<s>[INST] Calculate $(1+i)^{11}$ I'm studying complex numbers and I wanted to know if this solution is correct.
The problem is to calculate $(1+i)^{11}$, here's my attempt:
I can express $(1+i)^{11}$ using the argument notation: $z=\rho(\cos \phi +i\sin \phi)$
$$z=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$$
This is very helpful because I can use De Moivre's formula to calculate $z^{11}$
$$z^{11}=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{11}=\left(\cos \frac{11\pi}{2}+i\sin\frac{11\pi}{2}\right)=0-i=-i$$
Is this correct? Is there a better way to solve it?
 [/INST] There's another approach:
\begin{align}
(1+i)^{11}&=\sum_{k=0}^{11}\binom{11}{k}i^k
\\\\&=\sum_{\substack{0\le k\le 11 \\k\equiv 0\!\pmod 4}}\binom{11}{k} + \sum_{\substack{0\le k\le 11 \\k\equiv 1\!\pmod 4}}\binom{11}{k}\cdot i + \sum_{\substack{0\le k\le 11 \\k\equiv 2\!\pmod 4}}\binom{11}{k} \cdot (-1)+ \sum_{\substack{0\le k\le 11 \\k\equiv 3\!\pmod 4}}\binom{11}{k}\cdot (-i)&\scriptsize\text{by the binomial theorem, and computing each }i^k
\\\\&=\sum_{k=0,4,8}\binom{11}{k}+i\sum_{k=1,5,9}\binom{11}{k}-\sum_{k=2,6,10}\binom{11}{k}-i\sum_{k=3,7,11}\binom{11}{k}
\\\\&=\sum_{k=0,4,3}\binom{11}{k}+i\sum_{k=1,5,2}\binom{11}{k}-\sum_{k=2,5,1}\binom{11}{k}-i\sum_{k=3,4,0}\binom{11}{k}&\scriptsize\text{since }\binom{11}{k}=\binom{11}{11-k}
\\\\&=\left(\sum_{k=0,3,4}\binom{11}{k}-\sum_{k=1,2,5}\binom{11}{k}\right)(1-i)
\\\\&=\left(1 + \frac{11\cdot 10 \cdot 9}{3 \cdot 2 \cdot 1}+ \frac{11\cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} -11 -\frac{11\cdot 10}{2 \cdot 1}-\frac{11\cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\right)(1-i)
\\\\&=\left((1 + 165+ 330)-(11+55+462) \right)(1-i)
\\\\&=-32(1-i)
\\\\&=-32+32i
\end{align}
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