Quick Recall
Quick check — pick the correct option in each question before moving on.
$(z - 2)(z^2 - 4z + 3) = z^3 - 6z^2 + 11z - 6$, and $z^2 - 4z + 3 = (z - 1)(z - 3)$, giving roots $z = 1, 2, 3$.
The conjugate flips the sign of the imaginary part: $z^* = x - \mathrm{i}y$. On the Argand diagram it is the reflection of $z$ in the real axis.
With negative discriminant, $b^2 - 4c < 0$, the quadratic formula gives two non-real roots of the form $z = \dfrac{-b \pm \mathrm{i}\sqrt{4c - b^2}}{2}$, which are conjugates.
The Wall
Consider the cubic equation:
$$z^3 - z^2 - 7z + 15 = 0$$
Try the rational root theorem: test $z = \pm 1, \pm 3, \pm 5, \pm 15$. You find $z = -3$ works! Dividing gives $z^3 - z^2 - 7z + 15 = (z + 3)(z^2 - 4z + 5)$. But the quadratic $z^2 - 4z + 5 = 0$ has discriminant $16 - 20 = -4 < 0$ — the remaining two roots are complex: $z = 2 \pm \mathrm{i}$.
Now flip the problem: someone tells you that $z = 2 + \mathrm{i}$ is a root, but doesn't tell you the others. How can you use this single complex root to find all the roots—including the real one—without guessing?
The Need
We need a theorem that tells us: if a real-coefficient polynomial has a complex root $z = a + b\mathrm{i}$, then its conjugate $z^* = a - b\mathrm{i}$ is automatically a root too. This doubles our information and lets us build a real quadratic factor, divide it out, and find every remaining root.
💡 The Bridge (click to reveal)
The Conjugate Root Theorem states: if $z = a + b\mathrm{i}$ is a root of a polynomial with real coefficients, then $z^* = a - b\mathrm{i}$ is also a root.
From the conjugate pair, we form a real quadratic factor:
$$(x - z)(x - z^*) = x^2 - 2ax + (a^2 + b^2)$$
We then divide the original polynomial by this quadratic to find the remaining factors. Today we learn this three-step process: identify → conjugate → divide.
A polynomial with real coefficients has a remarkable property: its complex roots always come in conjugate pairs. If $z = a + b\mathrm{i}$ is a root, then $z^* = a - b\mathrm{i}$ is automatically a root too. This means we can turn one complex root into a real quadratic factor and use polynomial division to find all remaining roots.
If $z = a + b\mathrm{i}$ (with $b \neq 0$) is a root of a polynomial $P(x)$ with real coefficients, then its conjugate $z^* = a - b\mathrm{i}$ is also a root.
$$\boxed{\;P(a + b\mathrm{i}) = 0 \;\Longrightarrow\; P(a - b\mathrm{i}) = 0\;}$$Geometrically, the roots are mirror images across the real axis on the Argand diagram. This only holds for real-coefficient polynomials — if the coefficients are complex, the conjugate need not be a root.
If $z = a + b\mathrm{i}$ and $z^* = a - b\mathrm{i}$ are roots, the quadratic factor with these roots is:
$$\boxed{\;(x - z)(x - z^*) = (x - a - b\mathrm{i})(x - a + b\mathrm{i}) = x^2 - 2ax + (a^2 + b^2)\;}$$The coefficients $-2a$ and $a^2 + b^2$ are both real. The sum of the conjugate roots is $2a$ (real) and the product is $a^2 + b^2 = |z|^2$ (real and positive). This is why a conjugate pair always produces a real quadratic factor.
Given one complex root $z = a + b\mathrm{i}$ of a real polynomial $P(x)$:
Step 1: Write down the conjugate $z^* = a - b\mathrm{i}$ (also a root by the theorem).
Step 2: Form the real quadratic factor $x^2 - 2ax + (a^2 + b^2)$.
Step 3: Divide $P(x)$ by this quadratic factor to find the remaining factor(s), then solve.
Example: Dividing $z^3 - z^2 - 7z + 15$ by the quadratic factor $z^2 - 4z + 5$ (from roots $2 \pm \mathrm{i}$):
The teacher demonstrates the full process of finding all roots of a polynomial given one complex root, narrating each step. Students follow along and take notes on the three-step process.
Quick-response matching game: match each polynomial to its correct set of roots. Click a polynomial (left), then click the matching root-set (right). The root-sets are shuffled each time you load the page.
👁 Show Answer Key
$z^2 - 8z + 41 = 0$: sum $= 8$, product $= 41$ → roots $4 \pm 5\mathrm{i}$ (since $8/2 = 4$ and $4^2 + 5^2 = 41$).
$z^2 - 6z + 13 = 0$: sum $= 6$, product $= 13$ → roots $3 \pm 2\mathrm{i}$ (since $6/2 = 3$ and $3^2 + 2^2 = 13$).
$z^2 - 4z + 5 = 0$: sum $= 4$, product $= 5$ → roots $2 \pm \mathrm{i}$ (since $4/2 = 2$ and $2^2 + 1^2 = 5$).
$z^2 - 2z + 2 = 0$: sum $= 2$, product $= 2$ → roots $1 \pm \mathrm{i}$ (since $2/2 = 1$ and $1^2 + 1^2 = 2$).
$4 - 5\mathrm{i}$ is one root of a quadratic equation with real coefficients. Write down the second root of the equation and hence find the equation.
👁 Reveal Solution
Since the quadratic has real coefficients and one root is $z = 4 - 5\mathrm{i}$, the conjugate root theorem gives the second root:
$$z^* = 4 + 5\mathrm{i}$$
Sum: $(4 - 5\mathrm{i}) + (4 + 5\mathrm{i}) = 8$
Product: $(4 - 5\mathrm{i})(4 + 5\mathrm{i}) = 4^2 + 5^2 = 16 + 25 = 41$
(Using $(a - b\mathrm{i})(a + b\mathrm{i}) = a^2 + b^2$, since the cross terms cancel.)
A quadratic with roots $r_1, r_2$ is $z^2 - (r_1 + r_2)z + r_1 r_2 = 0$.
$$\boxed{z^2 - 8z + 41 = 0}$$
Verification: discriminant $= 64 - 164 = -100 < 0$, so the roots are indeed complex. Using the quadratic formula: $z = \dfrac{8 \pm \sqrt{-100}}{2} = \dfrac{8 \pm 10\mathrm{i}}{2} = 4 \pm 5\mathrm{i}$ ✓
Check that $2 + \mathrm{i}$ is a root of $z^3 - z^2 - 7z + 15 = 0$, and find the other roots.
👁 Reveal Solution
Compute the powers of $2 + \mathrm{i}$:
$(2 + \mathrm{i})^2 = 4 + 4\mathrm{i} + \mathrm{i}^2 = 3 + 4\mathrm{i}$
$(2 + \mathrm{i})^3 = (2 + \mathrm{i})(3 + 4\mathrm{i}) = 6 + 8\mathrm{i} + 3\mathrm{i} + 4\mathrm{i}^2 = 6 + 11\mathrm{i} - 4 = 2 + 11\mathrm{i}$
Substitute into $P(z) = z^3 - z^2 - 7z + 15$:
$P(2+\mathrm{i}) = (2 + 11\mathrm{i}) - (3 + 4\mathrm{i}) - 7(2 + \mathrm{i}) + 15$
$= (2 - 3 - 14 + 15) + (11 - 4 - 7)\mathrm{i} = 0 + 0\mathrm{i} = 0$ ✓
So $z = 2 + \mathrm{i}$ is indeed a root.
Since the polynomial has real coefficients, the conjugate root theorem applies:
$$z^* = 2 - \mathrm{i} \text{ is also a root.}$$
Sum of conjugate roots: $(2 + \mathrm{i}) + (2 - \mathrm{i}) = 4$
Product of conjugate roots: $(2 + \mathrm{i})(2 - \mathrm{i}) = 4 + 1 = 5$
Quadratic factor: $z^2 - 4z + 5$
Divide $z^3 - z^2 - 7z + 15$ by $z^2 - 4z + 5$ using polynomial long division:
The quotient is $z + 3$, so:
$$z^3 - z^2 - 7z + 15 = (z^2 - 4z + 5)(z + 3)$$
From the linear factor $z + 3 = 0$:
$$z = -3$$
Verification: $(-3)^3 - (-3)^2 - 7(-3) + 15 = -27 - 9 + 21 + 15 = 0$ ✓
$$\boxed{\text{All roots: } z = 2 + \mathrm{i}, \quad z = 2 - \mathrm{i}, \quad z = -3}$$
The cubic has one real root ($-3$) and a conjugate pair of complex roots ($2 \pm \mathrm{i}$).
One root of $z^3 - 15z^2 + 76z - 140 = 0$ is an integer. Solve the equation.
💡 Hint
Use the rational root theorem: test integer factors of $140$ ($\pm 1, \pm 2, \pm 4, \pm 5, \pm 7, \pm 10, \pm 14, \ldots$).
Once you find the integer root $r$, divide the cubic by $(z - r)$ to get a quadratic, then solve the quadratic.
If the quadratic has a negative discriminant, the other two roots are complex conjugates.
👁 Reveal Solution
Testing factors of $140$: $P(7) = 343 - 735 + 532 - 140 = 0$ ✓
So $z = 7$ is a root.
$z^3 - 15z^2 + 76z - 140 = (z - 7)(z^2 - 8z + 20)$
Check: $(z-7)(z^2 - 8z + 20) = z^3 - 8z^2 + 20z - 7z^2 + 56z - 140 = z^3 - 15z^2 + 76z - 140$ ✓
Discriminant: $64 - 80 = -16 < 0$, so complex roots.
$z = \dfrac{8 \pm \sqrt{-16}}{2} = \dfrac{8 \pm 4\mathrm{i}}{2} = 4 \pm 2\mathrm{i}$
$$\boxed{z = 7, \quad z = 4 + 2\mathrm{i}, \quad z = 4 - 2\mathrm{i}}$$
Verification: $P(4+2\mathrm{i}) = (4+2\mathrm{i})^3 - 15(4+2\mathrm{i})^2 + 76(4+2\mathrm{i}) - 140$. Computing: $(4+2\mathrm{i})^2 = 12+16\mathrm{i}$, $(4+2\mathrm{i})^3 = (4+2\mathrm{i})(12+16\mathrm{i}) = 16+88\mathrm{i}$. Then $P = (16+88\mathrm{i}) - 15(12+16\mathrm{i}) + 76(4+2\mathrm{i}) - 140 = (16-180+304-140) + (88-240+152)\mathrm{i} = 0 + 0\mathrm{i}$ ✓
Given that $1 - \mathrm{i}$ is a root of $z^3 + pz^2 + qz + 12 = 0$, find the real numbers $p$ and $q$, and the other roots.
💡 Hint
Since the coefficients are real, $1 + \mathrm{i}$ is also a root (conjugate root theorem).
Form the quadratic factor from the conjugate pair: sum $= 2$, product $= 2$.
Let the third root be $r$. Then $(z^2 - 2z + 2)(z - r) = z^3 + pz^2 + qz + 12$. Match the constant term to find $r$, then match the other coefficients to find $p$ and $q$.
👁 Reveal Solution
Since $p$ and $q$ are real, the conjugate root theorem applies: $1 + \mathrm{i}$ is also a root.
Sum: $(1-\mathrm{i}) + (1+\mathrm{i}) = 2$. Product: $(1-\mathrm{i})(1+\mathrm{i}) = 1 + 1 = 2$.
Quadratic factor: $z^2 - 2z + 2$
Let the third root be $r$. Then:
$(z^2 - 2z + 2)(z - r) = z^3 - (r+2)z^2 + (2r+2)z - 2r$
Matching the constant term: $-2r = 12 \Rightarrow r = -6$.
With $r = -6$:
$p = -(r + 2) = -(-6 + 2) = 4$
$q = 2r + 2 = 2(-6) + 2 = -10$
Verification: $(z^2 - 2z + 2)(z + 6) = z^3 + 6z^2 - 2z^2 - 12z + 2z + 12 = z^3 + 4z^2 - 10z + 12$ ✓
$$\boxed{p = 4, \quad q = -10, \quad \text{roots: } z = 1 - \mathrm{i}, \; z = 1 + \mathrm{i}, \; z = -6}$$
One root of $z^4 - 10z^3 + 42z^2 - 82z + 65 = 0$ is $3 + 2\mathrm{i}$. Solve the equation.
💡 Hint
The conjugate $3 - 2\mathrm{i}$ is also a root. Form the quadratic factor: sum $= 6$, product $= 9 + 4 = 13$.
Divide the quartic by $z^2 - 6z + 13$ to get a second quadratic, then solve it.
The second quadratic should also have a negative discriminant, giving another conjugate pair.
👁 Reveal Solution
Since the coefficients are real, $3 - 2\mathrm{i}$ is also a root.
Sum: $(3+2\mathrm{i}) + (3-2\mathrm{i}) = 6$. Product: $(3+2\mathrm{i})(3-2\mathrm{i}) = 9 + 4 = 13$.
Quadratic factor: $z^2 - 6z + 13$
$z^4 - 10z^3 + 42z^2 - 82z + 65 = (z^2 - 6z + 13)(z^2 - 4z + 5)$
Check by expanding: $(z^2 - 6z + 13)(z^2 - 4z + 5) = z^4 - 4z^3 + 5z^2 - 6z^3 + 24z^2 - 30z + 13z^2 - 52z + 65 = z^4 - 10z^3 + 42z^2 - 82z + 65$ ✓
Discriminant: $16 - 20 = -4 < 0$, so complex roots.
$z = \dfrac{4 \pm \sqrt{-4}}{2} = \dfrac{4 \pm 2\mathrm{i}}{2} = 2 \pm \mathrm{i}$
$$\boxed{z = 3 + 2\mathrm{i}, \quad z = 3 - 2\mathrm{i}, \quad z = 2 + \mathrm{i}, \quad z = 2 - \mathrm{i}}$$
The quartic has two conjugate pairs of complex roots and no real roots.
The equation $z^4 - 8z^3 + 20z^2 - 72z + 99 = 0$ has a pure imaginary root. Solve the equation.
💡 Hint 1
Let the pure imaginary root be $z = b\mathrm{i}$ where $b$ is real and $b \neq 0$.
Substitute $z = b\mathrm{i}$ into the equation and use $\mathrm{i}^2 = -1$, $\mathrm{i}^3 = -\mathrm{i}$, $\mathrm{i}^4 = 1$.
Separate the result into real and imaginary parts. Both must equal zero.
💡 Hint 2
The imaginary part gives: $8b^3 - 72b = 0 \Rightarrow 8b(b^2 - 9) = 0$.
Since $b \neq 0$, we get $b = \pm 3$. So the pure imaginary roots are $\pm 3\mathrm{i}$.
The real part must also be zero: $b^4 - 20b^2 + 99 = 0$. Verify this holds for $b = 3$.
Form the quadratic factor from $\pm 3\mathrm{i}$: $z^2 + 9$. Divide the quartic by $z^2 + 9$.
👁 Reveal Solution
$(b\mathrm{i})^4 - 8(b\mathrm{i})^3 + 20(b\mathrm{i})^2 - 72(b\mathrm{i}) + 99 = 0$
Using $\mathrm{i}^2 = -1$, $\mathrm{i}^3 = -\mathrm{i}$, $\mathrm{i}^4 = 1$:
$b^4 + 8b^3\mathrm{i} - 20b^2 - 72b\mathrm{i} + 99 = 0$
Real part: $b^4 - 20b^2 + 99 = 0$ (1)
Imaginary part: $8b^3 - 72b = 0$ (2)
$8b(b^2 - 9) = 0 \Rightarrow b = 0$ or $b = \pm 3$.
Since $b \neq 0$ (pure imaginary), we have $b = 3$ or $b = -3$.
Check in equation (1): $81 - 180 + 99 = 0$ ✓ (for $b = 3$).
So the pure imaginary roots are $z = 3\mathrm{i}$ and $z = -3\mathrm{i}$.
The roots $\pm 3\mathrm{i}$ give the quadratic factor $z^2 + 9$.
$z^4 - 8z^3 + 20z^2 - 72z + 99 = (z^2 + 9)(z^2 - 8z + 11)$
Check: $(z^2 + 9)(z^2 - 8z + 11) = z^4 - 8z^3 + 11z^2 + 9z^2 - 72z + 99 = z^4 - 8z^3 + 20z^2 - 72z + 99$ ✓
Discriminant: $64 - 44 = 20 > 0$, so two real roots.
$z = \dfrac{8 \pm \sqrt{20}}{2} = \dfrac{8 \pm 2\sqrt{5}}{2} = 4 \pm \sqrt{5}$
$$\boxed{z = 3\mathrm{i}, \quad z = -3\mathrm{i}, \quad z = 4 + \sqrt{5}, \quad z = 4 - \sqrt{5}}$$
The quartic has a conjugate pair of pure imaginary roots and two irrational real roots.
Exit Ticket
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