Multiplying is rotating
Tonight's idea
A complex number is \(z = a + bi\) with \(i^2 = -1\) (the grade 6 imaginary numbers sheet). Treat it as an arrow from the origin to the point \((a, b)\). Describe the arrow by its modulus \(r = |z| = \sqrt{a^2 + b^2}\), its length, and its argument \(\theta\), the angle it makes with the positive real axis. Since \(a = r\cos\theta\) and \(b = r\sin\theta\),
\(z = r(\cos\theta + i\sin\theta)\), abbreviated \(r\operatorname{cis}\theta\).
Now multiply \(z = r\operatorname{cis}\alpha\) by \(w = s\operatorname{cis}\beta\):
\[ zw = rs\big[(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)\big]. \]
The brackets are the angle-addition formulas for \(\cos(\alpha+\beta)\) and \(\sin(\alpha+\beta)\). So \(zw = rs\operatorname{cis}(\alpha + \beta)\).
To multiply complex numbers, multiply the lengths and add the angles.
Multiplying by \(i\) (length 1, angle 90°) is a quarter turn, which is why \(i^2 = -1\): two quarter turns point you backwards. Repeating the rule gives De Moivre's theorem: \((r\operatorname{cis}\theta)^n = r^n\operatorname{cis}(n\theta)\).
Run it backwards to take roots. For \(z^3 = -8\): \(r^3 = 8\) gives \(r = 2\), and \(3\theta\) must be \(180°\), or \(540°\), or \(900°\), which all point left, so \(\theta = 60°, 180°\) or \(300°\): three roots, equally spaced.
Watch one: a coincidence, and a big power
\((2 + i)(3 + i) = 6 + 5i + i^2 = 5 + 5i\). Angles: \(2 + i\) sits at \(\arctan\tfrac12\), \(3 + i\) at \(\arctan\tfrac13\), and \(5 + 5i\) at exactly 45°. So \(\arctan\tfrac12 + \arctan\tfrac13 = 45°\), proved by one multiplication.
Now \((1 + i)^8\). Length \(\sqrt2\), angle 45°. De Moivre: length \((\sqrt2)^8 = 16\), angle \(8 \times 45° = 360°\), the same direction as 0°. So \((1+i)^8 = 16\).
Your turn
- Find the modulus and argument of \(3 + 4i\), of \(-1 + i\), and of \(-2i\).
- Write \(\sqrt3 + i\) as \(r\operatorname{cis}\theta\), then compute \((\sqrt3 + i)^6\) with De Moivre. Give the answer as \(a + bi\).
- Rotate the point \((2, 1)\), treated as \(2 + i\), by 90° anticlockwise about the origin by multiplying by the right complex number. Then rotate it by 60° instead, using \(\operatorname{cis}60° = \tfrac12 + \tfrac{\sqrt3}{2}i\), and give the new coordinates to 3 decimal places.
- Solve \(z^3 = 8i\). (Hint: \(8i\) has length 8 and angle 90°, but also 450° and 810°. Divide each by 3 and compare with the figure.) Give all three answers as \(a + bi\) and check one of them by cubing it.
Drag \(z\) and watch \(iz\) and \(z^2\) rotate at mathdetours.com/demos/complex.html.
Stretch: the golden ratio in a pentagon
Let \(z = \operatorname{cis}72°\), so \(z^5 = 1\) and \(1 + z + z^2 + z^3 + z^4 = 0\) (because \((z-1)(1 + z + \cdots + z^4) = z^5 - 1 = 0\) and \(z \ne 1\)). Let \(w = z + z^4 = 2\cos72°\) and \(w' = z^2 + z^3\). Show \(w + w' = -1\) and \(ww' = -1\), so \(w\) is a root of \(t^2 + t - 1 = 0\). Deduce an exact value for \(\cos72°\).
Answers
- \(3 + 4i\): modulus 5, argument \(\arctan\tfrac43 \approx 53.13°\). \(-1 + i\): modulus \(\sqrt2\), argument 135°. \(-2i\): modulus 2, argument \(-90°\) (or 270°).
- \(r = 2\), \(\theta = 30°\), so \(\sqrt3 + i = 2\operatorname{cis}30°\). Then \((\sqrt3 + i)^6 = 2^6\operatorname{cis}180° = -64\).
- Multiply by \(i\): \(i(2 + i) = -1 + 2i\), the point \((-1, 2)\). By 60°: \((\tfrac12 + \tfrac{\sqrt3}{2}i)(2 + i) = (1 - \tfrac{\sqrt3}{2}) + (\tfrac12 + \sqrt3)i \approx 0.134 + 2.232i\), the point \((0.134, 2.232)\). Both have length \(\sqrt5\), as they must.
- Length 2; angles \(30°, 150°, 270°\): the three points of the figure turned by 30°, still 120° apart on the radius-2 circle. Solutions \(\sqrt3 + i\), \(-\sqrt3 + i\), \(-2i\). Check: \((\sqrt3 + i)^2 = 2 + 2\sqrt3\, i\); times \((\sqrt3 + i)\) gives \(2\sqrt3 + 8i - 2\sqrt3 = 8i\).
Stretch. \(z^4 = \operatorname{cis}(-72°)\) is the mirror image of \(z\), so \(z + z^4 = 2\cos72°\). \(w + w' = z + z^2 + z^3 + z^4 = -1\). \(ww' = z^3 + z^4 + z^6 + z^7 = z^3 + z^4 + z + z^2 = -1\) using \(z^5 = 1\). So \(w, w'\) are the roots of \(t^2 + t - 1 = 0\), namely \(\frac{-1 \pm \sqrt5}{2}\). \(w = 2\cos72°\) is positive, so \(\cos72° = \frac{\sqrt5 - 1}{4} \approx 0.309\). The other root is \(-\varphi\), minus the golden ratio: a pentagon's diagonals are \(\varphi\) times its sides for exactly this reason.