Prove that there do not exist natural $n$ such that $(1+i)^n=1$

Question

Prove that there do not exist natural $n$ such that $(1+i)^n=1$.

I try to prove with the binomial and proving by induction but it isn't working

Maybe there is another way to prove and I'm not thinking in the right way.

Answer 1

Note that for any $n\in\mathbb{Z}$, $$|(1+i)^n|=|(1+i)|^n=(\sqrt{2})^{n}.$$ What may we conclude about the equation $(1+i)^n=1$?

Answer 2

$$(1+i)^0=1$$

However, it is quite simple to show that 0 is the only integer exponent that solves the equation, e.g. by considering the polar form; the magnitude of $(1+i)^n$ is $\sqrt2^n$, which is 1 only for $n=0$.

Answer 3

Shortcut:

If indeed $z^n=1$ then also $|z|^n=|z^n|=1$.

Answer 4

Take modulus. $1=|(1+i)^{n}|=(\sqrt 2)^{n}$ which is impossible except when $n=0$.

Answer 5

Let's take $n$ a positive integer.

Compute $(1+i)^1=1+i; (1+i)^2=2i; (1+i)^3=2(i-1); (1+i)^4=-4$

You can then easily express $(1+i)^{4n+r}: 0\le r\le3$ as one of four simple expressions which depend on $r$, and this will extend to negative $n$ too. The only real values occur for $r=0$.