I'm currently reading through Ahlfors' "Complex Analysis" to clean up a lot of loose ends in my complex variables knowledge (my intro course was a disaster and I switched to engineering after undergrad :-) ).


An interesting exercise (the last part of his very first exercise, as a matter of fact) is to find the value of $(1+i)^n+(1-i)^n$ for arbitrary $n$. My solution used the Binomial theorem:

$$ (1+i)^n+(1-i)^n = \sum_{k=0}^nC(n,k)[i^k+(-i)^k] = \sum_{k=0}^nC(n,k)[1+(-1)^k]i^k, $$

where $C(n,k)$ is the alternative notation for $n$-choose-$k$.

If $k$ is odd then the term is clearly zero. Thus we need only to sum over even $k$. We can define $m = n/2$ if $n$ is even or $(n-1)/2$ if $n$ is odd and reindex $k$ by $2p = k$. This gives us

$$ 2\sum_{p=0}^mC(2m,2p)i^{2p} = 2\sum_{p=0}^mC(2m,2p)(-1)^{p} $$

if $n$ is even, or

$$ 2\sum_{p=0}^mC(2m+1,2p)i^{2p} = 2\sum_{p=0}^mC(2m+1,2p)(-1)^{p} $$

if $n$ is odd.


After getting my solution (you are free to point out if I am wrong --quite possible!) I became interested in the limit

$$ \lim_{n\rightarrow\infty}[(1+i)^n+(1-i)^n]. $$

From the above considerations, this is formally equivalent to the series whose $n$th partial sum is

$$ 2\{C(n,0)-C(n,2) + C(n,4) - \cdots \pm C(n,[n,n-1])\} $$

where the last term is $C(n,n) = 1$ if $n$ is even and $C(n,n-1)= n$ if $n$ is odd and plus or minus whenever the partial sum has an odd or even total number of terms respectively. You can check that the first few terms in the sqeuence defined by $a_n = (1+i)^n+(1-i)^n$ are $2,2,0,-4,-8,-48,\cdots$

Now, if this sequence converges (an assumption), the completeness of the complex numbers implies the sequence of partial sums is Cauchy, thus all subsequences of partial sums converge to the same limit as the original sequence, and I can take only the evens. This is the series whose $n$th partial sum is

$$ 1 - \frac{n(n-1)}{2} + \frac{n(n-1)(n-2)(n-3)}{4!}-\cdots + \frac{n(n-1)(n-2)\cdots (n-n+1)(n-n)}{n!} $$

which is always less than the partial sum

$$ 1 - \frac{n^2}{2} + \frac{n^4}{4!} -\cdots + \frac{n^{n}}{n!} $$ which approximates $\cos n$ for large $n$. So, if the limit exists, then it is located in $[-1,1]$, but the fact that $\lim_{n\rightarrow\infty}\cos n$ does not exist, paired with the assumption about subsequences, makes the existence of the limit itself far from evident.


The problem I present to the denizens of Math Stack Exchange is finding the status of the existence of $\lim_{n\rightarrow \infty}[(1+i)^n+(1-i)^n]$, and, supposing it does exist, computing its value.

In any case, I think it speaks volumes for the quality of Ahlfors' book that this sort of depth is capable of being extracted from the first problem!

  • $\begingroup$ WA says the result is $-\infty$ $\endgroup$ – Dr. Sonnhard Graubner Aug 26 '17 at 17:19
  • 3
    $\begingroup$ @Dr.SonnhardGraubner With Wolfie A, you get what you pay for! $\endgroup$ – Angina Seng Aug 26 '17 at 17:20
  • $\begingroup$ whelp, looks like I missed the obvious exponential representation. Fun anyway though; good way to spend a saturday morning. $\endgroup$ – JMJ Aug 26 '17 at 17:46
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    $\begingroup$ This line of thought is useful if turned around. Faced with a sum like $\sum_k (-1)^k\binom{n}{2k}$ then it's good to notice it's related to $(1+i)^n$. $\endgroup$ – Angina Seng Aug 26 '17 at 17:49
  • $\begingroup$ @LordSharktheUnknown true. $\endgroup$ – JMJ Aug 26 '17 at 17:50

$\forall n \in \mathbb{N}$,

$$(1+i)^n+(1-i)^n = (\sqrt2 e^{i\frac{\pi}{4}})^n + (\sqrt2e^{-i\frac{\pi}{4}})^n$$

$$ =2^{\frac{n}{2}}(e^{ni\frac{\pi}{4}}+e^{-ni\frac{\pi}{4}})$$

$$ = 2^{\frac{n}{2}}(2 \cos (n\frac{\pi}{4}))$$

As the cosine oscillates between $-1$ and $1$ and $2^{\frac{n}{2}}$ grows, there is no limit.

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If $(1+i)^n=a_n+b_n i$ with real $a_n$, $b_n$, then $(1-i)^n=a_n-b_n i$ (complex conjugation). So you are considering $2a_n$, twice the real part of $(1+i)^n$.

One can get a handle on this by using the polar form of $(1+i)$. This is $re^{it}$ where $r=|1+i|=\sqrt2$ and $t$ solves $\tan t=1/1=1$ so $t=\pi/4$. We get $(1+i)^n=2^{n/2}e^{n\pi i/4}$ and then $$2a_n=2\times2^{n/2}\cos(n\pi/4).$$ The factor $2^{n/2}$ grows exponentially, but the factor $\cos(n\pi/4)$ repeats with period $8$, going like $1/\sqrt2,0,-1/\sqrt2,-1,-1/\sqrt2,0,1/\sqrt2,1,\ldots$. So the sequence $(a_n)$ "oscillates infinitely" with some terms going to $\infty$, some to $-\infty$ and some remaining at zero.

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