Complex roots and equations | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Real coefficients force conjugate root pairs

If a polynomial has real coefficients and $a+bi$ is a root, then $a-bi$ is also a root. If r is a root, the factor theorem says $z-r$ is a factor.

Key idea

Conjugating the whole polynomial equation leaves every real coefficient unchanged. That is why the conjugate is another root. The rule is not guaranteed when a coefficient is complex.

Worked example

Use one non-real root to finish a cubic

The polynomial $z^3-5z^2+9z-5$ has the root $2+i$ . Therefore $2-i$ is also a root.

(z-2-i)(z-2+i)

=(z-2)^2+1

=z^2-4z+5

z^3-5z^2+9z-5

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

Answer

z=1,\quad2+i,\quad2-i

The quadratic formula also allows complex coefficients

If $Aw^2+Bw+C=0$ and $A\ne0$ , use the same formula as for a real quadratic:

w=\frac{-B\pm\sqrt{B^2-4AC}}{2A}

Keep every complex multiplication visible, then simplify any complex square root exactly.

Worked example

Solve a quadratic with complex coefficients

w^2(1+i)+2w=i-1

Move every term to the left. Then $A=1+i$ , $B=2$ and $C=1-i$ . The discriminant is

B^2-4AC

=4-4(1+i)(1-i)

=4-8=-4

\sqrt{-4}=2i

w=\frac{-2\pm2i}{2+2i}

Answer

w=i\quad\text{or}\quad w=-1

Examiner note

A conjugate pair is not expected here because the coefficients are not all real.

Square roots appear as opposite points

Square roots occur as an opposite pair

ReIm

w = 3 + 2i

−w = −3 − 2i

w² = z and (−w)² = z

If w squared equals z, then negative w squared also equals z.

Worked example

Find both square roots of 5 + 12i

Let $w=a+bi$ , where a and b are real, and square it.

w^2=(a^2-b^2)+2abi

a^2-b^2=5,\qquad2ab=12

Also $a^2+b^2=|w|^2=|w^2|=|5+12i|=13$ . Adding and subtracting now gives $a^2=9$ and $b^2=4$ . Since $ab=6$ , a and b have the same sign.

Answer

w=\pm(3+2i)

Form a quartic when the question asks for it

If the instruction says “by first forming a quartic”, do not replace that method with the modulus shortcut. Eliminate one unknown from the two real equations.

Worked example

Find the square roots of 7 − 24i

a^2-b^2=7

2ab=-24

b=-\frac{12}{a}

a^2-\frac{144}{a^2}=7

a^4-7a^2-144=0

(a^2-16)(a^2+9)=0

Because a is real, $a^2=16$ . Substituting into $b=-12/a$ pairs the signs correctly.

Answer

w=\pm(4-3i)

Your turnIndependent exact-root check [5 marks]

By first forming a quartic equation in a or b, find both square roots of $-9+40i$ in exact Cartesian form.

Show worked answer

a^2-b^2=-9

2ab=40

b=\frac{20}{a}

a^4+9a^2-400=0

(a^2-16)(a^2+25)=0

Answer

w=\pm(4+5i)

Common mistake

Do not report only one square root. If w is a root, −w is the second root because

(-w)^2=w^2

(Pure Mathematics 2 & 3 Coursebook, Ch. 11.4)

Real coefficients force conjugate root pairs

If a polynomial has real coefficients and $a+bi$ is a root, then $a-bi$ is also a root. If r is a root, the factor theorem says $z-r$ is a factor.

Key idea

Conjugating the whole polynomial equation leaves every real coefficient unchanged. That is why the conjugate is another root. The rule is not guaranteed when a coefficient is complex.

Worked example

Use one non-real root to finish a cubic

The polynomial $z^3-5z^2+9z-5$ has the root $2+i$ . Therefore $2-i$ is also a root.

(z-2-i)(z-2+i)

=(z-2)^2+1

=z^2-4z+5

z^3-5z^2+9z-5

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

Answer

z=1,\quad2+i,\quad2-i

The quadratic formula also allows complex coefficients

If $Aw^2+Bw+C=0$ and $A\ne0$ , use the same formula as for a real quadratic:

w=\frac{-B\pm\sqrt{B^2-4AC}}{2A}

Keep every complex multiplication visible, then simplify any complex square root exactly.

Worked example

Solve a quadratic with complex coefficients

w^2(1+i)+2w=i-1

Move every term to the left. Then $A=1+i$ , $B=2$ and $C=1-i$ . The discriminant is

B^2-4AC

=4-4(1+i)(1-i)

=4-8=-4

\sqrt{-4}=2i

w=\frac{-2\pm2i}{2+2i}

Answer

w=i\quad\text{or}\quad w=-1

Examiner note

A conjugate pair is not expected here because the coefficients are not all real.

Square roots appear as opposite points

Square roots occur as an opposite pair

ReIm

w = 3 + 2i

−w = −3 − 2i

w² = z and (−w)² = z

If w squared equals z, then negative w squared also equals z.

Worked example

Find both square roots of 5 + 12i

Let $w=a+bi$ , where a and b are real, and square it.

w^2=(a^2-b^2)+2abi

a^2-b^2=5,\qquad2ab=12

Also $a^2+b^2=|w|^2=|w^2|=|5+12i|=13$ . Adding and subtracting now gives $a^2=9$ and $b^2=4$ . Since $ab=6$ , a and b have the same sign.

Answer

w=\pm(3+2i)

Form a quartic when the question asks for it

If the instruction says “by first forming a quartic”, do not replace that method with the modulus shortcut. Eliminate one unknown from the two real equations.

Worked example

Find the square roots of 7 − 24i

a^2-b^2=7

2ab=-24

b=-\frac{12}{a}

a^2-\frac{144}{a^2}=7

a^4-7a^2-144=0

(a^2-16)(a^2+9)=0

Because a is real, $a^2=16$ . Substituting into $b=-12/a$ pairs the signs correctly.

Answer

w=\pm(4-3i)

Your turnIndependent exact-root check [5 marks]

By first forming a quartic equation in a or b, find both square roots of $-9+40i$ in exact Cartesian form.

Show worked answer

a^2-b^2=-9

2ab=40

b=\frac{20}{a}

a^4+9a^2-400=0

(a^2-16)(a^2+25)=0

Answer

w=\pm(4+5i)

Common mistake

Do not report only one square root. If w is a root, −w is the second root because

(-w)^2=w^2

(Pure Mathematics 2 & 3 Coursebook, Ch. 11.4)