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

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.

• Yes, your way is easier. The binomial theorem also allows to see directly which terms cancel out, albeit in a more complicated fashion. – Andreas Sep 17 '18 at 7:28
• Do you mean expand $(1+i)^8$? Struggling to find an equation here... if so, it’s more straightforward to just use the binomial expansion. – PhysicsMathsLove Sep 17 '18 at 7:29
• Technically $\,(1+i)^2=2i\,$ is using the binomial expansion as well ;-) – dxiv Sep 17 '18 at 7:31
• No, it was to Simplify $(1+i)^8$ – Mohammad Zuhair Khan Sep 17 '18 at 7:31
• I can't argue with that logic. Therefore technically I started learning Taylor series in grade $6$ – Mohammad Zuhair Khan Sep 17 '18 at 7:32

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: $$(1+i)^8=[(1+i)^4]^2=\require{cancel}\left[{4\choose 0}+\cancel{{4\choose 1}i}-{4\choose 2}-\cancel{{4\choose 3}i}+{4\choose 4}\right]^2=16.$$
• Just for curiosities sake, why did you use the matrices for? I do not know and would like to learn about that technique. Was that using $^4C_N$ for the Binomial Theorem? – Mohammad Zuhair Khan Oct 12 '18 at 13:54
• Yes, it is a standard notation for combination: $C^n_k=C(n,k)={n\choose k}$. – farruhota Oct 12 '18 at 13:56
• Sorry for the inconvenience, but how is $P^n_k$ represented. Is it $k {n \choose k}$? – Mohammad Zuhair Khan Oct 12 '18 at 13:58
• Unfortunately, there is no shorthand notation for permutation. Yes, it is $P^n_k=P(n,k)=\frac{n!}{(n-k)!}=\frac{n!}{k!(n-k)!}\cdot k!={n\choose k}\cdot k!$. – farruhota Oct 12 '18 at 14:03
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?