What do I learn if a real polynomial has an imaginary root?

Its conjugate is also a root. If 1 + 2i is a root, so is 1 - 2i, and x^2 - 2x + 5 is a factor.

Math II

Complex numbers and equations: practice problems

With the imaginary unit $i$ ( $i^2 = -1$ ), every quadratic equation has solutions. Practise the rules of complex arithmetic and the tools for polynomial equations.

Basic, Standard: Grade 11 · Term 1 / Advanced: Grade 12 · Exam prep

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Arithmetic with complex numbers

A complex number has the form $a + bi$ with $a$ , $b$ real. Treat $i$ like a variable and replace $i^2$ by $-1$ :

(2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 = 5 + i

To divide, multiply the numerator and denominator by the conjugate of the denominator:

\frac{1 + 3i}{1 + i} = \frac{(1 + 3i)(1 - i)}{(1 + i)(1 - i)} = \frac{4 + 2i}{2} = 2 + i

Watch out

Rewrite square roots of negative numbers first: $\sqrt{-2}\sqrt{-8} = \sqrt{2}\,i \cdot 2\sqrt{2}\,i = -4$ , not $\sqrt{16} = 4$ .

The discriminant and Vieta formulas

For $ax^2 + bx + c = 0$ , the discriminant $D = b^2 - 4ac$ decides the type of roots: two real roots ( $D > 0$ ), a double root ( $D = 0$ ) or two imaginary roots ( $D < 0$ ).

Sum and product of roots

\alpha + \beta = -\frac{b}{a},\qquad \alpha\beta = \frac{c}{a}

Symmetric expressions can be written in terms of the sum and product, e.g. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ . A quadratic with roots $p$ and $q$ is $x^2 - (p + q)x + pq = 0$ .

Remainder and factor theorems

The remainder when $P(x)$ is divided by $x - a$ is $P(a)$ . In particular, $P(a) = 0$ exactly when $x - a$ is a factor of $P(x)$ .

A remainder on division by $(x - a)(x - b)$ has the form $px + q$ ; find $p$ and $q$ from $P(a)$ and $P(b)$ .

To solve a cubic, find a root $a$ among the divisors of the constant term, factor out $x - a$ and solve the remaining quadratic. For $x^4 + px^2 + q = 0$ , substitute $t = x^2$ .

Cube roots of unity

If $\omega$ is an imaginary root of $x^3 = 1$ , then $x^3 - 1 = (x - 1)(x^2 + x + 1)$ gives

\omega^3 = 1,\qquad \omega^2 + \omega + 1 = 0.

Reduce powers using $\omega^3 = 1$ (for example $\omega^{100} = \omega$ ) and simplify with $\omega^2 + \omega = -1$ .

Q1BasicGr 11 · Term 1Complex numbers and equations

Let $\alpha$ and $\beta$ be the solutions of $2x^{2} - 2x + 3 = 0$ . Find the value of each expression.

(1) $\alpha^2 + \beta^2$
(2) $(\alpha - \beta)^2$

Answer

(1) $-2$
(2) $-5$

Solution

By the relations between roots and coefficients,
$\alpha + \beta = 1,\quad \alpha\beta = \frac{3}{2}$
(1)
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = -2$
(2)
$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = -5$

Q2StandardGr 11 · Term 1Complex numbers and equations

Find the real numbers $x$ and $y$ that satisfy the equation.

(2 + 3i)(x + yi) = 10 + 2i

Answer

x = 2,\ y = -2

Solution

Rearranging the left side,
$2x - 3y + (3x + 2y)i = 10 + 2i$
Since $x$ and $y$ are real, comparing real and imaginary parts gives
$2x - 3y = 10,\quad 3x + 2y = 2$
Solving,
$x = 2,\quad y = -2$

Q3AdvancedExam prepComplex numbers and equations

Let $\alpha$ , $\beta$ and $\gamma$ be the solutions of $x^{3} + x^{2} + 2x - 6 = 0$ . Find the value of each expression.

(1) $(\alpha + 1)(\beta + 1)(\gamma + 1)$
(2) $\alpha^3 + \beta^3 + \gamma^3$

Answer

(1) $8$
(2) $23$

Solution

By the relations between roots and coefficients,
$\alpha + \beta + \gamma = -1,\quad \alpha\beta + \beta\gamma + \gamma\alpha = 2,\quad \alpha\beta\gamma = 6$
(1)
$(\alpha + 1)(\beta + 1)(\gamma + 1) = \alpha\beta\gamma + (\alpha\beta + \beta\gamma + \gamma\alpha) + (\alpha + \beta + \gamma) + 1 = 8$
(2)
$\alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)^3 - 3(\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) + 3\alpha\beta\gamma = 23$

Q1BasicGr 11 · Term 1Complex numbers and equations

Solve the quadratic equation.

x^{2} - 4x + 5 = 0

Q2StandardGr 11 · Term 1Complex numbers and equations

Let $\alpha$ and $\beta$ be the solutions of $x^{2} - 4x - 6 = 0$ . Find a quadratic equation with integer coefficients whose solutions are $\alpha^2$ and $\beta^2$ .

Q3AdvancedExam prepComplex numbers and equations

Let $a$ and $b$ be real constants. The cubic equation $x^3 + ax^2 + bx + 15 = 0$ has the solution $-1 + 2i$ . Find $a$ , $b$ and the other solutions.

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