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:


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

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

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

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    $\begingroup$ Yes, your way is easier. The binomial theorem also allows to see directly which terms cancel out, albeit in a more complicated fashion. $\endgroup$ – Andreas Sep 17 '18 at 7:28
  • $\begingroup$ 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. $\endgroup$ – PhysicsMathsLove Sep 17 '18 at 7:29
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    $\begingroup$ Technically $\,(1+i)^2=2i\,$ is using the binomial expansion as well ;-) $\endgroup$ – dxiv Sep 17 '18 at 7:31
  • $\begingroup$ No, it was to Simplify $(1+i)^8$ $\endgroup$ – Mohammad Zuhair Khan Sep 17 '18 at 7:31
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    $\begingroup$ I can't argue with that logic. Therefore technically I started learning Taylor series in grade $6$ $\endgroup$ – 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: $$(1+i)^8=[(1+i)^4]^2=\require{cancel}\left[{4\choose 0}+\cancel{{4\choose 1}i}-{4\choose 2}-\cancel{{4\choose 3}i}+{4\choose 4}\right]^2=16.$$

  • $\begingroup$ 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? $\endgroup$ – Mohammad Zuhair Khan Oct 12 '18 at 13:54
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    $\begingroup$ Yes, it is a standard notation for combination: $C^n_k=C(n,k)={n\choose k}$. $\endgroup$ – farruhota Oct 12 '18 at 13:56
  • $\begingroup$ Sorry for the inconvenience, but how is $P^n_k$ represented. Is it $k {n \choose k}$? $\endgroup$ – Mohammad Zuhair Khan Oct 12 '18 at 13:58
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    $\begingroup$ Unfortunately, there is no shorthand notation for permutation. Yes, it is $P^n_k=P(n,k)=\frac{n!}{(n-k)!}=\frac{n!}{k!(n-k)!}\cdot k!={n\choose k}\cdot k!$. $\endgroup$ – 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?


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