2

The question:

$$\forall n \in \Bbb N$$

is the number

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

in the real numbers ($\Bbb R$)?

My try for solution using Newton binom:

$$(2+i)^n = \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k}$$

$$(2-i)^n = \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k}$$

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

$$\Downarrow$$

$$\forall k - odd: \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k} + \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k} = 0$$

$$\forall k - even: \sum_{k=0}^{n}\binom{n}{k}2^{n-k}i^{k} + \sum_{k=0}^{n}\binom{n}{k}2^{n-k}(-i)^{k} = Real \ number$$

So the answer is YES, the sum $(2+i)^n + (2-i)^n \in \Bbb R.$

Correct? any better/smarter/more efficient way?

*I tried to solve it trigonometricaly, didnt realy work, stuck...: $$(2+i) \Rightarrow r = \sqrt{2^2+1^2} =\sqrt{5}, \tan \theta = \frac{1}{2} \Rightarrow \theta = 26.565$$

$$(2+i) = \sqrt{5}(\cos 26.565 + i \sin 26.565)$$ $$(2+i)^n = \sqrt{5}^n(\cos 26.565n + i \sin 26.565n)$$ $$(2-i)^n = \sqrt{5}^n(\cos 26.565n - i \sin 26.565n)$$

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

J. W. Tanner
  • 60,406
Alon
  • 1,647

3 Answers3

5

More efficient way:

A complex number plus its complex conjugate is real, so

$(2+i)^n+(2-i)^n=(2+i)^n+(\overline{2+i})^n=(2+i)^n+\overline{(2+i)^n}\in\mathbb R.$

J. W. Tanner
  • 60,406
4

$2+i$ and $2-i$ are conjugates. Then $(2+i)^n$ and $(2-i)^n$ are conjugates for all natural numbers $n$. Hence their sum is real.

2

A good method to try to prove that a complex number is real is to use the complex conjugate, which takes a complex number $z=a+bi$ to $\bar z = a-bi$. A complex number is real if and only if it equals its conjugate; similar tricks prove helpful in some more general contexts (e.g. to show that Binet's formula for Fibonnaci numbers yields rational numbers or to show all sorts of things in Galois theory).

If you prove two relationships about conjugation, you're pretty much good to solve your problem: $$\overline{z_1 \cdot z_2} = \overline{z_1}\cdot \overline{z_2}$$ $$\overline{z_1 + z_2} = \overline{z_1}+ \overline{z_2}$$ which say that complex conjugation is a symmetry of arithmetic over the complex numbers; there are also rules such as $\overline{e^z} = e^{\overline{z}}$, though those aren't necessary here. If you repeatedly apply the first rule, you get $\overline{z^n}=\overline{z}^n$ for $n\in \mathbb N$, and then you can see $$\overline{(2+i)^n+(2-i)^n}=\overline{(2+i)}^n+\overline{(2-i)}^n=(2-i)^n+(2+i)^n$$ which shows that your expression is equal to its conjugate, hence real.

Somewhat unrelatedly, your trigonometric approach can work; you can observe that $2+i=\sqrt{5}\cdot e^{i\tan^{-1}(1/2)}$ by working in polar coordinates (note $e^{ix}=\cos(x)+i\sin(x)$) and $2-i=\sqrt{5}\cdot e^{-i\tan^{-1}(1/2)}$. Then, $$(2+i)^n + (2-i)^n = \sqrt{5}^n \cdot \left(e^{ni\tan^{-1}(1/2)} + e^{-ni\tan^{-1}(1/2)}\right).$$ Observing that $e^{ix}+e^{-ix}=2\cos(x)$ we get $$(2+i)^n + (2-i)^n = 2\sqrt{5}^n \cos(n\tan^{-1}(1/2)).$$

Milo Brandt
  • 60,888
  • Very nice solution and thanks for the explanation. – Alon Dec 28 '19 at 23:54
  • Alittle bit unrealated, but could you explain what is: $re^{2\pi i\theta}$? i see that alot and dont understand. – Alon Dec 28 '19 at 23:56
  • 1
    @Alon Intuitively, the notation $re^{i\theta}$ refers to a point which is a distance of $r$ from the origin and an angle of $\theta$ (in radians) counterclockwise from the positive real axis - it's like polar coordinates in the plane. It generalizes how multiplying by $2$ doubles the size of something and multiplying by $i$ (you can check) takes a point in the complex plane and rotates it by a quarter-turn around the origin... – Milo Brandt Dec 29 '19 at 00:02
  • ...basically, complex numbers are some amount of scaling ($r$) and some amount of rotation ($\theta$) - where the scalings multiply and rotations add when we multiply two complex numbers together. The particular notation $e^{i\theta}$ to refer to a point a distance of $1$ from the origin at an angle of $\theta$ is not so important - it just turns out that this has the same properties as exponentiation, so is notated that way. – Milo Brandt Dec 29 '19 at 00:02
  • Thank you, thats nice, i start to understand, does it have a name (so i could read about)? – Alon Dec 29 '19 at 00:04
  • 1
    @Alon It has various names - but the most common, to my knowledge, is that this is called the polar form of a complex number, representing the modulus/absolute value ($r$) and argument ($\theta$). – Milo Brandt Dec 29 '19 at 00:10
  • Ok, thank you again for the help – Alon Dec 29 '19 at 00:13