Why should I use the binomial theorem to solve $(1+i)^8$?

Question

I have recently started on Edexcel AS and A Level Modular Mathematics FP$1$.

We are tasked to solve $(1+i)^8$ and beside the question they have given the hint to use the binomial theorem.

However I solved it in a much easier way:

$(1+i)^2=1+2i-1=2i$

$(1+i)^4= (2i)^2=-4$

$(1+i)^8= (-4)^2=16$

So I was wondering why exactly was the hint giving a relatively lengthy technique? Thank you in advance.

Answer 1

You did use the Binomial theorem, not for $8$-th degree, but for $2$nd. The hint does not insist on using it directly to the $8$-th degree, does it?

As an alternative: $$(1+i)^8=[(1+i)^4]^2=\require{cancel}\left[{4\choose 0}+\cancel{{4\choose 1}i}-{4\choose 2}-\cancel{{4\choose 3}i}+{4\choose 4}\right]^2=16.$$

Answer 2

Because this is essentially a special case of de Moivre's formula: $$(\cos(\theta)+i\sin(\theta))^n = \cos(n\theta)+i\sin(n\theta)$$ with $\theta = 0$ (and also a $2^n$ inserted). If you expand the LHS with the binomial formula and you consider real/imaginary parts, then you can find formulas such as $\cos(3\theta) = 4 \cos^3(\theta) - 3 \cos(\theta)$ etc.

It was (probably) meant as an easy example of application. If you expand the LHS with the binomial formula and you identify real and imaginary parts, you find $\sum_{k=0}^{4/2} \binom{4}{2k} (-1)^k = 2^4$ and $\sum_{k=0}^{4/2-1} \binom{4}{2k+1} (-1)^k = 0$. Spoiler alert, it works for other numbers than $4$ (for odd numbers you have to think on how to adapt the formula). Nifty uh?