Conjugate pairs: non-real roots arrive in twos

By the end of this lesson

By the end you can explain why non-real roots of a polynomial with real coefficients come in conjugate pairs, and use one given complex root to find every root of a cubic or quartic.

Key to the symbols

Symbol	Meaning
$P(x)$	a polynomial, such as $x^3 - 3x^2 + 4x - 12$
root	a number $r$ with $P(r) = 0$
coefficient	a number multiplying a power of $x$
real coefficients	every coefficient is an ordinary real number, with no $i$ in it
$z^*$	the conjugate of $z$ : the imaginary part's sign is flipped
degree	the highest power of $x$ in the polynomial

A cubic has degree 3 and a quartic has degree 4. Counting repeats, a polynomial of degree $n$ has exactly $n$ roots, which is why finding one root of a cubic leaves two more to find.

The result you will use

Suppose $P(x)$ has real coefficients, and $z$ is a root that is not real.

Then its conjugate $z^*$ is also a root.

So non-real roots come in conjugate pairs: if $2 + 3i$ is a root, then $2 - 3i$ is a root too.

Real roots are not paired up. A real number is its own conjugate, so a real root can appear on its own.

Why a lone complex root is impossible

The numbers $i$ and $-i$ behave identically: both of them square to $-1$ . Nothing in ordinary arithmetic can tell them apart.

A polynomial with real coefficients has no $i$ written anywhere in it, so it can't tell them apart either.

So suppose $1 + 2i$ makes the polynomial equal zero. Swap every $i$ for $-i$ in that calculation. Every step stays true, which means $1 - 2i$ makes it equal zero too.

That is the idea. The next block makes it precise.

Why it is true: conjugate both sides

Three facts about conjugating, all easy to check by writing $z = x + iy$ :

Rule	Example
The conjugate of a sum is the sum of the conjugates	$(z_1 + z_2)^* = z_1^* + z_2^*$
The conjugate of a product is the product of the conjugates	$(z_1z_2)^* = z_1^z_2^$
A real number is its own conjugate, and $(z^k)^* = (z^*)^k$	$5^* = 5$

Now suppose $P(z) = 0$ , where

$P(x) = a_n x^n + \dots + a_1x + a_0$

and every $a_k$ is real. Take the conjugate of both sides of $P(z) = 0$ .

Each coefficient stays the same, because each one is real. Each power of $z$ becomes the same power of $z^*$ . The zero on the right stays zero. So

$a_n(z^*)^n + \dots + a_1z^* + a_0 = 0$

But the left-hand side is exactly $P(z^*)$ . Therefore $P(z^*) = 0$ , so $z^*$ is a root.

This is the sentence examiners look for: since the coefficients are real, the conjugate is also a root.

Understanding the theorem

Multiple choiceQuestion 1 · Recall

Answer the question above to continue.