Prove that if $Re(z^n)\geq 0$ for every positive $n$ then $z\in \mathbb C$ is nonnegative real number If $Re(z^n)\geq 0$ for every positive $n$ then $z\in \mathbb C$ is positive
This is what I got so far:
Start with 
$$z^n = r^n (\cos\theta+i\sin\theta)^n=r^n\cos(n\theta)+i r^n \sin (n\theta)$$
by the assumption, $Re(z^n)=r^n\cos(n\theta)\geq 0$. We need to assume that $r \not = 0$ and $\theta \not = \frac {1}{n} (\frac{\pi}{2}+2k\pi)$ otherwise this will be vacuously true. 
Note that $r$ is always positive, so in order for $r^n\cos(n\theta)\geq 0$ is true, $\cos(n\theta)\geq 0$, so $\theta \in [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2},2\pi)$.
Now I'm stucked.
 A: As pointed out, you have to replace $\ge$ with $>$ in your inequality. Think in terms of geometry. If $\Re (z^n) > 0$ for all $n$, then $z$ is a real number. Why is this true?
One could try to prove this by contraposition, that means: if $z = r(\cos \phi+i \sin \phi)$ is not a real number, then $\Re (z^n) < 0$ for some natural $n$.
Observe that $\arg (z^n) = n\arg (z)$.
We know that $\phi \neq 0$ (otherwise $z$ would be real) and $\phi \in (-\pi/2, \pi/2)$ (because $\Re(z) > 0$).
Now you just have to choose an appropriate $n$ for which $n \phi$ lies outside this interval: $$n = \min\{n \in \mathbb N : \pi/2 < |n\phi| \le \pi\}$$
Besides that, the notion of being positive isn't defined for complex numbers. Here's short proof: assume that there is an order on $\mathbb C$ compatible with addition and multiplication. No matter if $x < 0$ or $x > 0$, then $x^2 > 0$ (that's true for any ordered field) and we have $0 < i^2 + 1^2 = -1 + 1 = 0$, contradiction.
A: Here's a hint:
We need to show two things:
$$
cos\theta>0, sin\theta=0
$$
The first is easy: $Re(z^n) > 0$ for every positive $n$, therefore for $n=1, Re(z) = rcos\theta > 0 \implies cos\theta>0$
I'm leaving the rest for you. Assume that $sin\theta \ne 0$ and show that there must be $n$ such that $cosn\theta < 0$
(for intuition, keep in mind that $cos\theta > 0$ and think about what happens to $z$ when you multiply it by itself)
A: Here is a proof in view of geometry of complex plane. Remember what is the geometric meaning of multiplication in complex field, rotation and dilation. So a complex number with angle other than 0 must locate in the left half complex plane after some power.
Here comes the analytic proof following the idea stated above.

Without loss of generality, assume $\mbox{Im}(z)\ge0$ or we can apply the same argument to conjugate of $z$.
Thus we can take $\theta\in[0,\frac{\pi}{2}]$ such that $z=re^{i\theta}$. If $\theta>0$, choose $n\in \mathbb{N}$, the smallest number such that $n\theta>\frac{\pi}{2}$,
$$
n\theta=(n-1)\theta+\theta\le\frac{\pi}{2}+\frac{\pi}{2}=\pi
$$
holds from the smallestness of $n$, which then implies,
$$
n\theta\in(\frac{\pi}{2},\pi].
$$
Therefore, $\mbox{cos}(n\theta)<0$, that is, $\mbox{Re}(z^n)<0$, which is a contradiction.
Hence $\theta=0$ as desired.
