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?
 A: It is wrong when you write $z=\left(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2}\right)$ (actually this number is $z=i$), indeed $z=\sqrt2\left(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$.
Finally you get :
$$z^{11}=\sqrt2^{11}\left(\cos \frac{11\pi}{4}+i\sin\frac{11\pi}{4}\right)$$
$$=2^5\sqrt2\left(\cos \frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)$$
$$=2^5\sqrt2\left(\frac{-\sqrt2}{2}+i\frac{\sqrt2}{2}\right)$$
$$=2^5(i-1)=-32+32i$$
A: It should be
$$z=\sqrt{2}e^{i\pi/4}$$
So
$$z^{11}=2^{11/2}\left(\cos\frac{11\pi}{4}+i\sin\frac{11\pi}{4}\right)$$
A: We can do that without trigonometry.  Just notice that:
$$(1+i)^{11} = [(1+i)^2]^5 \cdot (1+i)$$
So: 
$$ (i+1) \cdot (i^2 + 2i + 1)^5$$
From that: 
$$(1+i)\cdot (2i)^5$$
But $i^5 = (- 1)\cdot (-1)\cdot i$
So:
$$[(2^5)\cdot i] \cdot (i+1) =  - 32+32i$$
That equals:
$$32(i-1)$$
I find that way easier to solve,  but is up to you. Good luck! 
A: You might note that $(1+i)^2 = 2i$, so that $(1+i)^{10} = 2^{5}i^{5}
=2^{5}i$.  Then $(1+i)^{11} = -2^{5}i(1+i)= 32 - 32i.$
A: 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}
A: it is too simple,
$z=(i+1)^11$
$= [(i+1)^2]^5 (i+1);$
$= [2i]^5 x (i+1)$
$=  32i  (i+1)$
$=   -32+32i$
