Calculating $\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$ 
Calculate: $$\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$$

The solution of this exercise:
Let $$S_1=\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{8}+\cdots$$ 
$$S_2=\binom{n}{1}-\binom{n}{3}+\binom{n}{5}-\cdots$$
$$S_3=\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots$$
$$S_4=\binom{n}{2}+\binom{n}{6}+\binom{n}{10}+\cdots$$ 
And we consider $$(1+i)^n=S_1+iS_2=\sqrt2^n\left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right)$$ and $$2^{n-1}+S_1=2S_3$$
The problem is that i didn't get the part with $(1+i)^n$.. from here i got lost.I saw more exercises like this with combinatorial sums whose solution was about complex numbers and i wish that someone explain me that method.Thanks!
 A: As we need every fourth term,
Calculate $$a(1+1)^n+b(1-1)^n+c(1+i)^n+d(1-i)^n=\binom n0+\binom n4+\cdots$$
Compare the coefficients of $\binom n0=1$ and those of $\binom nr,1\le r\le3$ 
to find $a,b,c,d$
A: The part with $(1+i)^n$ is explained by De Moivre:
$$(1+i)^n=\sqrt2^ne^{i\frac{n\pi}{4}}=\sqrt2^n(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})^n=\sqrt2^n(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4})$$
Now comparing the $\Re$ parts of the LHS and RHS of the given equation:
$$\Re(\sqrt2^n(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}))=\Re(S_1+iS_2)$$
$$\iff\sqrt2^n\cos\frac{n\pi}{4}=S_1$$
$$\iff \sqrt2^n\cos\frac{n\pi}{4}=2^{n-1}-2S_3.$$  Thereby,
$$S_3=\binom{n}{0}+\binom{n}{4}+\dots=\frac{1}{2}\left(2^{n-1}-\sqrt2^n\cos\frac{n\pi}{4}\right).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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With $\ds{n \in \mathbb{N}_{\ \geq\ 0}}$:

\begin{align}
\sum_{k = 0}^{\infty}{n \choose 4k} & =
\sum_{k = 0}^{\infty}{n \choose k}{1^{k} + \pars{-1}^{k} + \ic^{k} + \pars{-\ic}^{k} \over 4}
\\[5mm] & =
{1 \over 4}\
\underbrace{\sum_{k = 0}^{\infty}{n \choose k}1^{k}}_{\ds{2^{n}}}\ +\
{1 \over 4}\
\underbrace{\sum_{k = 0}^{\infty}{n \choose k}\pars{-1}^{k}}
_{\ds{\delta_{n0}}} +
{1 \over 2}\,\Re\
\underbrace{\sum_{k = 0}^{\infty}{n \choose k}\ic^{k}}
_{\ds{\pars{1 + \ic}^{n}}}
\\[5mm] & =
2^{n - 2} + {\delta_{n0} \over 4} +
{1 \over 2}\,\Re\bracks{2^{n/2}\expo{n\pi\ic/4}}
\\[5mm] & =
\bbx{2^{n - 2} + {\delta_{n0} \over 4} +
2^{n/2 - 1}\cos\pars{n\pi \over 4}}
\end{align}
