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I am trying to figure out the follwoing

\begin{align*} 2^{\frac{n}{2}}\Big[\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\Big]&=2^{\frac{n}{2}}\Big[\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\Big]^n\\ &=(1+i)^n\\ &=1+\begin{pmatrix}n\\1\end{pmatrix}i-\begin{pmatrix}n\\2\end{pmatrix}-\begin{pmatrix}n\\3\end{pmatrix}i+\begin{pmatrix}n\\4\end{pmatrix}+........ \end{align*}

I am trying to find the real and imaginary part in sigma notation. Any help would be appreciated.

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In $2^{\frac{n}{2}}\Big[\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\Big]$ the real and imaginary part are respectively : $R=2^{\frac{n}{2}}\cos(\frac{n\pi}{4})$ and $I=2^{\frac{n}{2}}\sin(\frac{n\pi}{4})$.

Then you can study the values of $R$ ad $I$ for different values of $n \mod 8$. $$n=0\pmod8 \Rightarrow R=2^{\frac{n}{2}}, I=0 \\ n=1 \pmod 8 \Rightarrow R=2^{\frac{n}{2}}\frac{\sqrt2}{2}, I=2^{\frac{n}{2}}\frac{\sqrt2}{2} \\ n=2 \pmod 8 \Rightarrow R=0, I=2^{\frac{n}{2}} \\ n=3 \pmod 8 \Rightarrow R=-2^{\frac{n}{2}}\frac{\sqrt2}{2}, I=2^{\frac{n}{2}}\frac{\sqrt2}{2} \\ n=4 \pmod 8 \Rightarrow R=-2^{\frac{n}{2}}, I=0 \\ n=5 \pmod 8 \Rightarrow R=-2^{\frac{n}{2}}\frac{\sqrt2}{2}, I=-2^{\frac{n}{2}}\frac{\sqrt2}{2} \\ n=6 \pmod 8 \Rightarrow R=0, I=-2^{\frac{n}{2}}\\ n=7 \pmod 8 \Rightarrow R=2^{\frac{n}{2}}\frac{\sqrt2}{2}, I=-2^{\frac{n}{2}}\frac{\sqrt2}{2} $$


In the sigma notation : \begin{align*} 2^{\frac{n}{2}}\Big[\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\Big]&=2^{\frac{n}{2}}\Big[\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\Big]^n\\ &=(1+i)^n\\ &=1+\begin{pmatrix}n\\1\end{pmatrix}-\begin{pmatrix}n\\2\end{pmatrix}+\begin{pmatrix}n\\3\end{pmatrix}-\begin{pmatrix}n\\4\end{pmatrix}+........ \end{align*} The last line i wrong, it should be : $$1+i\begin{pmatrix}n\\1\end{pmatrix}-\begin{pmatrix}n\\2\end{pmatrix}-i\begin{pmatrix}n\\3\end{pmatrix}+\begin{pmatrix}n\\4\end{pmatrix}+........ \\ =\sum_{j=0}^ni^j\binom{n}{j}$$ So $$R=\sum_{j=0}^{\lfloor{n/2}\rfloor}(-1)^{j}\binom{n}{2j}$$ and $$I=\sum_{j=0}^{\lfloor{(n-1)/2}\rfloor}(-1)^{j}\binom{n}{2j+1}$$

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  • $\begingroup$ Thanks for that, but I was looking for $2^{\frac{n}{2}}\cos(\frac{n\pi}{4})=\sum_{k=0}^b (-1)^k\begin{pmatrix}n\\2k\end{pmatrix}$ where I couldn't figure out b. $\endgroup$ – mint Sep 11 '16 at 7:59
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We know $$e^{ix}=\exp(ix)=\cos(x)+i\sin(x)$$ so \begin{align*} 2^{\frac{n}{2}}\Big[\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\Big] &=2^{\frac{n}{2}}\Big[\exp\Big(i\frac{n\pi}{4}\Big)\Big]\\ &=2^{\frac{n}{2}}\Big[\exp\Big(i\frac{\pi}{4}\Big)\Big]^{n}\\ &=2^{\frac{n}{2}}\Big[\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\Big]^n\\ &=\Big(2^{\frac{1}{2}}\Big)^{n}\Big[\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4})\Big]^n\\ &=\Big(\sqrt2\ \Big)^{n}\Big[\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}\Big]^n\\ &=\Big(\sqrt2\cdot\Big[\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}\Big]\Big)^{n}\\ &=(1+i)^n \end{align*}

Now I use Binomial Theorem $$(a+b)^n=\sum_{k=0}^{n}{\binom{n}{k}a^k b^{n-k}}$$ if $a=i$ and $b=1$ then $$ \begin{align} (a+b)^n&=(i+1)^n\\ &=\sum_{k=0}^{n}{\binom{n}{k}i^k 1^{n-k}}\\ &=\sum_{k=0}^{n}{\binom{n}{k}i^k}\\ &=\binom{n}{0}i^0+\binom{n}{1}i^1+\binom{n}{2}i^2+\binom{n}{3}i^3+ \dots \end{align} $$ $i^k$ has 4 values for k, be aware '$k \equiv 1 \pmod 4$' means remainder of division of $k$ by $4$ is $1$. $$ \begin{align} k \equiv 0 \pmod 4 & \Rightarrow i^k=i^0=i^4=\dots=1 \\ k \equiv 1 \pmod 4 & \Rightarrow i^k=i^1=i^5=\dots=i\\ k \equiv 2 \pmod 4 & \Rightarrow i^k=i^2=i^6=\dots=-1\\ k \equiv 3 \pmod 4 & \Rightarrow i^k=i^3=i^7=\dots=-i \end{align} $$ so $$ \begin{align} \sum_{k=0}^{n}{\binom{n}{k}i^k}&=\binom{n}{0}+\binom{n}{1}i-\binom{n}{2} -\binom{n}{3}i+\binom{n}{4}+\dots \end{align} $$

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