Prove that $(1+i)^{n} +(1-i)^{n}$ is Real for all $n$.

Question

Show that $$(1+i)^n +(1-i)^n$$ is real for all $n$.

There is a property that says $z$ is real if and only if $\bar{z}=z$. I am not sure how this will help in proving the above statement.

Answer 1

Let $w=(1+i)^n+(1-i)^n$. Then $$\begin{align} \overline{w}&= \overline{(1+i)^n+(1-i)^n} \\ &= \overline{(1+i)^n}+\overline{(1-i)^n} \\ &= (\overline{1+i})^n+(\overline{1-i})^n \\ &= (1-i)^n+(1+i)^n \\ &= w \end{align}$$ Thus $w\in \mathbb R$ for all $n$.

Answer 2

Hint: Simplify $\overline{(1+i)^n+(1-i)^n}$.

Answer 3

Geometrically, note that multiplying a number by $(1+i)$ multiplies the size by $\sqrt{2}$ and rotates it 45 degrees counterclockwise, and multiplying by $(1-i)$ does the same thing to the size but rotates it clockwise by 45 degrees. Then try to justify why the imaginary component must cancel out by using $e^{i \theta}=cos(\theta)+isin(\theta)$.

Answer 4

An idea, using that for any $\;z\in\Bbb C\;$ we have that $\;z+\overline z=2\,\text{Re}\,z\;$:

$$1-i=\overline{1+i}\implies(1-i)^n=\overline{(1+i)^n}\implies(1+i)^n+(1-i)^n=2\,\text{Re}\,((1+i)^n)$$

Answer 5

$$a_n=(1+i)^n+(1-i)^n$$ satisfies the recurrence $$a_{n+2}=2a_{n+1}-2a_n.$$ So if you can show that both $a_0$ and $a_1$ are real, all subsequent $a_n$ will be real too.

Answer 6

This first suggestion is not an elegant method, but is designed to show that this can easily be done by brute force, and that you shouldn't give up - sometimes brute force progress can suggest an easier way through which you may not have noticed.

So it is easy to show by direct calculation that $(1+i)^4=(1-i)^4=-4$.

Then if $n=4m+r$ with $0\le r\le 3$ then you have $(-4)^m\left((1+i)^r+(1-i)^r\right)$ and the four cases for $r$ are easily checked.

I note also that expressing $1+i=\sqrt 2 e^{i\theta}; 1-i=\sqrt 2 e^{-i\theta}$ for a suitable $\theta$ gives you an easy formula for the sum - and this form is suggested by plotting $1+i$ and $1-i$ on the argand diagram (or simply noting that they are complex conjugates). To prove that the expression is real does not require the identification of $\theta$.

Answer 7

If $n=2m$, then: $$(1+i)^n +(1-i)^n=(2i)^m+(-2i)^m=(2^m+(-2)^m)i^m=\begin{cases}0,m-odd\\ real,m-even\end{cases}$$ If $n=2m+1$, then: $$(1+i)^n +(1-i)^n=(1+i)(2i)^m+(1-i)(-2i)^m=\\ ((2^m-(-2)^m)i^{m+1}+((2^m+(-2)^m)i^m=\begin{cases}real+0,m-odd\\ 0+real,m-even\end{cases}$$