Calculating $B^{10}$ 
Calculate $B^{10}$ when $$B = \begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix}$$


The way I did it was
$$ B = I + A $$
where
$$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$
and $A^2=-I$. Since $A$ and $I$ are commutative,
$$\begin{aligned} B^2 &= (I+A)^2 = 2A \\ B^3 &= (I+A)2A = 2A-2I\\ B^4 &= (I+A)(2A-2I) = -4I\\ \vdots \\ B^{10} &= 32A \end{aligned}$$
Is there a simpler method or a smarter approach if you want to do this for, e.g., $B^{100}$?
 A: I would use the fact that$$B=\sqrt2\begin{bmatrix}\cos\left(\frac\pi4\right)&-\sin\left(\frac\pi4\right)\\\sin\left(\frac\pi4\right)&\cos\left(\frac\pi4\right)\end{bmatrix},$$and that therefore$$(\forall n\in\Bbb N):B^n=2^{n/2}\begin{bmatrix}\cos\left(\frac{n\pi}4\right)&-\sin\left(\frac{n\pi}4\right)\\\sin\left(\frac{n\pi}4\right)&\cos\left(\frac{n\pi}4\right)\end{bmatrix}.$$
A: If you use matrix representation of complex numbers:
$$
a+bi \leftrightarrow\begin{pmatrix}
a&-b\\b&a
\end{pmatrix}\tag{1}
$$
then using also the polar form of complex numbers gives:
$$
w=(1+i)^n=(\sqrt{2}e^{i\pi/4})^n=(\sqrt{2})^n\exp(in\pi/4)=(\sqrt{2})^n(\cos(n\pi/4)+i\sin(n\pi/4))
$$
which by (1) can be translated into a matrix (see also José's answer).

Let $z=\sqrt{2}\omega$ with $\omega=e^{i\pi/4}$. Notice that $\omega^4=-1$. Then for any integer $k$
$$
z^{4k}=2^{2k}(-1)^k
$$
(see also J.W. Tanner's answer).
A: $B^{100}$ $=(B^4)^{25}$ $=(-4I)^{25}=(-4)^{25}I$
A: $$
B^2 = 2A, A^2=-I \implies
B^{100}= (B^2)^{50} = (2A)^{50} = 2^{50} A^{50} = 2^{50} (A^{2})^{25} = 2^{50} (-I)^{25} = -2^{50} I
$$
