Complex number identity question

Question

If $n$ is even and $z=\cos\theta+i\sin\theta$. By expressing $z^n$ in two ways, show that $$\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\cdots+(-1)^{\frac{n}{2}} \binom{n}{n}=2^{\frac{n}{2}}\cos\left(\dfrac{n\pi}{4}\right)$$

Answer 1

We can express $z$ using the binomial expansion:

\begin{align} z^n = & (\cos\theta + i\sin \theta)^n \\ = & C^n_0(\cos\theta)^n(i\sin\theta)^0+ C^n_1(\cos\theta)^{n-1}(i\sin\theta)^1 + \cdots + C^n_n(\cos\theta)^0(i\sin\theta)^n \\ = & C^n_0\cos^n\theta+ iC^n_1\cos^{n-1}\theta\sin\theta +i^2C^n_2 \cos^{n-2}\theta \sin^2\theta \cdots + i^nC^n_n\sin^n\theta \\ = & \big(C^n_0\cos^n\theta - C^n_2\cos^{n-2}\theta\sin^2\theta +C^n_4\cos^{n-4}\theta\sin^4\theta + \cdots + (-1)^{\frac n2}C^n_n\sin^n\theta \big) \\ & +i\big(C^n_1\cos^{n-1}\theta\sin\theta - C^n_3\cos^{n-3}\theta\sin^3\theta + \cdots + (-1)^{\frac {n-2}{2}}C^n_{n-1}\cos\theta\sin^{n-1}\theta \big) \end{align}

On the other hand, we can use de Moivre's theorem:

$$z^n = (\cos\theta + i\sin\theta)^n = \cos(n\theta)+i\sin(n\theta)$$

Equating the expressions and taking real parts, we get

$$C^n_0\cos^n\theta - C^n_2\cos^{n-2}\theta\sin^2\theta +C^n_4\cos^{n-4}\theta\sin^4\theta + \cdots + (-1)^{\frac n2}C^n_n\sin^n\theta = \cos(n\theta) \tag{1}$$

That is, this must hold for all $\theta $ as we have not made any assumptions about $\theta$ when deriving this. In particular, we choose $\theta = \frac \pi 4$. Note that

$$\cos \frac \pi 4 = \sin \frac \pi 4 = \frac{\sqrt{2}}{2}$$

so that equation $(1)$ above becomes

$$C^n_0\bigg(\frac{\sqrt 2}{2}\bigg)^n - C^n_2\bigg(\frac{\sqrt 2}{2}\bigg)^n +C^n_4\bigg(\frac{\sqrt 2}{2}\bigg)^n + \cdots + (-1)^{\frac n2}C^n_n\bigg(\frac{\sqrt 2}{2}\bigg)^n = \cos\bigg(\frac{n\pi}{4}\bigg)$$

$$\implies C^n_0 - C^n_2 + C^n_4 + \cdots + (-1)^{\frac n2}C^n_n = 2^{-\frac n2} \cos \bigg(\frac{n\pi}{4}\bigg)$$