomials that have real coefficients, complex roots always come in conjugate pairs, i.e. if $a + bi$ is a root of a polynomial, then $a - bi$ is also a root of the same polynomial.

This is because of the properties of the sum and product of a polynomial's roots, which are covered further in Topic 2. To summarize, if the coefficients are all real, the sum and product of the roots must be real. If there is a complex root, the only way to make the sum and product of all roots positive is to also include its complex conjugate as a root.

For example, if we have the quadratic $x^2 + 1 = 0$:

• The roots are $i$ and $-i$ (conjugate pairs)
• We can verify: $(x-i)(x+i) = x^2 + 1$

De Moivre's Theorem

De Moivre's theorem is a result formalizing the properties of complex numbers when raised to a power. To derive it, we can start with a complex number in Euler form:

$$re^{i\theta}$$

If we raise it to a power $n$, we get:

$$(re^{i\theta})^n$$

We can apply normal exponent rules to expand this into:

$$r^ne^{in\theta}$$

If we then replace $e^{i\theta}$ with $\cos\theta + i\sin\theta$, we get de Moivre's theorem. For any complex number in polar form:

$$(r(\cos θ + i\sin θ))^n = r^n(\cos(nθ) + i\sin(nθ))$$

Or alternatively, in $\text{cis}$ notation:

$$(r\text{ cis }\theta)^n = r^n\text{ cis }n\theta$$

Using De Moivre's for Powers

Finding nth Roots of Complex Numbers

To find the nth roots of a complex number:

1. Convert to polar form: $r(\cos θ + i\sin θ)$
2. Use the formula: $\sqrt[n]{r}(\cos(\frac{θ + 2\pi k}{n}) + i\sin(\frac{θ + 2\pi k}{n}))$ where $k = 0, 1, 2, ..., n-1$

Applications to Polynomial Equations

A polynomial of degree n will always have exactly n roots, counting both real and complex (counting duplicate roots). This is called the fundamental theorem of algebra.

This theory helps us solve higher-degree polynomial equations: