Determine all complex number for which: $ \arg(Z^6) = \arg(-Z^2),\ \mathrm{Re}(Z^3) = 2 $ While preparing for the next semester, I stumbled upon this complex number problem which kind of confuses me. I know it has something to do with this - but I simply can't think of any proper way to solve it.
Here it is:
Determine all complex number for which following is true:
$$
\arg(Z^6) = \arg(-Z^2),\ \mathrm{Re}(Z^3) = 2,
$$
so, I was thinking
$$
Z^2 = |Z|^2\mathrm{c}i\mathrm{s}(2\varphi)
$$
$$
-Z^2 = -|Z|^2\mathrm{c}i\mathrm{s}(2\varphi)
$$
thus
$$
\arg(Z^6) = \arg(-Z^2) \Rightarrow 6\varphi = 2\varphi \Rightarrow \varphi = 0,
$$
but obviously there's a trick to this $-Z^2$ since correct solutions are
$$
\varphi_1 = \frac{3\pi}{4}
$$
$$
\varphi_2 = \frac{5\pi}{4}
$$
$$
r = |Z| = \sqrt 2
$$
If I go with $\varphi = 0$, and include it in 
$$
\mathrm{Re}\left(Z^3\right) = 2
$$
I get $$\sqrt[3]{2}$$
 A: Suppose the angle/argument of $Z$ is $\varphi$.  Then the angle of $Z^6$ is $\arg(Z^6) = 6\varphi$, the angle of $Z^2$ is $2\varphi$, and the angle of $-Z^2$ is $\arg(-Z^2) = -2\varphi$.  But angles are equivalent only up to an integer multiple of $2\pi$; e.g., the angle $-\pi/2$ is the same as the angle $3\pi/2$, which is the same as $7\pi/2$, etc.  So if $\arg(Z^6) = \arg(-Z^2)$, we must write $6\varphi = -2\varphi + 2\pi k$ for some integer $k$.  Hence $\varphi = \pi k/4$, for which there are $8$ essentially distinct values, corresponding to $k = 0, 1, \ldots, 7$.
Now if $\Re(Z^3) = 2$, this means that $r^3 \cos 3\varphi = 2$, where $r = |Z|$.  But since we must have $r > 0$, it follows that $\cos 3\varphi = 2/r^3 > 0$.  If $\varphi \in \{0, \pi/4, 2\pi/4, 3\pi/4, \ldots, 7\pi/4\}$, then which of these satisfy $\cos 3\varphi > 0$?  These are the possible angles for $Z$.
A: The two errors you make and the one made by the remaining answers are:


*

*The minussign must be included when calculating the argument.

*The argument is cyclic, so you have to include an arbitrary number of periods when you multiply them.


and the one that was wrong in the other answers:


*

*The argument of a negated number is not the negation of the argument.


So on to solving these errors.
To begin with, as $-z = (-1) \cdot z$, $\arg(-z) = \arg(z) + \pi$.
It is not $-{\arg(z)}$ as stated in other answers. That is what you get from $\arg(\bar z)$.
Secondly, because of the argument's cyclic nature, $\arg(z^n)$ is not just $n\cdot\arg(z)$, but $n\cdot\arg(z) + 2\pi\cdot \mathbb{Z}$.
In general you can however just include one arbitrary integer.
You can check the other answers for more about this problem.
A: The rub is that $\arg Z$ is multi-valued. I.e., you get that $\arg Z = \varphi + 2 k \pi$ for arbitrary $k \in \mathbb{Z}$ and $0 \le \varphi < 2\pi$ is the principal value of the argument (sometimes noted $\operatorname{Arg} Z$).
