Finding smallest n satisfying $z^n$ is real and positive.

Question

Let $z = 1 + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$. Find smallest positive integer $n$ such that $z^n$ is real and positive.

Answer 1

Hint:

Write $z$ in complex exponential form $\,r\,\mathrm e^{i\theta}$.

Another hint:

You'll need the linearisation formulæ: $$\sin^2\theta=\tfrac12(1-\cos 2\theta),\qquad \cos^2\theta=\tfrac12(1+\cos 2\theta).$$

Answer 2

Hint:

By observation

If $2y=45^\circ,$

$$z=1+\cos2y+i\sin2y=2\cos^2y+2i\sin y\cos y=2\cos y(e^{iy})$$ using Intuition behind euler's formula

Answer 3

For $\theta=\arg(z)$, we have $$ \begin{align} \tan(\theta) &=\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\\[3pt] &=\sqrt2-1 \end{align} $$ Furthermore, $\theta$ is in the first quadrant and $$ \begin{align} \tan(2\theta) &=\frac{2\left(\sqrt2-1\right)}{1-\left(\sqrt2-1\right)^2}\\ &=1\\[3pt] &=\tan\left(\frac\pi4\right) \end{align} $$