e to the i pi
Before this sheet: Multiplying is rotating and the radians half of the unit circle.
Tonight's idea
Three functions you know can each be written as an endless polynomial. Take these as facts tonight; they are derived in a first calculus course, not here. Angles are in radians: an angle is measured by the arc it cuts from a circle of radius 1, so a full turn is \(2\pi\).
| radians | \(\pi/4\) | \(\pi/2\) | \(\pi\) | \(2\pi\) |
|---|---|---|---|---|
| degrees | 45° | 90° | 180° | 360° |
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \qquad \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \qquad \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]
Now put \(x = i\theta\) into the series for \(e^x\). Powers of \(i\) cycle \(i, -1, -i, 1, \ldots\), so
\[ e^{i\theta} = 1 + i\theta - \frac{\theta^2}{2!} - i\frac{\theta^3}{3!} + \frac{\theta^4}{4!} + i\frac{\theta^5}{5!} - \cdots \]
Collect the real terms: \(1 - \theta^2/2! + \theta^4/4! - \cdots = \cos\theta\). Collect the imaginary ones: \(i(\theta - \theta^3/3! + \cdots) = i\sin\theta\). That is Euler's formula:
\(e^{i\theta} = \cos\theta + i\sin\theta\)
It fits last sheet: angles add when you multiply, exactly as exponents do. Put \(\theta = \pi\): \(\cos\pi = -1\), \(\sin\pi = 0\), so
\(e^{i\pi} + 1 = 0\)
Five of the most important numbers in mathematics, each once. It says that half a turn points you backwards.
Watch one: two trig identities for the price of one
\(e^{i(a+b)} = e^{ia}e^{ib}\), so
\[ \cos(a+b) + i\sin(a+b) = (\cos a + i\sin a)(\cos b + i\sin b) = (\cos a\cos b - \sin a\sin b) + i(\sin a\cos b + \cos a\sin b). \]
Equal complex numbers have equal real parts and equal imaginary parts. Real: \(\cos(a+b) = \cos a\cos b - \sin a\sin b\). Imaginary: \(\sin(a+b) = \sin a\cos b + \cos a\sin b\).
Your turn
- Write each as \(a + bi\): \(e^{i\pi/2}\), \(e^{i\pi}\), \(e^{2\pi i}\), \(e^{i\pi/4}\), \(e^{-i\pi/2}\).
- Multiply \(e^{i\theta}\) by \(e^{-i\theta}\) in two ways: as exponentials, and as \((\cos\theta + i\sin\theta)(\cos\theta - i\sin\theta)\). What identity drops out, and why does it say \(e^{i\theta}\) always lies on the unit circle?
- Square Euler's formula: \((e^{i\theta})^2 = e^{2i\theta}\). Expand the left side and compare parts to get formulas for \(\cos2\theta\) and \(\sin2\theta\). Then show \(\cos2\theta = 2\cos^2\theta - 1\).
- What is \(i^i\)? A complex power is defined by \(z^w = e^{w \log z}\), and since \(i = e^{i\pi/2}\) the usual choice is \(\log i = i\pi/2\). Is the answer real, imaginary or complex? Evaluate it to 4 decimal places.
Drag \(z = e^{i\theta}\) round the unit circle at mathdetours.com/demos/complex.html.
Stretch: why you cannot trisect an angle
Cube Euler's formula, expand with \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\), take real parts, and show \(\cos3\theta = 4\cos^3\theta - 3\cos\theta\). Put \(\theta = 20°\) and \(x = \cos20°\): show \(8x^3 - 6x - 1 = 0\), and check \(x \approx 0.9397\). That this cubic has no rational root is why a 60° angle cannot be trisected with ruler and compass.
Answers
- \(e^{i\pi/2} = i\). \(e^{i\pi} = -1\). \(e^{2\pi i} = 1\). \(e^{i\pi/4} = \tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2}i \approx 0.7071 + 0.7071i\). \(e^{-i\pi/2} = -i\).
- As exponentials: \(e^0 = 1\). As brackets: \(\cos^2\theta - i^2\sin^2\theta = \cos^2\theta + \sin^2\theta\). So \(\cos^2\theta + \sin^2\theta = 1\): the modulus of \(e^{i\theta}\) is 1, so the point is always at distance 1 from the origin.
- \((\cos\theta + i\sin\theta)^2 = \cos^2\theta - \sin^2\theta + 2i\sin\theta\cos\theta\). So \(\cos2\theta = \cos^2\theta - \sin^2\theta\) and \(\sin2\theta = 2\sin\theta\cos\theta\). With \(\sin^2\theta = 1 - \cos^2\theta\): \(\cos2\theta = 2\cos^2\theta - 1\).
- \(i^i = e^{i \cdot i\pi/2} = e^{-\pi/2} \approx 0.2079\). A real number. (The angles \(\pi/2 + 2\pi k\) also give \(i\), and they give other real values; \(0.2079\) is the principal one. The rule \((e^a)^b = e^{ab}\) is only safe when \(b\) is a whole number.)
Stretch. The real part of \((c + is)^3\) is \(c^3 - 3cs^2\), so \(\cos3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = 4\cos^3\theta - 3\cos\theta\). With \(x = \cos20°\) and \(\cos60° = \tfrac12\): \(\tfrac12 = 4x^3 - 3x\), so \(8x^3 - 6x - 1 = 0\). At \(x = 0.9397\): \(8(0.8298) - 5.638 - 1 \approx 0.000\). The only rational candidates are \(\pm1, \pm\tfrac12, \pm\tfrac14, \pm\tfrac18\), and none works. Every ruler-and-compass length comes from square roots, and such a number is never a root of a rational cubic with no rational root. So \(\cos20°\) is not constructible, and 60° cannot be trisected. (That last step is a theorem from a later course.)