Complex and Irrational Roots: Practice

Every coefficient of $p$ is real, so the conjugate root theorem applies and the conjugate of any root is a root.

$\overline{2 + 5i} = 2 - 5i$

Only the sign of the imaginary part flips.

Write the root in the form $a + bi$ first. Here $-4i = 0 - 4i$, so its real part is $0$ and its imaginary part is $-4$.

$\overline{0 - 4i} = 0 + 4i = 4i$

The conjugate root theorem applies because the coefficients are real, so $4i$ is a root as well.

Real coefficients force the conjugate of a root to be a root, so the second root is the conjugate of the first.

$\overline{1 + i} = 1 - i$

The other root cannot be $1 + i$ a second time. That would make the quadratic $(x - (1+i))^2 = x^2 - (2 + 2i)x + 2i$, whose coefficients are not real.

Multiply the two linear factors, using the fact that a conjugate pair has a real sum and a real product.

The sum of the roots is $(2 + 3i) + (2 - 3i) = 4$, and the product is $(2 + 3i)(2 - 3i) = 2^2 + 3^2 = 13$.

$\big(x - (2+3i)\big)\big(x - (2-3i)\big) = x^2 - 4x + 13$

The coefficients are real, so the nonreal root $i$ cannot stand alone. Its conjugate must be a root as well.

$\overline{i} = \overline{0 + 1i} = 0 - 1i = -i$

A cubic has exactly three roots counted with multiplicity, and $1$, $i$, $-i$ already fill all three.

A cubic has exactly three roots counted with multiplicity. Real coefficients force the nonreal roots into conjugate pairs, so the number of nonreal roots is even.

$3 = (\text{real roots}) + (\text{an even number})$

An even number subtracted from $3$ leaves an odd number, so the count of real roots is $3$ or $1$. Either way it is at least one.

The root $2 + i$ is nonreal, and the coefficients are real, so its conjugate must be a root too.

$\overline{2 + i} = 2 - i$

The fourth root cannot be another copy of $2 + i$, because then $2 - i$ would be missing from the list of four, and the theorem forbids that.

Solve $x - i = 0$ directly. A degree-$1$ polynomial has exactly one root.

$p(i) = i - i = 0, \qquad p(-i) = -i - i = -2i \neq 0$

So $i$ is a root and $-i$ is not. This is not a violation of the conjugate root theorem, because the coefficient $-i$ is not real, so the theorem never applied. It is the standard example that the hypothesis cannot be dropped.

The conjugate $1 - 2i$ is a root as well, and two distinct roots give a quadratic factor. Multiply the pair.

The sum is $(1 + 2i) + (1 - 2i) = 2$ and the product is $(1+2i)(1-2i) = 1^2 + 2^2 = 5$.

$\big(x - (1+2i)\big)\big(x - (1-2i)\big) = x^2 - 2x + 5$

Rational coefficients force the partner $1 + \sqrt{3}$ to be a root too, so multiply the two linear factors.

The sum is $(1 - \sqrt{3}) + (1 + \sqrt{3}) = 2$ and the product is $(1 - \sqrt{3})(1 + \sqrt{3}) = 1 - 3 = -2$.

$\big(x - (1 - \sqrt{3})\big)\big(x - (1 + \sqrt{3})\big) = x^2 - 2x - 2$

Nonreal roots come in conjugate pairs, so their number is even. The possible counts are $0$, $2$, and $4$; the odd numbers $3$ and $5$ are ruled out at once.

The largest even number that fits inside a degree of $5$ is $4$, and it really happens.

$(x - 1)(x^2 + 1)(x^2 + 4)$

That polynomial has degree $5$ with the four nonreal roots $i$, $-i$, $2i$, $-2i$ and the single real root $1$.

Check each list for nonreal roots that are missing their conjugate partner.

The list $2, 3, i, -i$ pairs $i$ with $-i$. The list $1+i, 1-i, 2+i, 2-i$ contains two complete pairs. The list $i, -i, i, -i$ gives $i$ and $-i$ the same multiplicity, which the theorem also requires.

$1,\ 2,\ 3,\ i \quad\text{has } i \text{ with no } -i$

A real polynomial cannot do that, so this list is the impossible one.