4j+3}i^3\\
&=&\displaystyle \left(\sum_{j=0}^m{4m\choose 4j}-\sum_{j=0}^m{4m\choose 4j+2}\right)+\left(\sum_{j=0}^m{4m\choose 4j+1}-\sum_{j=0}^m{4m\choose 4j+3}\right)i.\end{aligned}$$ It leads to that $$\label{eq-x-1}\left\{\begin{array}{l}
\displaystyle \sum_{j=0}^m{4m\choose 4j}-\sum_{j=0}^m{4m\choose 4j+2}=2^{2m}\cos{m\pi}=(-1)^m2^{2m}\\
\displaystyle \sum_{j=0}^m{4m\choose 4j+1}-\sum_{j=0}^m{4m\choose 4j+3}=2^{2m}\sin{m\pi}=0.
\end{array}\right.$$ From Lemma \[lem-z-5\], we have $$\label{eq-x-2}
\sum_{j=0}^m{4m\choose 4j}+\sum_{j=0}^m{4m\choose 4j+2}=\sum_{j=0}^m{4m\choose 4j+1}+\sum_{j=0}^m{4m\choose 4j+3}=2^{4m-1}.$$ Thus, the result follows from (\[eq-x-1\]) and (\[eq-x-2\]).

Similarly, by respectively considering $(1+i)^{4m+1}$, $(1+i)^{4m+2}$ and $(1+i)^{4m+3}$, we get the following result.

\[lem-z-7\] For a positive integer $m$, we have\
(a) $\displaystyle \sum_{j=0}^m{4m+1\choose 4j}=\sum_{j=0}^m{4m+1\choose 4j+1}=2^{4m-1}+(-1)^m2^{2m-1}$;\
(b) $\displaystyle \sum_{j=0}^m{4m+1\choose 4j+2}=\sum_{j=0}^m{4m+1\choose 4j+3}=2^{4m-1}-(-1)^m2^{2m-1}$;\
(c) $\displaystyle \sum_{j=0}^m{4m+2\choose 4j}=\sum_{j=0}^m{4m+2\choose 4j+2}=2^{4m}$;\
(d) $\displaystyle \sum_{j=0}^m{4m+2\choose 4j+1}=2^{4m}+(-1)^m2^{2m}$;\
(e) $\displaystyle \sum_{j=0}^m{4m+2\choose 4j+3}=2^{4m}-(-1)^m2^{2m}$;\
(f) $\displaystyle \sum_{j=0}^m{4m+3\choose 4j}=\sum_{j=0}^m{4m+3\choose 4j+3}=2^{4m+1}-(-1)^m2^{2m}$;\
(g) $\displaystyle \sum_{j=0}^m{4m+3\choose 4j+1}=\sum_{j=0}^m{4m+3\choose 4j+2}=2^{4m+1}+(-1)^m2^{2m}$.

As similar as the proof of Lemma \[lem-z-6\], we have $$\begin{aligned}
&&\displaystyle 2^{2m}\cos{m\pi}+(2^{2m}\cos{m\pi}) i=(\sqrt{2}e^{\pi i/4})^{4m+1}=(1+i)^{4m+1}\\
&=&\displaystyle \sum_{j=0}^m{4m+1\choose 4j}+\sum_{j=0}^m{4m+1\choose 4j+2}i^2+\sum_{j=0}^m{4m+1\choose 4j+1}i+\sum_{j=0}^m{4m+1\choose 4j+3}i^3\\
&=&\displaystyle \left(\sum_{j=0}^m{4m+1\choose 4j}-\sum_{j=0}^m{4m+1\choose 4j+2}\right)+\left(\sum_{j=0}^m{4m+1\choose 4j+1}-\sum_{j=0}^m{4m+1\choose 4j+3}\right)i.\end{aligned}$$ It leads to that $$\left\{\begin{array}{l}
\displaystyle \sum_{j=0}^m{4m+1\choose 4j}-\sum_{j=0}^m{4m+1\choose 4j+2}=2^{2m}\cos{m\pi}=(-1)^m2^{2m}\\
\displaystyle\sum_{j=0}^m{4m+1\choose 4j+1}-\sum_{j=0}^m{4m+1\choose 4j+3}=2^{2m}\cos{m\pi}=(-1)^m2^{2m}.
\end{array}\right.$$ From Lemma \[lem-z-5\], we have $$\sum_{j=0}^m{4m+1\choose 4j}+\sum_{j=0}^m{4m+1\choose 4j+2}=\sum_{j=0}^m{4m+1\choose 4j+1}+\sum_{j=0}^m{4m+1\choose 4j+3}=2^{4m}.$$ Therefore, we have $\sum_{j=0}^m{4m+1\choose 4j}=\sum_{j=0}^m{4m+1\choose 4j}=2^{4m-1}+(-1)^m2^{2m-1}$ and $\sum_{j=0}^m{4m+1\choose 4j+2}=\sum_{j=0}^m{4m+1\choose 4j+3}=2^{4m-1}-(-1)^m2^{2m-1}$. It follows (a) and (b).

Similarly, we get (c) (d) and (e) when we consider $(1+i)^{4m+2}$ and get (f) and (g) when we consider $(1+i)^{4m+3}$.

By applying Lemmas \[lem-z-6\] and \[lem-z-7\], we get the following combination identity.

\[thm-z-2\] For two positive integers $k$ and $m$ such 