Given that $z=1+i$, find the value of $n\in\mathbb{Z^+}$ such that $z^n$ is real.

Question

Given that $z=1+i$, find the smallest value of $n\in\mathbb{Z^+}$ such that $z^n$ is real.

I'm wondering if there's an algebraic way of solving this question, aside from the obvious trial and error method.

Using the trial and error method: $$(1+i)^2=2i$$ $$(1+i)^3=-2+2i$$ $$(1+i)^4=-4$$

Hence, $n=4$.

Answer 1

Convert to polar coordinate, and note that $~\displaystyle \left[re^{(i\theta)}\right]^n = r^n e^{in\theta}.$

For this particular problem, when converting $(1 + i)$ into polar coordinates, the $r$ factor is irrelevant. Within a modulus of $(2\pi)$, you have that $\theta = \pi/4.$

Further, $re^{(i\alpha)}$ will be an integer, if and only if $\alpha$ has form $k \times \pi ~: k \in \Bbb{Z}.$

Therefore, the first thing to notice is that since $\theta = \pi/4$, the value of $n=4$, will work, since $4 \times (\pi/4) = \pi.$

In fact, the result will be an integer if and only if $n$ is any multiple of $4$. That is, $n$ must have form $4k : ~k \in \Bbb{Z}.$

This follows, because you need $n \times (\pi/4)$ to have form $k\pi ~: k \in \Bbb{Z}.$

Answer 2

Hint:

Use the polar form of $z$ and apply de Moivre's theorem

Answer 3

Alt. hint (without trig): $\,z^n\,$ is real iff $\,z^n=\overline{z^n}$, so $(1+i)^n=(1-i)^n \iff 1=\left(\frac{1+i}{1-i}\right)^n=i^n\,$.