Dividing by a complex number
Tonight's idea
In grade 10 a complex number \(a + bi\) was an arrow, and multiplying turned it. Tonight is the arithmetic. Adding and multiplying are ordinary algebra, with \(i^2\) replaced by \(-1\):
\[(3 + 2i)(1 - 5i) = 3 - 15i + 2i - 10i^2 = 13 - 13i.\]
Division is harder. What is \(\dfrac{1}{3 + 2i}\) in the form \(a + bi\)? The tool is the conjugate. The conjugate of \(z = a + bi\) is \(\bar z = a - bi\), its mirror image in the real axis.
\(z\bar z = (a + bi)(a - bi) = a^2 + b^2\), a real number.
So to divide, multiply top and bottom by the conjugate of the bottom, as when you rationalised a denominator. The bottom becomes real:
\[\frac{1}{3 + 2i} = \frac{3 - 2i}{(3 + 2i)(3 - 2i)} = \frac{3 - 2i}{13}.\]
Now quadratics. When \(b^2 - 4ac\) is negative the formula still works, because \(\sqrt{-36} = 6i\). For \(x^2 - 4x + 13 = 0\): \(x = \dfrac{4 \pm \sqrt{-36}}{2} = 2 \pm 3i\). The two roots are conjugates of each other; next year you will see why.
Watch one: \((5 + 5i) \div (3 + i)\)
The conjugate of the bottom is \(3 - i\). Multiply top and bottom by it:
\[\frac{5 + 5i}{3 + i} = \frac{(5 + 5i)(3 - i)}{(3 + i)(3 - i)} = \frac{15 - 5i + 15i - 5i^2}{9 + 1} = \frac{20 + 10i}{10} = 2 + i.\]
Check by multiplying back: \((2 + i)(3 + i) = 6 + 5i + i^2 = 5 + 5i\). Division undoes multiplication, as it should.
Drag the point \(z\) and watch \(iz\) and \(z^2\) turn with it: mathdetours.com/demos/complex.html.
Your turn
- Let \(z = 4 - 3i\) and \(w = 1 + 2i\). Find \(z + w\), \(zw\) and \(z\bar z\).
- Write each in the form \(a + bi\): \(\dfrac{1 + 7i}{2 - i}\) and \(\dfrac{1}{i}\).
- Solve \(x^2 + 2x + 5 = 0\). Then add your two roots and multiply them. Compare with the coefficients of the equation.
- Let \(z = 0.6 + 0.8i\). Show that \(z\bar z = 1\), so that \(\dfrac1z = \bar z\). Use this to find \(\dfrac{i}{z}\). Multiplying by \(z\) turns an arrow anticlockwise through the angle of \(z\). What does dividing by \(z\) do?
Stretch: the square root of a complex number
Find both complex numbers \(z\) with \(z^2 = -3 + 4i\). Write \(z = a + bi\) with \(a\) and \(b\) real, square it, and match the real parts and the imaginary parts.
Answers
- \(z + w = 5 - i\). \(zw = 4 + 8i - 3i - 6i^2 = 10 + 5i\). \(z\bar z = 16 + 9 = 25\).
- \(\dfrac{(1 + 7i)(2 + i)}{(2 - i)(2 + i)} = \dfrac{2 + i + 14i + 7i^2}{5} = \dfrac{-5 + 15i}{5} = -1 + 3i\). And \(\dfrac1i = \dfrac{-i}{i \times (-i)} = \dfrac{-i}{1} = -i\).
- \(x = \dfrac{-2 \pm \sqrt{4 - 20}}{2} = \dfrac{-2 \pm 4i}{2} = -1 \pm 2i\). Sum: \(-2\). Product: \((-1)^2 + 2^2 = 5\). These are \(-b/a\) and \(c/a\), the same rule as for real roots (grade 10).
- \(z\bar z = 0.36 + 0.64 = 1\). Divide both sides by \(z\): \(\bar z = 1/z\). So \(\dfrac{i}{z} = i\bar z = i(0.6 - 0.8i) = 0.8 + 0.6i\). The conjugate has the opposite angle, so dividing by \(z\) turns an arrow clockwise through the angle of \(z\). In general, dividing divides the lengths and subtracts the angles.
Stretch. \((a + bi)^2 = (a^2 - b^2) + 2abi\), so \(a^2 - b^2 = -3\) and \(2ab = 4\). Put \(b = 2/a\): \(a^2 - 4/a^2 = -3\), so \(a^4 + 3a^2 - 4 = 0\), which factors as \((a^2 + 4)(a^2 - 1) = 0\). Since \(a\) is real, \(a = \pm 1\) and \(b = \pm 2\). The two square roots are \(1 + 2i\) and \(-1 - 2i\). Check: \((1 + 2i)^2 = 1 + 4i - 4 = -3 + 4i\).