Try solving \(x^2 = -1\) with the numbers you know: every real number squared is zero or positive, so there is no solution. Mathematicians fixed this by inventing a new number, and it turned out to be one of the most useful ideas ever: electrical engineers, game designers, and signal analysts use it every day. In this chapter you will learn to calculate with complex numbers, solve every quadratic equation, and picture these numbers as points in a plane.
1. The imaginary unit i
The imaginary unit is the number \(i\) defined by \(i^2 = -1\). We also write \(i = \sqrt{-1}\).
A complex number is a number that can be written in standard form \(a + bi\), where \(a\) and \(b\) are real numbers. We call \(a\) the real part and \(b\) the imaginary part. For example, in \(5 - 4i\) the real part is \(5\) and the imaginary part is \(-4\) (the number \(-4\), not \(-4i\)). If \(b = 0\) the number is real, so every real number is also a complex number. If \(a = 0\), the number \(bi\) is called pure imaginary. Two complex numbers are equal exactly when their real parts are equal and their imaginary parts are equal.
For any positive real number \(n\), \(\sqrt{-n} = i\sqrt{n}\).
Write \(\sqrt{-36}\) and \(\sqrt{-45}\) in terms of \(i\).
\(\sqrt{-36} = i\sqrt{36} = 6i\).
\(\sqrt{-45} = i\sqrt{45} = i\sqrt{9 \cdot 5} = 3i\sqrt{5}\).
The rule \(\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}\) only works when \(a\) and \(b\) are not negative. Always rewrite each radical with \(i\) first: \(\sqrt{-4}\cdot\sqrt{-9} = 2i \cdot 3i = 6i^2 = -6\), not \(6\).
2. Powers of i
Start multiplying by \(i\) again and again: \(i^1 = i\), \(i^2 = -1\), \(i^3 = i^2 \cdot i = -i\), \(i^4 = i^2\cdot i^2 = 1\), and then \(i^5 = i\) again. The pattern repeats every four steps.
- Divide the exponent by \(4\) and keep only the remainder \(r\).
- Replace \(i^n\) by \(i^r\), with \(i^0 = 1\).
- Use the cycle \(1,\ i,\ -1,\ -i\).
Simplify \(i^{58}\) and \(i^{103}\).
\(58 = 4 \cdot 14 + 2\), so \(i^{58} = i^2 = -1\).
\(103 = 4 \cdot 25 + 3\), so \(i^{103} = i^3 = -i\).
3. Adding, subtracting, and multiplying
To add or subtract complex numbers, combine the real parts together and the imaginary parts together, just like combining like terms: \((a + bi) + (c + di) = (a + c) + (b + d)i\). To multiply, distribute (FOIL works well) and then replace \(i^2\) by \(-1\).
Compute \((3 + 2i)(4 - 5i)\).
\((3 + 2i)(4 - 5i) = 12 - 15i + 8i - 10i^2 = 12 - 7i - 10(-1) = 22 - 7i\).
On my home planet we say: “every time an \(i^2\) shows up, it turns into a \(-1\)”. After that, the answer must have only one \(i\) in it, never \(i^2\).
4. Conjugates and division
The conjugate of \(z = a + bi\) is \(\overline{z} = a - bi\): only the sign of the imaginary part changes.
The product of a complex number and its conjugate is always a non-negative real number: \((a + bi)(a - bi) = a^2 - b^2 i^2 = a^2 + b^2\). This is exactly what we need to divide: multiplying the top and the bottom of a fraction by the conjugate of the denominator removes \(i\) from the denominator.
- Write the conjugate of the denominator.
- Multiply the numerator and the denominator by it.
- Expand the numerator; the denominator becomes \(a^2 + b^2\).
- Write the answer in the form \(a + bi\).
Write \(\dfrac{5 + i}{2 - 3i}\) in standard form.
\(\dfrac{5 + i}{2 - 3i} = \dfrac{(5 + i)(2 + 3i)}{(2 - 3i)(2 + 3i)} = \dfrac{10 + 15i + 2i + 3i^2}{4 + 9} = \dfrac{7 + 17i}{13} = \dfrac{7}{13} + \dfrac{17}{13}i\).
5. Complex solutions of quadratics
For \(ax^2 + bx + c = 0\), the quadratic formula gives \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). When the discriminant \(b^2 - 4ac\) is negative, the radical is the square root of a negative number, and the equation now has solutions: two complex ones.
If a quadratic equation has real coefficients and a negative discriminant, it has exactly two non-real solutions, and they are complex conjugates of each other.
Solve \(x^2 + 6x + 25 = 0\).
The discriminant is \(6^2 - 4(1)(25) = 36 - 100 = -64 < 0\).
\(x = \dfrac{-6 \pm \sqrt{-64}}{2} = \dfrac{-6 \pm 8i}{2} = -3 \pm 4i\).
Graphically, a negative discriminant means that the parabola never touches the \(x\)-axis. For instance, \(y = x^2 + 2x + 5\) has its lowest point at \((-1,\,4)\), so it stays above the axis. Its complex roots are \(-1 \pm 2i\).
6. The complex plane
Every complex number \(a + bi\) matches exactly one point \((a, b)\). In the complex plane the horizontal axis is the real axis and the vertical axis is the imaginary axis. The conjugate of a point is its mirror image across the real axis.
Adding complex numbers moves points like adding vectors. Multiplying by \(i\) turns a point a quarter turn (\(90^\circ\)) counterclockwise around the origin: \(i(a + bi) = -b + ai\). That is why the powers of \(i\) hop around the plane in a square.
7. Absolute value of a complex number
The absolute value of \(z = a + bi\) is its distance from the origin in the complex plane: \[ |a + bi| = \sqrt{a^2 + b^2} \]
The formula comes from the Pythagorean theorem in the right triangle with legs \(|a|\) and \(|b|\). Some useful facts: \(|z| = |\overline{z}|\), \(z\cdot\overline{z} = |z|^2\), \(|zw| = |z|\,|w|\), and the distance between two complex numbers \(z\) and \(w\) is \(|z - w|\).
Find \(|-5 + 12i|\), and the distance between \(1 + 2i\) and \(4 + 6i\).
\(|-5 + 12i| = \sqrt{25 + 144} = \sqrt{169} = 13\).
The difference is \((4 + 6i) - (1 + 2i) = 3 + 4i\), so the distance is \(\sqrt{9 + 16} = 5\) units.
Key takeaways
- \(i^2 = -1\), and \(\sqrt{-n} = i\sqrt{n}\) for \(n > 0\). Rewrite radicals with \(i\) before multiplying.
- Powers of \(i\) repeat in a cycle of four: \(i,\ -1,\ -i,\ 1\). Use the remainder when the exponent is divided by \(4\).
- Add and subtract the real and imaginary parts separately; multiply with FOIL and replace \(i^2\) by \(-1\).
- The conjugate of \(a + bi\) is \(a - bi\), and \((a + bi)(a - bi) = a^2 + b^2\). Use it to divide.
- A quadratic with real coefficients and a negative discriminant has two conjugate complex solutions.
- \(a + bi\) is the point \((a, b)\) in the complex plane, and \(|a + bi| = \sqrt{a^2 + b^2}\) is its distance from the origin.
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