What is Complex exponential of an imaginary argument?

Example 3.6 (First values, and one famous identity)

Reading the definition at the cardinal angles:

The second one, rearranged as $\eu^{\iu\pi} + 1 = 0$, is Euler’s celebrated identity linking $\eu$, $\iu$, $\pi$, $1$ and $0$; at this stage of the book it is a definition unwinding rather than a theorem, and its real content — why the analytic exponential function of Chapter 4, extended to imaginary arguments, deserves the same name — is settled by the power series of Chapter 17. Meanwhile the display above is worth memorizing as a conversion table: it is used silently every time an argument is read off a picture.

Example 3.9 (Expanding $\cos 3\theta$ by de Moivre)

Write $c = \cos\theta$, $s = \sin\theta$. De Moivre and the binomial theorem give

and identifying real parts, then substituting $s^2 = 1 - c^2$:

The imaginary part yields $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ for free: one complex identity always carries two real ones. Read backwards, the boxed identity is the key to the classical trisection equation: constructing $\cos(\theta/3)$ from $\cos\theta$ means solving the cubic $4x^3 - 3x = \cos\theta$, which is where the algebra of Chapter 8 takes over.

Example 3.13 (A binomial trigonometric sum)

For $n \in \N$ and $\theta \in \R$, evaluate $S = \sum_{k=0}^{n}\binom nk \cos k\theta$. Recognize the real part of a binomial expansion:

then factor the half-angle (Method 3.11 (4)): $1 + \eu^{\iu\theta} = 2\cos\frac\theta2\,\eu^{\iu\theta/2}$, so

The imaginary part yields $\sum_k\binom nk\sin k\theta = 2^n\cos^n\frac\theta2\sin\frac{n\theta}2$ for free. Sanity checks: $\theta = 0$ recovers $\sum\binom nk = 2^n$, and $\theta = \pi$ gives $S = 0$ for $n \geq 1$ (each factor $\cos\frac\pi2$ vanishes), i.e. the alternating row sum of Example 2.17. The method — “see the cosine sum as the shadow of a complex power, then factor half-angles” — is exactly that of Exercise 3.6, with the binomial theorem replacing the geometric series.