combinatorics proof with i - Peter J. Cameron book I'm working through Peter J. Cameron's combinatorics book and I'm having trouble understanding one of his proofs.
In proposition 3.3.3, he states: "If n is a multiple of $8$, then the number of sets of size divisible by $4$ is 
$$
\ 2^{n-2}+2^{(n-2)/2}\
$$
As proof, he starts by saying: "Let $A$ be the required number and $B$ the number of sets whose size is congruent to $2$ (mod $4$). Then $A+B=2^{n-1}$. 
Now substitute $\def\i{\mathrm i}t=\i$ in the Binomial Theorem and note that 
$1+\i=\sqrt2e^{\i\pi/4}$ and so (since $n$ is a multiple of $8$) 
$ (1+\i)^n =2^{n/2}$.
Thus $$ 2^{n/2}=\sum \binom nk\i^k.
$$
Take the real part of the right-hand side, noting that $\i^k=1,\i,-1, -\i $ according as $k=0,1,2$ or $3$ (mod $4$). We obtain  $ A-B=2^{n/2}$. From this and the expression for $A+B$ above, we obtain the value of $A$ (and that of $B$)."
I understand why $$ A+B=2^{n-1} $$ I don't understand where he gets that 
$$
(1+\i)=\sqrt2e^{\i\pi/4}
$$
or why imaginary numbers are even necessary here at all. Any help is greatly appreciated. Thank you.
ETA: Corrected typos and included entire proof. The Cameron book I'm referring to is here: http://books.google.ca/books?id=_aJIKWcifDwC&pg=PR9&lpg=PR9&dq=combinatorics+peter+j+cameron&source=bl&ots=Nr591nVyxM&sig=RO7M38lseBVE1blYHVtuF9lcLoI&hl=en&sa=X&ei=0de3UrbjE-Sm2gWiroHoCA&ved=0CF0Q6AEwBQ#v=onepage&q=combinatorics%20peter%20j%20cameron&f=false and the proof in question is on page 26.
 A: I do not have the book in front of me, so the way I will do it is undoubtedly a little different from the way it is done there. Expanding $(1+1)^n$ we get
$$2^n= \binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\binom{n}{4}+\binom{n}{5}+\cdots +\binom{n}{n}.$$
Similarly, expanding $(1-1)^n$, we get
$$0= \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3}+\binom{n}{4}-\binom{n}{5}+\cdots +(-1)^n\binom{n}{n}.$$
Similarly, expanding $(1+i)^n$, we get
$$(1+i)^n= \binom{n}{0}+i\binom{n}{1}-\binom{n}{2}-i\binom{n}{3}+\binom{n}{4}+i\binom{n}{5}+\cdots +i^n\binom{n}{n}.$$
Similarly, expanding $(1-i)^n$, we get
$$(1-i)^n= \binom{n}{0}-i\binom{n}{1}-\binom{n}{2}+i\binom{n}{3}+\binom{n}{4}-i\binom{n}{5}+\cdots +(-i)^n\binom{n}{n}.$$
Add. Note the very nice cancellations. We get
$$2^n+(1+i)^n+(1-i)^n=4\left(\binom{n}{0}+\binom{n}{4}+\binom{n}{8}+\cdots                                  \right).$$
For an easy simplification of $(1+i)^n$ and $(1-i)^n$, please see the comment by Marc van Leeuwen below. On the right we get $4$ times  the number of subsets of size a multiple of $4$. 
A: One arrives at $(1+i)^n=(\sqrt2e^{i\pi/4})^n$ by converting $1+i$ from cartesian to polar form. The statement as given, or as you transcribed, is not correct unless $n=1$ (which it isn't since it's a multiple of $8$).
I also have no idea why complex numbers were used here; is there some additional text in this proof?
