The sum $(1+i)^x + (1-i)^x = 0$

Question

I recently saw that $(1+i)^x + (1-i)^x = 0$ for $x = 2\pmod 4$, but can't figure out how its true. Any help/hints would be greatly appreciated!

Answer 1

Notice that

$$1+i =\sqrt{2}e^{i\frac{\pi}{4}}$$ $$1-i =\sqrt{2}e^{-i\frac{\pi}{4}}$$

for $x=4n+2$, $n\in\mathbb{N}$, we have

$$ (1+i)^x+(1-i)^x=2^{2n+1}\Big(e^{in\pi+i\frac{\pi}{2}} + e^{-in\pi-i\frac{\pi}{2}} \Big)=2^{2n+1}\Big((-1)^ni-(-1)^ni\Big)$$

Answer 2

If $x \equiv 2\pmod 4$, then let $x = 4n + 2$ for some integer $n$.

$$\begin{align*} (1+i)^x + (1-i)^x &= (1+i)^{4n+2} + (1-i)^{4n+2}\\ &= \left[(1+i)^2\right]^{2n+1} + \left[(1-i)^2\right]^{2n+1}\\ &= \left(1+2i + i^2\right)^{2n+1} + \left(1-2i + i^2\right)^{2n+1}\\ &= (2i)^{2n+1} + (-2i)^{2n+1}\\ &= (2i)^{2n+1} - (2i)^{2n+1}\\ &= 0 \end{align*}$$

Answer 3

Just to give a variation on Oliver Diaz' answer. Recall that the modulus of a product is the product of the moduli, and argument of a product is the sum of arguments. You can easily see here that $$ \arg(1+i) = \frac\pi4, |i+1| = \sqrt2 $$ and $$ \arg(1-i) = -\frac\pi4, |i+1| = \sqrt2 $$ Therefore $$ \arg(1+i)^x = \frac\pi 4 x, |(i+1)^x| = (\sqrt2)^x $$ and $$ \arg(1-i)^x = -\frac\pi 4 x, |(i-1)^x| = (\sqrt2)^x $$ Now think about when the sum of these will cancel out.

They need to have equal modulus, and there argument must differ by an odd multiple of $\pi$.

Answer 4

If $x\equiv 2\pmod 4$ is an integer, then $i^x=-1$ and hence $$(1-i)^x=-i^x(1-i)^x=-(i(1-i))^x=-(i+1)^x.$$ The conclusion follows.