# Show that $\forall n \in \mathbb{N} \left ( \left [(2+i)^n + (2-i)^n \right ]\in \mathbb{R} \right )$

Show that $\forall n \in \mathbb{N} \left ( \left [(2+i)^n + (2-i)^n \right ]\in \mathbb{R} \right )$

My Trig is really rusty and weak so I don't understand the given answer:

$(2+i)^n + (2-i)^n$

$= \left ( \sqrt{5} \right )^n \left (\cos n\theta + i \sin n\theta \right ) + \left ( \sqrt{5} \right )^n \left (\cos (-n\theta) + i \sin (-n\theta) \right )$

$= \left ( \sqrt{5} \right )^n \left ( \cos n\theta + \cos (-n\theta) + i \sin n\theta + i \sin (-n\theta) \right )$

$= \left ( \sqrt{5} \right )^n 2\cos n\theta$

• You have $z^n=|z|^n\exp(ni\arg\,z)=|z|^n(\cos(n\arg\,z)+i\sin(n\arg\,z))$ for starters... – J. M. is a poor mathematician May 14 '12 at 8:40
• This gives a neat formula. Another way of proving this is to show that (if we call your expression $a_n$) it satisfies the equation $a_n=4a_{n-1}-5a_{n-2}$ and work from there. – Mark Bennet May 14 '12 at 8:45
• Where did Mark get that recursion relation, you ask? Note that $(z-(2+i))(z-(2-i))=z^2-4z+5$... it's the same theory behind Fibonacci sequences. – J. M. is a poor mathematician May 14 '12 at 8:52
• ...and the high-brow route is Newton-Girard: $x^n+y^n$ for integer $n$ is always expressible as a combination of $x+y$ and $xy$; for your particular case, $x+y=4$ and $xy=5$ (notice a pattern?) – J. M. is a poor mathematician May 14 '12 at 8:55
• The binomial expansion works, because the odd powers of $i$ are attached to odd powers of $b$ and $-b$ respectively, so will cancel. – Mark Bennet May 14 '12 at 9:15

There are two ways to write a complex number: rectangular form, e.g., $x+iy$, and polar form, e.g., $re^{i\theta}$. The conversion between them uses trig functions: $$re^{i\theta}=r\cos\theta+ir\sin\theta\;.\tag{1}$$ Going in the other direction, $$x+iy=\sqrt{x^2+y^2}\,e^{i\theta}\;,$$ where $\theta$ is any angle such that $$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}\;\text{ and }\sin\theta=\frac{y}{\sqrt{x^2+y^2}}\;.$$ The important thing for your argument is that $r=\sqrt{x^2+y^2}$.

The $r$ corresponding to $2+i$ is therefore $\sqrt{2^2+1^2}=\sqrt5$, and that corresponding to $2-i$ is $\sqrt{2^2+(-1)^2}=\sqrt5$ as well. The angles for $2+i$ is an angle $\theta$ whose cosine is $\frac2{\sqrt5}$ and whose sine is $\frac1{\sqrt5}$, while the angle for $2-i$ is an angle whose cosine is $\frac2{\sqrt5}$ and whose sine is $-\frac1{\sqrt5}$. It doesn’t matter exactly what they are; the important thing is that if we let the first be $\theta$, the second is $-\theta$, since $$\cos(-\theta)=\cos\theta\;\text{ and }\sin(-\theta)=-\sin\theta\;.$$

Substituting into $(1)$ gives you $$2+i=\sqrt5\cos\theta+i\sqrt5\sin\theta=\sqrt5(\cos\theta+i\sin\theta)=\sqrt5 e^{i\theta}$$ and $$2-i=\sqrt5\cos(-\theta)+i\sqrt5\sin(-\theta)=\sqrt5(\cos\theta-i\sin\theta)=\sqrt5 e^{-i\theta}\;.$$

Now use the fact that it’s easy to raise an exponential to a power:

\begin{align*} (2+i)^n+(2-i)^n&=(\sqrt5)^n\left(e^{i\theta}\right)^n+(\sqrt5)^n\left(e^{-i\theta}\right)^n\\ &=(\sqrt5)^n\left(e^{in\theta}+e^{-in\theta}\right)\\ &=(\sqrt5)^n\Big(\big(\cos n\theta+i\sin n\theta\big)+\big(\cos(-n\theta)+i\sin(-n\theta)\big)\Big)\\ &=(\sqrt5)^n\Big(\cos n\theta+i\sin n\theta+\cos n\theta-i\sin n\theta\Big)\\ &=(\sqrt5)^n 2\cos n\theta\;. \end{align*}

• Thanks, it's the fact that $\cos \theta = \cos (- \theta)$ and $\sin - \theta = - \sin \theta$ that threw me. – Robert S. Barnes May 14 '12 at 9:21
• @Robert: I wasn’t sure just where the problem was, so I took you at your word and went back to basics; I’m glad that it helped. – Brian M. Scott May 14 '12 at 9:23

If you believe that complex conjugation respects products (hence also powers), then the simple way is: $$\overline{x}=\overline{(2+i)^n+(2-i)^n}=(\overline{2+i})^n+(\overline{2-i})^n=(2-i)^n+(2+i)^n=x.$$ So $\overline{x}=x$, and hence $x$ is real.

The binomial formula gives an alternative route: $$x=(2+i)^n+(2-i)^n=\sum_{k=0}^n{n\choose k}2^ki^{n-k}+\sum_{k=0}^n{n\choose k}2^ki^{n-k}(-1)^{n-k}.$$ Here the terms where $n-k$ is odd cancel each other, so we get $$x=2\sum_{k=0,\ k\equiv n\pmod2}^n{n\choose k}2^ki^{n-k}.$$ Here everywhere $i^{n-k}$ is real, because $(n-k)$ is even in all the terms remaining in the sum.

• +1 Even though it doesn't explain the given answer, that's really nice. – Robert S. Barnes May 14 '12 at 9:16
• @Robert, sorry about that. I simply looked at the title, and didn't read your question to the end. I did see your own suggestion of using the binomial formula, so I added that. – Jyrki Lahtonen May 14 '12 at 9:21
• Still a very nice answer. If I could mark it up more than once I would. :-) Glad to see that intuition on the binomial formula was right. – Robert S. Barnes May 14 '12 at 9:31
• So you're notation there $2|n-k$ means that the summation is only over even values of k? Is that a common notation? – Robert S. Barnes May 14 '12 at 9:35
• @Robert, I mean that the summation is over only such values of $k$ that $n-k$ is even. A better way of expressing that would be $k\equiv n\pmod2$. Will edit. – Jyrki Lahtonen May 14 '12 at 9:47

Hint $\$ Scaling the equation by $\sqrt{5}^{\:-n}$ and using Euler's $\: e^{{\it i}\:\!x} = \cos(x) + {\it i}\: \sin(x),\$ it becomes

$$\smash[b]{\left(\frac{2+i}{\sqrt{5}}\right)^n + \left(\frac{2-i}{\sqrt{5}}\right)^n} =\: (e^{{\it i}\:\!\theta})^n + (e^{- {\it i}\:\!\theta})^n$$ But $$\smash[t]{ \left|\frac{2+i}{\sqrt{5}}\right| = 1\ \Rightarrow\ \exists\:\theta\!:\ e^{{\it i}\:\!\theta} = \frac{2+i}{\sqrt{5}} \ \Rightarrow\ e^{-{\it i}\:\!\theta} = \frac{1}{e^{i\:\!\theta}} = \frac{\sqrt{5}}{2+i} = \frac{2-i}{\sqrt 5}}$$

Remark $\$ This is an example of the method that I describe here, of transforming the equation into a simpler form that makes obvious the laws or identities needed to prove it. Indeed, in this form, the only nontrivial step in the proof becomes obvious, viz. for complex numbers on the unit circle, the inverse equals the conjugate: $\: \alpha \alpha' = 1\:\Rightarrow\: \alpha' = 1/\alpha.$