Complex Numbers — Part C
Quadratic equations and polynomials
🔢 Square root of a negative number
\(\sqrt{-a} = \sqrt{a} \cdot i\)
(when \(a > 0\))
examples:
\(\sqrt{-4} = \sqrt{4} \cdot i = 2i\)
\(\sqrt{-9} = \sqrt{9} \cdot i = 3i\)
\(\sqrt{-5} = \sqrt{5} \cdot i\)
\(\sqrt{-12} = \sqrt{12} \cdot i = 2\sqrt{3} \cdot i\)
⚠️ Caution! \(\sqrt{-a} \cdot \sqrt{-b} \neq \sqrt{ab}\)
For example: \(\sqrt{-2} \cdot \sqrt{-2} = i\sqrt{2} \cdot i\sqrt{2} = i^2 \cdot 2 = -2\)
And not: \(\sqrt{(-2)(-2)} = \sqrt{4} = 2\) ❌
📐 Reminder: quadratic equation
For the equation \(ax^2 + bx + c = 0\), the solutions are:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
the discriminant: \(\Delta = b^2 - 4ac\)
| \(\Delta > 0\) | two different real solutions |
| \(\Delta = 0\) | one real solution (double root) |
| \(\Delta < 0\) | two conjugate complex solutions |
⭐ solution when Δ < 0
When the discriminant is negative, use complex numbers:
\(x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-b \pm i\sqrt{|\Delta|}}{2a}\)
💡 Important result:
The two solutions are always conjugate numbers!
If \(z_1 = a + bi\) solution, then also \(z_2 = a - bi = \bar{z_1}\) solution.
✏️ Example 1: simple equation
solve: \(x^2 + 4 = 0\)
Solution:
\(x^2 = -4\)
\(x = \pm\sqrt{-4} = \pm 2i\)
Answer: \(x = 2i\) or \(x = -2i\)
✏️ Example 2: using the formula
solve: \(x^2 - 4x + 13 = 0\)
Solution:
step 1: identify the coefficients
\(a = 1, \, b = -4, \, c = 13\)
step 2: calculate discriminant
\(\Delta = b^2 - 4ac = 16 - 52 = -36\)
The discriminant negative → complex solutions
step 3: substitute into the formula
\(x = \frac{-(-4) \pm \sqrt{-36}}{2 \cdot 1} = \frac{4 \pm 6i}{2} = 2 \pm 3i\)
Answer: \(x = 2 + 3i\) or \(x = 2 - 3i\)
💡 Note: the solutions are conjugates of each other!
✏️ Example 3: leading coefficient ≠ 1
solve: \(2x^2 + 2x + 5 = 0\)
Solution:
\(a = 2, \, b = 2, \, c = 5\)
\(\Delta = 4 - 40 = -36\)
\(x = \frac{-2 \pm \sqrt{-36}}{4} = \frac{-2 \pm 6i}{4} = \frac{-2}{4} \pm \frac{6i}{4}\)
\(x = -\frac{1}{2} \pm \frac{3}{2}i\)
Answer: \(x = -\frac{1}{2} + \frac{3}{2}i\) or \(x = -\frac{1}{2} - \frac{3}{2}i\)
🔨 Building an equation from complex solutions
If \(z_1\) and \(z_2\) are solutions, the equation is:
\((x - z_1)(x - z_2) = 0\)
Example: build an equation whose solutions are \(z_1 = 1 + 2i\) and \(z_2 = 1 - 2i\)
\((x - (1+2i))(x - (1-2i)) = 0\)
\(((x-1) - 2i)((x-1) + 2i) = 0\)
This is of the form \((A - B)(A + B) = A^2 - B^2\)
\((x-1)^2 - (2i)^2 = 0\)
\((x-1)^2 - 4i^2 = 0\)
\((x-1)^2 + 4 = 0\)
\(x^2 - 2x + 1 + 4 = 0\)
Answer: \(x^2 - 2x + 5 = 0\)
📐 Vieta's formulas (also in complex numbers!)
For the equation \(x^2 + px + q = 0\) with solutions \(z_1, z_2\):
\(z_1 + z_2 = -p\)
\(z_1 \cdot z_2 = q\)
💡 Important:
If the solutions are conjugates (\(z_1 = a+bi, z_2 = a-bi\)):
- \(z_1 + z_2 = 2a\) (always real!)
- \(z_1 \cdot z_2 = a^2 + b^2 = |z_1|^2\) (always real positive!)
🔗 Polynomial factorisation
If \(z_1, z_2\) are roots of \(ax^2 + bx + c\), then:
\(ax^2 + bx + c = a(x - z_1)(x - z_2)\)
Example: factor \(x^2 + 4\)
Roots: \(x^2 = -4 \Rightarrow x = \pm 2i\)
\(x^2 + 4 = (x - 2i)(x + 2i)\)
check:
\((x - 2i)(x + 2i) = x^2 - (2i)^2 = x^2 - 4i^2 = x^2 + 4\) ✓
✏️ Example 4: factorisation
factor: \(x^2 - 6x + 13\)
Solution:
step 1: find roots
\(\Delta = 36 - 52 = -16\)
\(x = \frac{6 \pm 4i}{2} = 3 \pm 2i\)
step 2: write factorisation
\(x^2 - 6x + 13 = (x - (3+2i))(x - (3-2i))\)
Answer: \((x - 3 - 2i)(x - 3 + 2i)\)
📋 Summary table — Part C
| topic | formula/rule |
|---|---|
| root negative | \(\sqrt{-a} = i\sqrt{a}\) |
| Δ < 0 | conjugate complex solutions |
| sum roots | \(z_1 + z_2 = -\frac{b}{a}\) |
| product of roots | \(z_1 \cdot z_2 = \frac{c}{a}\) |
| factorisation | \(a(x-z_1)(x-z_2)\) |
💡 exam tips
1️⃣ Δ is negative
No panic! the solutions are conjugate complex numbers
2️⃣ conjugates
If \(a+bi\) solution, also \(a-bi\) solution
3️⃣ check
Substitute back and verify that 0 is obtained
4️⃣ root negative
\(\sqrt{-a} = i\sqrt{a}\) do not get confused!
📝 Summary — Part C
Quadratic equation with \(\Delta < 0\) → conjugate complex solutions
\(\sqrt{-a} = i\sqrt{a}\)
This is the basis of complex numbers!