CP 2.1, 2.3 - Complex roots of polynomial equations | Edexcel A-Level Further Mathematics

Solve quadratics even when their discriminant is negative, then use a known root or factor to solve cubics and quartics completely (CP 2.1). The conjugate-pair result (CP 2.3) turns one non-real root into a real quadratic factor, often saving a great deal of algebra.

Quadratics with a negative discriminant

How can a quartic have four roots when its graph never reaches the horizontal axis? For example, $z^4 + 5z^2 + 4 = 0$ has no real solution: every term on the left is non-negative and the constant is positive. By the end you will find all four complex roots.

Start with $z^2 + 4 = 0$ . The two roots are $2\mathrm{i}$ and $-2\mathrm{i}$ , since both square to $-4$ . A negative discriminant means “no real roots”, rather than “no roots”. Complex numbers finish the job.

You already know

From A-level Mathematics, the discriminant is $b^2 - 4ac$ , the quadratic formula solves $az^2 + bz + c = 0$ , and a repeated root counts twice. The factor theorem says $f(k) = 0$ exactly when $z-k$ is a factor. From complex arithmetic, $\sqrt{-m} = \mathrm{i}\sqrt m$ for $m > 0$ .

The same quadratic formula works over the complex numbers. Only the interpretation of the square root changes.

Quadratic formula over the complex numbers

For real $a$ , $b$ , $c$ , with $a \ne 0$ ,

az^2 + bz + c = 0 \quad\Longrightarrow\quad z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

If $b^2-4ac < 0$ , write the square root as $\mathrm{i}\sqrt{4ac-b^2}$ . The quadratic formula is assumed knowledge and is not printed in the supplied formula booklet.

Not in the formula booklet: learn it.

Completing the square gives a useful picture of the algebra: $z^2 - 6z + 13 = 0$ becomes $(z-3)^2 = -4$ . The two roots share a real part and have opposite imaginary parts.

Worked example 1: two non-real roots

Question. Solve $z^2 - 6z + 13 = 0$ .

Plan. Calculate the discriminant, write its square root using $\mathrm{i}$ , and simplify both roots.

\begin{aligned} b^2-4ac &= 36-52=-16 && \text{calculate discriminant} \\ z &= \frac{6\pm\sqrt{-16}}{2} && \text{quadratic formula} \\ &= \frac{6\pm4\mathrm{i}}2 && \text{use }\mathrm{i}^2=-1 \\ &= 3\pm2\mathrm{i} && \text{divide both terms} \end{aligned}

Why this step? The symbol $\pm$ supplies both square roots of $-16$ , so neither solution disappears.

Answer. $z = 3+2\mathrm{i}$ or $z = 3-2\mathrm{i}$ .

Check. For $z=3+2\mathrm{i}$ , $(z-3)^2+4=(2\mathrm{i})^2+4=0$ . The negative choice gives the same result.

The second example changes the leading coefficient. Keep the entire numerator over $2a$ : the imaginary part must be divided too.

Worked example 2: a non-monic quadratic

Question. Find the exact roots of $2z^2-4z+5=0$ .

Plan. Use the quadratic formula and simplify the negative square root before dividing.

\begin{aligned} z &= \frac{4\pm\sqrt{16-40}}4 && \text{substitute }a=2 \\ &= \frac{4\pm2\mathrm{i}\sqrt6}{4} && \sqrt{-24}=2\mathrm{i}\sqrt6 \\ &= 1\pm\frac{\mathrm{i}\sqrt6}{2} && \text{simplify both terms} \end{aligned}

Why this step? Simplifying $\sqrt{24}$ before dividing makes the exact imaginary coefficient easy to see.

Answer. $z=1+\frac{\mathrm{i}\sqrt6}{2}$ or $z=1-\frac{\mathrm{i}\sqrt6}{2}$ .

Check. The original expression is $2(z-1)^2+3$ ; substituting either root gives $2(-6/4)+3=0$ .

The denominator is a particularly easy place to lose a factor of two.

Watch out

Wrong: $\frac{4+2\mathrm{i}\sqrt6}{4}=1+2\mathrm{i}\sqrt6$ . The denominator divides both terms. Right: $1+\frac12\mathrm{i}\sqrt6$ . Keep exact surds when the question asks for exact roots.

Try the same decision with a different leading coefficient.

Why non-real roots come in pairs

A real polynomial cannot have a non-real root without also having its conjugate: changing $\mathrm{i}$ to $-\mathrm{i}$ preserves every real coefficient and every valid algebraic step.

For example, $(z-(2+3\mathrm{i}))(z-(2-3\mathrm{i}))=(z-2)^2+9=z^2-4z+13$ . The imaginary coefficients cancel. By contrast, pairing $2+3\mathrm{i}$ with $-2-3\mathrm{i}$ gives a polynomial with a non-real constant term; negating is not conjugating.

Conjugate pair of roots

If a polynomial has real coefficients and $a+b\mathrm{i}$ is a non-real root, then $a-b\mathrm{i}$ is also a root. They form a conjugate pair. The condition “real coefficients” is essential: $f(z)=z-\mathrm{i}$ has root $\mathrm{i}$ but not $-\mathrm{i}$ .

Here is the reason in a line. For $f(z)=a_nz^n+\cdots+a_1z+a_0$ with every $a_j$ real,

f(\alpha^*)=\bigl(f(\alpha)\bigr)^*=0^*=0

Conjugation passes through sums and products, while a real coefficient stays unchanged. Thus $\alpha^*$ is a root too. This reasoning also pairs repeated non-real roots with the same multiplicity.

The pair immediately supplies a factor with real coefficients.

The real factor from a conjugate pair

If $a\pm b\mathrm{i}$ , with real $a,b$ and $b\ne0$ , are roots of a real-coefficient polynomial, their quadratic factor is

(z-a-b\mathrm{i})(z-a+b\mathrm{i})=(z-a)^2+b^2 =z^2-2az+a^2+b^2

Not in the formula booklet: learn it.

A useful surprise follows: every cubic with real coefficients has at least one real root. Non-real roots use up places in pairs, but a cubic has three roots counting multiplicity. You cannot fill three places using only pairs. This is also consistent with the opposite end behaviour of an odd-degree real graph.

Key idea

Use the conjugate to build a real quadratic factor; the remaining algebra can then stay entirely real.

The next check tests the condition of the theorem, not just the sign change.

Reducing a cubic to a quadratic

A given real root removes one linear factor from a cubic, leaving a quadratic that you can solve exactly.

Suppose $f(-1)=0$ . The factor is $z+1$ , not $z-1$ . Divide by that factor, or compare coefficients in $f(z)=(z+1)(Az^2+Bz+C)$ . The leading and constant terms often reveal $A$ and $C$ immediately.

Worked example 3: a given linear factor

Question. Given that $z+1$ is a factor of $f(z)=z^3-3z^2+4z+8$ , use algebra to solve $f(z)=0$ completely.

Plan. Find the quadratic factor by coefficient comparison, then solve it by completing the square.

\begin{aligned} f(z)&=(z+1)(z^2+Bz+8) && \text{leading and constant terms} \\ B+1&=-3\quad\Longrightarrow\quad B=-4 && \text{compare }z^2 \\ f(z)&=(z+1)(z^2-4z+8) && \text{remaining factor} \\ z^2-4z+8&=(z-2)^2+4 && \text{complete the square} \end{aligned}

Why this step? Finding the real quadratic avoids substituting an unknown complex number into all four terms of the cubic.

Answer. $z=-1$ , $z=2+2\mathrm{i}$ or $z=2-2\mathrm{i}$ .

Check. Expansion gives the stated cubic; also $B+8=4$ checks its $z$ coefficient. Each quadratic root makes $(z-2)^2+4=0$ .

The next example changes the leading coefficient. The quadratic factor must now begin with $2z^2$ , and its roots need not have integer imaginary parts.

Worked example 4: a non-monic cubic

Question. Given that $f(2)=0$ , where $f(z)=2z^3-3z-10$ , solve $f(z)=0$ completely.

Plan. Factor out $z-2$ , retaining the missing $z^2$ coefficient as zero, then use the quadratic formula.

\begin{aligned} f(z)&=(z-2)(2z^2+Bz+5) && \text{leading and constant terms} \\ B-4&=0\quad\Longrightarrow\quad B=4 && \text{zero }z^2\text{ coefficient} \\ f(z)&=(z-2)(2z^2+4z+5) && \text{check }5-8=-3 \\ z&=\frac{-4\pm\sqrt{16-40}}4 && \text{solve the quadratic} \\ &=-1\pm\frac{\mathrm{i}\sqrt6}{2} && \text{simplify exactly} \end{aligned}

Why this step? A missing power is a coefficient of zero, so it still contributes a comparison equation.

Answer. $z=2$ , $z=-1+\frac{\mathrm{i}\sqrt6}{2}$ or $z=-1-\frac{\mathrm{i}\sqrt6}{2}$ .

Check. $f(2)=16-6-10=0$ , and expansion of the displayed factors recovers all coefficients.

A complete solution must include the root that supplied the factor.

Watch out

Wrong: after dividing by $z-2$ , listing only the two quadratic roots. Division removes a factor, not a solution. Right: list the given root and both roots of the remaining quadratic.

The following cubic has no $z^2$ term either.

Reducing a quartic with a conjugate pair

One known non-real root of a real quartic gives two roots and a quadratic factor, leaving just one more quadratic to solve.

You may be given the whole polynomial, or some coefficients may be unknown. You do not always need to find those coefficients: the leading term, cubic term and constant term can determine the second factor directly. This is a useful reading habit in a “determine the other roots” question.

Worked example 5: a known complex root

Question. Given that $1+2\mathrm{i}$ is a root of $z^4+3z^2+6z+10=0$ , determine the other roots.

Plan. Use the conjugate root, build its quadratic factor, and compare coefficients to obtain the second factor.

\begin{aligned} 1-2\mathrm{i}&\text{ is also a root} && \text{real coefficients} \\ (z-1)^2+4&=z^2-2z+5 && \text{factor from the pair} \\ f(z)&=(z^2-2z+5)(z^2+pz+q) && \text{factorise quartic} \\ 5q=10&\quad\Longrightarrow\quad q=2 && \text{constant term} \\ p-2=0&\quad\Longrightarrow\quad p=2 && \text{cubic term} \\ z^2+2z+2&=(z+1)^2+1 && \text{remaining quadratic} \end{aligned}

Why this step? Conjugate roots give a real factor, so two simple coefficient comparisons replace complex polynomial division.

Answer. The other roots are $1-2\mathrm{i}$ , $-1+\mathrm{i}$ and $-1-\mathrm{i}$ .

Check. $(z^2-2z+5)(z^2+2z+2)=z^4+3z^2+6z+10$ . All four roots have been accounted for.

Now change the information supplied: two middle coefficients are unknown. The same comparison still determines the roots, without first finding those coefficients.

Worked example 6: unknown coefficients need not be found

Question. The polynomial $f(z)=z^4-6z^3+az^2+bz+50$ has real coefficients and root $1+3\mathrm{i}$ . Determine its other roots.

Plan. Form the conjugate-pair factor and use only the cubic and constant coefficients to find the other quadratic.

\begin{aligned} (z-(1+3\mathrm{i}))(z-(1-3\mathrm{i}))&=z^2-2z+10 && \text{conjugate pair} \\ f(z)&=(z^2-2z+10)(z^2+pz+q) && \text{remaining monic quadratic} \\ 10q=50&\quad\Longrightarrow\quad q=5 && \text{constant terms} \\ p-2=-6&\quad\Longrightarrow\quad p=-4 && \text{cubic terms} \\ z^2-4z+5=0&\quad\Longrightarrow\quad (z-2)^2=-1 && \text{complete square} \end{aligned}

Why this step? The question asks for roots. The two supplied coefficients already determine the remaining factor; solving for $a$ and $b$ first would add work.

Answer. The other roots are $1-3\mathrm{i}$ , $2+\mathrm{i}$ and $2-\mathrm{i}$ .

Check. Expansion gives $z^4-6z^3+23z^2-50z+50$ , so real values $a=23$ , $b=-50$ do exist.

Check the sign in the conjugate-pair factor before comparing the remaining coefficients.

Watch out

Wrong: $(z-(1+3\mathrm{i}))(z-(1-3\mathrm{i}))=(z-1)^2-9$ . The product subtracts $(3\mathrm{i})^2=-9$ , so the correct factor is $(z-1)^2+9=z^2-2z+10$ .

The opening quartic is even simpler: set $u=z^2$ , or factor by inspection,

z^4+5z^2+4=(z^2+1)(z^2+4)

Its four roots are $\mathrm{i}$ , $-\mathrm{i}$ , $2\mathrm{i}$ and $-2\mathrm{i}$ . None appears as a crossing on the real graph because none is real. Complex roots have escaped the real line, not the equation.

Exam tip

For a real quartic, state the conjugate root explicitly, show the real quadratic factor, then show how you obtain and solve the remaining quadratic. These are separate creditworthy steps; a bare list hides the method.

Here the known root has a different real part. Keep the signs in its factor under control.

Quick Recap

A negative discriminant gives two non-real quadratic roots; simplify its square root with $\mathrm{i}$ .
For real coefficients, non-real roots occur in conjugate pairs. Their real factor is $(z-a)^2+b^2$ .
A given root reduces a cubic to a quadratic; retain that given root in the final list.
A conjugate pair reduces a quartic to a quadratic. Compare only the coefficients you need, and count repeated roots with their multiplicities.

The opening quartic has roots $\pm\mathrm{i}$ and $\pm2\mathrm{i}$ : four roots, although its real graph never meets the horizontal axis.

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CP 2.1, 2.3 - Complex roots of polynomial equations

Quadratics with a negative discriminant

Why non-real roots come in pairs

Reducing a cubic to a quadratic

Reducing a quartic with a conjugate pair

Quick Recap

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