Prove that (1+j)^n + (1-j)^n is a real number for every n>0

Question

I personally tried to use DeMoivre's Theorem to solve this. Since (1+i)^n can be written in it's polar form as:

z^n = r^n (cos(nθ)+isin(nθ))

= √(2)^n (cos(nπ/4)+isin(nπ/4))

and (1-i)^n as:

z^n = √(2)^n (cos(n(-π/4))+isin(n(-π/4)))

so (1+i)^n + (1-i)^n can be written as

(1+i)^n + (1-i)^n = (√(2)^n (cos(nπ/4)+isin(nπ/4))) + (√(2)^n (cos(n(-π/4))+isin(n(-π/4))))

              = √(2)^n (cos(nπ/4)+isin(nπ/4) + cos(n(-π/4))+isin(n(-π/4)))

Although once I got here, I was unsure as to where to go. If DeMoivre's Theorem the right approach here?

Answer 1

You seem to have lost track of where you're going.

You've partially carried out a calculation of separating this quantity into real and imaginary parts — you should finish this process.

And if you succeed, this gives an easy way to answer the given question: a complex number is real if and only if its imaginary part is zero.

Answer 2

You just have to chcek that the imaginary part vanishes.

$$i \sin\left( \frac{n\pi}4 \right) + i \sin \left( - \frac{n\pi}4\right)=i \sin\left( \frac{n\pi}4 \right) - i \sin \left( \frac{n\pi}4\right)= 0$$

Answer 3

It suffices to observe that

$$\bar w=\overline {(1+j)^n+(1-j)^n}=(1-j)^n+(1+j)^n=w$$

and

$$\bar w=w\iff \Im(w)=0$$

Indeed recall that for any $z=re^{i\theta}$

• $\overline {(z^n)}=r^ne^{-in\theta}=(\bar z)^n$