Mathematics 9709 · Pure Mathematics 3

Solving equations with complex roots

This topic has one rule: a polynomial with real coefficients has its complex roots in conjugate pairs.
If $a+bi$ is a root, then $a-bi$ is a root too. Multiply the pair and you get a real quadratic. Divide that out to find the other roots.

Real coefficients $\Rightarrow$ complex roots come in pairs $a\pm bi$ . Multiply the pair and you get the real quadratic $z^2-2az+(a^2+b^2)$ . That is the factor you divide the polynomial by.

Pair, build a real quadratic, divide out

The pair $a\pm bi$ multiplies to $(z-(a+bi))(z-(a-bi))=z^2-2az+(a^2+b^2)$ : the imaginary parts cancel and you are left with real coefficients.
Divide that into the polynomial. The factor left over gives the other roots.

Roots of a real polynomial come in conjugate pairs. Each pair sits mirrored across the real axis.

Why real coefficients force the conjugate to be a root too

Conjugating works smoothly with the real operations: $(z+w)^*=z^*+w^*$ and $(zw)^*=z^*w^*$ , and a real number is its own conjugate. So if $p(z)=0$ with real coefficients, take the conjugate of the whole equation: every coefficient stays the same, every power of $z$ turns into the same power of $z^*$ , and you get $p(z^*)=0$ . So the conjugate has to be a root. (Drop the real-coefficient condition and this falls apart — which is exactly the trap below.)

Worked examples

Worked example

Real question: verify $-1+\sqrt5\,i$ is a root of $2z^3+z^2+6z-18=0$ , then find the others

9709/31 Oct/Nov 2020 Q7. Check the given root, pair it up, then divide out.

Verify by substituting $z=-1+\sqrt5\,i$ . Using $z^2=1-2\sqrt5\,i-5=-4-2\sqrt5\,i$ and $z^3=z\cdot z^2=14-2\sqrt5\,i$ , the substitution collapses to $0$ — so it is a root.
Real coefficients, so the conjugate $-1-\sqrt5\,i$ is also a root. Their quadratic factor is $z^2-2(-1)z+\big((-1)^2+(\sqrt5)^2\big)=z^2+2z+6$ .
Divide: $2z^3+z^2+6z-18=(z^2+2z+6)(2z-3)$ , so the last root is $z=\tfrac32$ .

Answer

z=-1\pm\sqrt5\,i,\quad z=\tfrac32

Worked example

Given $z=2i$ is a root of $z^4+13z^2+36=0$ , solve it

Conjugate $-2i$ is a root too, giving the factor $(z-2i)(z+2i)=z^2+4$ .
Factorise the quartic: $z^4+13z^2+36=(z^2+4)(z^2+9)$ .
So $z^2=-4$ or $z^2=-9$ .

Answer

z=\pm2i,\ \pm3i

Square roots of a complex number

To find $\sqrt{a+bi}$ , set $(x+iy)^2=a+bi$ with $x,y$ real, expand, then match real and imaginary parts. That gives two equations to solve together.
There are always two roots, equal and opposite ( $\pm(x+iy)$ ).

Worked example

Real question: find the square roots of $6-8i$ in exact Cartesian form

9709/32 Oct/Nov 2024 Q3: “by first forming a quartic in x”.

Set $(x+iy)^2=6-8i$ . Expanding: $(x^2-y^2)+2xyi=6-8i$ .
Match parts: $x^2-y^2=6$ and $2xy=-8$ .
From the second, $y=-\dfrac{4}{x}$ . Substitute: $x^2-\dfrac{16}{x^2}=6$ , so the quartic is $x^4-6x^2-16=0$ .
Let $u=x^2$ : $(u-8)(u+2)=0$ . Since $x$ is real, take $x^2=8$ , giving $x=\pm2\sqrt2$ and $y=\mp\sqrt2$ .

Answer

\pm\big(2\sqrt2-\sqrt2\,i\big)

Worked example

Find the square roots of $-2-2\sqrt3\,i$

Textbook Worked Example 11.12.

$(x+iy)^2=-2-2\sqrt3\,i$ gives $x^2-y^2=-2$ and $2xy=-2\sqrt3$ .
Sub $y=-\dfrac{\sqrt3}{x}$ : $x^4+2x^2-3=0$ , so $(x^2+3)(x^2-1)=0$ . Real $x$ needs $x^2=1$ .

Answer

\pm\big(1-\sqrt3\,i\big)

Where the marks go

Common mistake

The pair rule only works for real coefficients. If the polynomial has complex coefficients, the conjugate need not be a root — don't assume it is.

Common mistake

When asked to “solve”, give every root, and don't miss the conjugate (flip only the imaginary sign). For square roots, keep only the positive value of

x^2

— throw out the negative one, since

x

must be real.

Now you try

$z=-2i$ is a root of $z^4+2z^3+6z^2+8z+8=0$ . Find all four roots. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

The conjugate $2i$ gives the factor $z^2+4$ ; dividing leaves $z^2+2z+2$ , whose roots are $-1\pm i$ :

Answer

z=\pm2i,\ -1\pm i

PreviousPolar and exponential form; multiply and divide NextLoci in the complex plane

Solving equations with complex roots

This topic has one rule: a polynomial with real coefficients has its complex roots in conjugate pairs.
If $a+bi$ is a root, then $a-bi$ is a root too. Multiply the pair and you get a real quadratic. Divide that out to find the other roots.

Real coefficients $\Rightarrow$ complex roots come in pairs $a\pm bi$ . Multiply the pair and you get the real quadratic $z^2-2az+(a^2+b^2)$ . That is the factor you divide the polynomial by.

Pair, build a real quadratic, divide out

The pair $a\pm bi$ multiplies to $(z-(a+bi))(z-(a-bi))=z^2-2az+(a^2+b^2)$ : the imaginary parts cancel and you are left with real coefficients.
Divide that into the polynomial. The factor left over gives the other roots.

Roots of a real polynomial come in conjugate pairs. Each pair sits mirrored across the real axis.

Why real coefficients force the conjugate to be a root too

Worked examples

Worked example

Real question: verify $-1+\sqrt5\,i$ is a root of $2z^3+z^2+6z-18=0$ , then find the others

9709/31 Oct/Nov 2020 Q7. Check the given root, pair it up, then divide out.

Verify by substituting $z=-1+\sqrt5\,i$ . Using $z^2=1-2\sqrt5\,i-5=-4-2\sqrt5\,i$ and $z^3=z\cdot z^2=14-2\sqrt5\,i$ , the substitution collapses to $0$ — so it is a root.
Real coefficients, so the conjugate $-1-\sqrt5\,i$ is also a root. Their quadratic factor is $z^2-2(-1)z+\big((-1)^2+(\sqrt5)^2\big)=z^2+2z+6$ .
Divide: $2z^3+z^2+6z-18=(z^2+2z+6)(2z-3)$ , so the last root is $z=\tfrac32$ .

Answer

z=-1\pm\sqrt5\,i,\quad z=\tfrac32

Worked example

Given $z=2i$ is a root of $z^4+13z^2+36=0$ , solve it

Conjugate $-2i$ is a root too, giving the factor $(z-2i)(z+2i)=z^2+4$ .
Factorise the quartic: $z^4+13z^2+36=(z^2+4)(z^2+9)$ .
So $z^2=-4$ or $z^2=-9$ .

Answer

z=\pm2i,\ \pm3i

Square roots of a complex number

To find $\sqrt{a+bi}$ , set $(x+iy)^2=a+bi$ with $x,y$ real, expand, then match real and imaginary parts. That gives two equations to solve together.
There are always two roots, equal and opposite ( $\pm(x+iy)$ ).

Worked example

Real question: find the square roots of $6-8i$ in exact Cartesian form

9709/32 Oct/Nov 2024 Q3: “by first forming a quartic in x”.

Set $(x+iy)^2=6-8i$ . Expanding: $(x^2-y^2)+2xyi=6-8i$ .
Match parts: $x^2-y^2=6$ and $2xy=-8$ .
From the second, $y=-\dfrac{4}{x}$ . Substitute: $x^2-\dfrac{16}{x^2}=6$ , so the quartic is $x^4-6x^2-16=0$ .
Let $u=x^2$ : $(u-8)(u+2)=0$ . Since $x$ is real, take $x^2=8$ , giving $x=\pm2\sqrt2$ and $y=\mp\sqrt2$ .

Answer

\pm\big(2\sqrt2-\sqrt2\,i\big)

Worked example

Find the square roots of $-2-2\sqrt3\,i$

Textbook Worked Example 11.12.

$(x+iy)^2=-2-2\sqrt3\,i$ gives $x^2-y^2=-2$ and $2xy=-2\sqrt3$ .
Sub $y=-\dfrac{\sqrt3}{x}$ : $x^4+2x^2-3=0$ , so $(x^2+3)(x^2-1)=0$ . Real $x$ needs $x^2=1$ .

Answer

\pm\big(1-\sqrt3\,i\big)

Where the marks go

Common mistake

The pair rule only works for real coefficients. If the polynomial has complex coefficients, the conjugate need not be a root — don't assume it is.

Common mistake

When asked to “solve”, give every root, and don't miss the conjugate (flip only the imaginary sign). For square roots, keep only the positive value of

x^2

— throw out the negative one, since

x

must be real.

Now you try

$z=-2i$ is a root of $z^4+2z^3+6z^2+8z+8=0$ . Find all four roots. (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

The conjugate $2i$ gives the factor $z^2+4$ ; dividing leaves $z^2+2z+2$ , whose roots are $-1\pm i$ :

Answer

z=\pm2i,\ -1\pm i