The sum of $(1+i)^x + (1-i)^x$ in a proof

Question

My friend sent me an interesting question: For all non-negative integers $x$, show that:

$$(1+i)^x + (1-i)^x = 2 \sum_{j=0}^{\lfloor x/2\rfloor}\binom{x}{2j}(-1)^j$$

I've tried induction on $x$, but that didn't seem to do the trick. Some hints as to how to approach this proof would be greatly appreciated.

Answer 1

Hint: Use $$ (1+x)^n=\sum_{j=0}^n\binom{n}{j}x^j. $$ Using $$ i^{2j}+(-i)^{2j}=2(-1)^j, i^{2j+1}+(-i)^{2j+1}=0 $$ one has \begin{eqnarray} &&(1+i)^x + (1-i)^x \\ &=& \sum_{j=0}^x\binom{x}{j}\bigg(i^j+(-i)^j\bigg)\\ &=&2\sum_{j=0}^{\lfloor x/2\rfloor}\binom{x}{2j}(-1)^j \end{eqnarray}

Answer 2

The explicit steps in processing the above relation -- which @xpaul has left out, as they are quite simple -- are as follows:

$$\begin{align} &(1+\mathrm{i})^n+(1-\mathrm{i})^n\\ &=\sum_{k=0}^n\binom{n}{k}\left(\mathrm{i}^k+(-\mathrm{i})^k\right)\\ &=\sum_{\substack{0 \leqslant k \leqslant n\\k \in 2\mathbb{N}}}\binom{n}{k}\left(\mathrm{i}^k+(-\mathrm{i})^k\right)+\sum_{\substack{0 \leqslant k \leqslant n\\k \in 2\mathbb{N}+1}}\binom{n}{k}\left(\mathrm{i}^k+(-\mathrm{i})^k\right)\\ &=\sum_{h=0}^{\left[\frac{n}{2}\right]}\binom{n}{2h}\left(\mathrm{i}^{2h}+(-\mathrm{i})^{2h}\right)\\ &=\sum_{h=0}^{\left[\frac{n}{2}\right]}\binom{n}{2h}\left((-1)^h+(-1)^h\right)\\ &=2\sum_{h=0}^{\left[\frac{n}{2}\right]}(-1)^h\binom{n}{2h}, \end{align}$$ where we need to bear in mind two important justifications for the above:

• for odd $k \in \mathbb{N}$ we obviously have $\mathrm{i}^k+(-\mathrm{i})^k=\mathrm{i}^k-\mathrm{i}^k=0$
• the map: $$\begin{align} \left\{h \in \mathbb{N} \mid h \leqslant \left[\frac{n}{2}\right]\right\} &\to \{k \in \mathbb{N} \mid 2 \mid k \wedge k \leqslant n \}\\ h&\mapsto 2h \end{align}$$ is a bijection on the grounds of which we can perform a bijective change of summation index.

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P.S. The relation that you seek to prove is valid in a more general setting, as it relies on the polynomial identity: $$(X+\mathrm{i}Y)^n+(X-\mathrm{i}Y)^n=2\sum_{k=0}^{\left[\frac{n}{2}\right]}(-1)^k\binom{n}{2k}X^{n-2k}Y^{2k},$$ considered in the polynomial ring $\mathbb{C}[X, Y]$.

However, given your particular choice of complex numbers $1\pm \mathrm{i}$ it would be a pity to pass the occasion of noticing a remarkable property they exhibit and which enables us to give yet another description of the general sum you are studying. For simplicity let us write $\varepsilon=\frac{1+\mathrm{i}}{\sqrt{2}}$ and notice that: $$\varepsilon^2=\frac{1+\mathrm{i}^2+2\mathrm{i}}{2}=\mathrm{i},$$ which furthermore entails the relations $\varepsilon^4=-1$ respectively $\varepsilon^8=1$. In other words, $\varepsilon$ is a so-called primitive $8$-th root of unity.

With this in mind we infer that: $$\begin{align*} (1+\mathrm{i})^n+(1-\mathrm{i})^n &=\left(\sqrt{2}\varepsilon\right)^n+\left(\sqrt{2}\overline{\varepsilon}\right)^n\\ &=\sqrt{2}^n\left(\varepsilon^n+\frac{1}{\varepsilon^n}\right)\\ &=\sqrt{2}^n\frac{\varepsilon^{2n}+1}{\varepsilon^n}\\ &=\sqrt{2}^n\frac{\mathrm{i}^n+1}{\varepsilon^n}\\ &=\sqrt{2}^n\begin{cases} 2, &n \equiv 0 \pmod{8}\\ \sqrt{2}, &n \equiv 1, 7 \pmod{8}\\ 0, &n \equiv 2, 6 \pmod{8}\\ -\sqrt{2}, &n \equiv 3, 5 \pmod{8}\\ -2, &n \equiv 4 \pmod{8} \end{cases}\\ &=\begin{cases} 2^{\frac{n}{2}+1}, &n \equiv 0 \pmod{8}\\ 2^{\frac{n+1}{2}}, &n \equiv 1, 7 \pmod{8}\\ 0, &n \equiv 2, 6 \pmod{8}\\ -2^{\frac{n+1}{2}}, & n \equiv 3, 5 \pmod{8}\\ -2^{\frac{n}{2}+1}, & n \equiv 4 \pmod{8}, \end{cases} \end{align*}$$

relations from which you can in particular obtain an answer to the related question you posted separately on this forum.