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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 $$\sqrt[3]{2}$$

Brad
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Jinx
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3 Answers3

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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 $Z$.

heropup
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  • But than why we don't write $6\varphi + 2k\pi = -2\varphi + 2k\pi$ – Jinx Feb 01 '14 at 20:48
  • Well, if you write that, then the $2\pi k$ terms cancel on both sides, unless what you mean is $$6\varphi + 2\pi m = -2\varphi + 2\pi k,$$ for integers $k, m$. In that case, you will note that this becomes $$8 \varphi = 2\pi(k-m),$$ and since $k, m$ are integers, so is $k-m$. So it suffices to write $$6\varphi = -2\varphi + 2\pi k,$$ because you lose no generality. – heropup Feb 01 '14 at 20:50
  • I think I understand. I could've written $-2\varphi = 6\varphi + 2k\pi$ if I were to choose $arg(-Z^2)$ as a reference? – Jinx Feb 01 '14 at 20:53
  • Yes...it doesn't matter. What matters is that the angles $-2\varphi$ and $6\varphi$ are equivalent up to an integer multiple of $2\pi$. In other words, their difference is an integer multiple of $2\pi$; i.e., $6\varphi - (-2\varphi) = 2\pi k$ for some integer $k$. This is the same idea as integer congruence modulo some number $n$: we say $a \equiv b \pmod n$ if $n$ divides $a-b$; i.e., $a-b$ is a multiple of $n$. – heropup Feb 01 '14 at 20:58
  • Damn, sorry to bother you again. I'm trying to solve this now and I'm simply not getting it. Solutions are supposed to be $\frac{3\pi}{4} and \frac{5\pi}{4}$, but $\varphi = 0$ is also valid, since $\cos(0) = 1$. – Jinx Feb 01 '14 at 21:27
  • Recall that we found $\varphi = \pi k/4$ for integers $k = 0, 1, 2, \ldots, 7$ satisfies the requirement $\arg(Z^6) = \arg(-Z^2)$. Now we must also have $\Re(Z^3) = 2$, which would require $\cos 3\varphi > 0$. Note that $\cos 3(0) = 1 > 0$, $\cos 3(3\pi/4) = \cos 9\pi/4 = 1/\sqrt{2} > 0$, and $\cos 3(5\pi/4) = \cos 15\pi/4 = 1/\sqrt{2} > 0$. All the other choices for $k$ (namely, $k = 1, 2, 4, 6, 7$) do not satisfy $\cos 3\varphi > 0$, so they cannot give you $\Re(Z^3) = 2$. – heropup Feb 01 '14 at 21:32
  • Aaaaa... Yes, sorry I'm a bit tired. I was practicing whole day :) – Jinx Feb 01 '14 at 21:34
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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.

PassingStranger
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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 $\operatorname{Arg} Z$).

vonbrand
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