Solving Polynomial Equations (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Properties of complex conjugates

Given that $z = 5 - 3i$, find:

• a) $z^*$
• b) $zz^*$
• c) $z + z^*$

Solution:

a) Write the complex conjugate:

$z^* = 5 + 3i$

b) Calculate the product:

$zz^* = (5 - 3i)(5 + 3i)$

$= 25 + 15i - 15i - 9i^2$

Since $i^2 = -1$:

$= 25 + 9 = 34$

Notice that $zz^*$ gives a real number (34).

c) Calculate the sum:

$z + z^* = (5 - 3i) + (5 + 3i)$

$= 10$

Notice that $z + z^*$ also gives a real number (10).

Worked Example 2: Finding a quadratic equation from one complex root

Find a quadratic equation with real coefficients that has one root of $2 + 7i$.

Solution:

Since $2 + 7i$ is a root and the equation has real coefficients, the complex conjugate $2 - 7i$ must also be a root.

The two roots are: $2 + 7i$ and $2 - 7i$

Therefore, the factors are: $(x - (2 + 7i))$ and $(x - (2 - 7i))$

Multiply the factors together: $(x - (2 + 7i))(x - (2 - 7i)) = 0$

Expand:

$x^2 - x(2 - 7i) - x(2 + 7i) + (2 + 7i)(2 - 7i) = 0$

$x^2 - 2x + 7xi - 2x - 7xi + (4 - 49i^2) = 0$

$x^2 - 4x + 4 + 49 = 0$

Since $-49i^2 = -49 \times (-1) = 49$:

$x^2 - 4x + 53 = 0$

Worked Example 3: Finding a quartic equation from given roots

The quartic equation $3x^4 + ax^3 + bx^2 + cx + d = 0$ has solutions $-3$, $\frac{2}{3}$, $5 + i$ and $5 - i$. Find the values of the real constants $a$, $b$, $c$ and $d$.

Solution:

Since $-3$ and $\frac{2}{3}$ are solutions, the other two roots must be the complex conjugate pair.

Notice that $5 + i$ and $5 - i$ are already given as complex conjugates.

Step 1: Write the factors from the roots:

• $x + 3$ and $3x - 2$ are factors of $3x^4 + ax^3 + bx^2 + 242x + c$
• Therefore $(x + 3)(3x - 2) = 3x^2 + 7x - 6$ is a factor

Step 2: For the complex roots:

• $x - (5 + i)$ and $x - (5 - i)$ are factors of $3x^4 + ax^3 + bx^2 + 242x + c$
• Therefore $(x - (5 + i))(x - (5 - i)) = x^2 - 10x + 26$ is a factor

Step 3: Multiply the two quadratic factors:

$(3x^2 + 7x - 6)(x^2 - 10x + 26)$

Expand:

$= 3x^4 - 30x^3 + 78x^2 + 7x^3 - 70x^2 + 182x - 6x^2 + 60x - 156$

$= 3x^4 - 23x^3 + 2x^2 + 242x - 156$

Therefore: $a = -23$, $b = 2$, $c = 242$, $d = -156$

Worked Example 4: Proving the sum and product formula for quadratic roots

A quadratic equation has roots $\alpha$ and $\beta$. Show that the equation is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.

Solution:

Since the roots are $\alpha$ and $\beta$, the equation in factored form is:

$(x - \alpha)(x - \beta) = 0$

Expand the brackets:

$x^2 - \alpha x - \beta x + \alpha\beta = 0$

$x^2 - (\alpha + \beta)x + \alpha\beta = 0$

This proves the result. You can use this formula (or the factor theorem) whenever working with quadratic roots.

Worked Example 5: Finding constants in a quartic equation

Two of the roots of the equation $ax^4 + bx^3 + cx^2 + dx + e = 0$ are $3 - 2i$ and $4i - 1$. Find the values of $a$, $b$, $c$, $d$ and $e$.

Solution:

Step 1: If $3 - 2i$ and $4i - 1$ are roots, their complex conjugates must also be roots:

• $3 + 2i$ and $-4i - 1$ are also roots

Step 2: Form quadratic factors:

• Roots $3 - 2i$ and $3 + 2i$ give the quadratic factor: $(x - (3 - 2i))(x - (3 + 2i)) = x^2 - 6x + 13$

• Similarly, roots $4i - 1$ and $-4i - 1$ give the quadratic factor: $(x - (4i - 1))(x - (-4i - 1)) = x^2 + 2x + 17$

Step 3: Multiply the two quadratic factors:

$(x^2 - 6x + 13)(x^2 + 2x + 17)$

$= x^4 - 4x^3 + 18x^2 - 76x + 221$

Therefore: $a = 1$, $b = -4$, $c = 18$, $d = -76$, $e = 221$

Worked Example 6: Using a known root to find unknown constants

$f(z) = z^4 + az^3 - 15z^2 + bz - 18 = 0$ where $a$ and $b$ are integers.

Use the fact that $f(1 - i) = 0$ to solve the equation $f(z) = 0$ and find the values of $a$ and $b$.

Solution:

Step 1: Since $1 - i$ is a solution, $1 + i$ will also be a solution (complex conjugate).

Therefore, both $z = 1 - i$ and $z = 1 + i$ are solutions.

Step 2: Form the quadratic factor from these roots: $(z - (1 - i))(z - (1 + i)) = z^2 - 2z + 2$

This is a factor of $z^4 + az^3 - 15z^2 + bz - 18 = 0$.

Step 3: Factorise by dividing:

$z^4 + az^3 - 15z^2 + bz - 18 = (z^2 - 2z + 2)(z^2 + kz - 9)$ for some real $k$

Expand the right side:

$= z^4 + kz^3 - 9z^2 - 2z^3 - 2kz^2 + 18z + 2z^2 + 2kz - 18$

$= z^4 + (k - 2)z^3 + (-7 - 2k)z^2 + (18 + 2k)z - 18$

Step 4: Equate coefficients:

For $z^3$: $a = k - 2$, but we need to find $k$ first

For the constant term, it's already $-18$, which matches.

For $z^2$: $-15 = -7 - 2k$, so $2k = 8$, giving $k = 4$

For $z$: $b = 18 + 2k = 18 + 2(4) = 26$

For $z^3$: $a = k - 2 = 4 - 2 = 2$

Step 5: Now solve $(z^2 - 2z + 2)(z^2 + 4z - 9) = 0$

From $z^2 + 4z - 9 = 0$:

$z = \frac{-4 \pm \sqrt{16 + 36}}{2} = \frac{-4 \pm \sqrt{52}}{2} = -2 \pm \sqrt{13}$

So the quartic equation has:

• Two real solutions: $z = -2 \pm \sqrt{13}$
• Two complex solutions: $z = 1 \pm i$

Therefore: $a = 2$ and $b = 26$

Worked Example 7: Proving conjugate of a sum

Prove that $(z + w)^* = z^* + w^*$ for all $z, w \in \mathbb{C}$.

Solution:

Step 1: Write in the form $a + bi$:

Let $z = a + bi$ and $w = c + di$ where $a, b, c, d \in \mathbb{R}$

Step 2: Find the complex conjugate of the sum:

$(z + w)^* = ((a + c) + (b + d)i)^*$

The conjugate is:

$= (a + c) - (b + d)i$

Step 3: Rearrange the equation:

$= (a - bi) + (c - di)$

$= z^* + w^*$

This proves the result as required.