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The question I have been given is
Given that $z=2e^{i\frac{\pi}{7}}$, find the smallest positive integer of $k$ such that $z\times{z^2}\times{z^3}\times{...}\times{z^k}$ is real, and state the value of $|z\times{z^2}\times{z^3}\times{...}\times{z^k}|$ in this case.

A previous part of the question had me show that for any complex number $z=re^{i\theta}$, $z\times{z^2}\times{z^3}\times{...}\times{z^k}=(re^{i\theta})^{\frac{k(k+1)}{2}}$, which I achieved using de Moivre's Theorem

Here's what I've tried doing;
Using Euler's Identity, I expanded the product to

$\cos{\left(\frac{k(k+1)\frac{\pi}{7}}{2}\right)}+i\sin{\left(\frac{k(k+1)\frac{\pi}{7}}{2}\right)}$

For a number to be real, I know that it's imaginary component must equal zero, so I figured that I would have to find the smallest positive integer of $k$ that satisfies the equation

$$\sin{\left(\frac{k(k+1)\frac{\pi}{7}}{2}\right)}=0$$

From here, I tried the following; $$\therefore \frac{k(k+1)\frac{\pi}{7}}{2}=\arcsin{0}=0+a\pi\qquad a\in\mathbb{Z} $$ $$\therefore k(k+1)=\frac{2a\pi}{\frac{\pi}{7}}$$ $$\therefore k^2+k=14a$$ $$\therefore k^2+k-14a=0$$ $$\therefore k=\frac{-1\pm\sqrt{1-56a}}{2}$$ I'm not sure where to go from here. If $a$ was a defined constant, I could easily find $k$, however, because it isn't and I'm trying to find the smallest positive integer. According to the textbook, the solution is $k=6$, giving a modulus of $2^{21}$, but I do not understand how they came to this answer. I think I've gone about this the wrong way and there is probably a simpler method.

Edit: Fixed my dumb $\arcsin{0}$ mistake..

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  • $\begingroup$ No the angles with sine $0$ are not $a\frac\pi2$ for $a$ integer. Please revise this step. $\endgroup$ – Did Jun 18 '17 at 11:41
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You have $z=2w$ where $w=\exp(2\pi i/14)$. Then $z^m=2^m w^m$. Then $z^m$ is real iff $w^m$ is real, but that is the case iff $m$ is divisible by $7$. The values of $m$ that arise are $1,3,6,10,\ldots$, the triangular numbers. What is the first triangular number that is divisible by $7$?

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HINT:

The general value of $\arcsin0$ is $n\pi$ where $n$ is any integer

All we need is $$\dfrac{k(k+1)}{2\cdot7}$$ to be integer

Clearly, $2|k(k+1)$ for all integer $k$

So, we need $7|k(k+1)$

As $(k+1,k)=1,$ either $7$ divides $k$ or $k+1$

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You are correct.

We have that $Arg((z)(z^2)(z^3)(z^4)...)=Arg(z)+2Arg(z)+3Arg(z)...\frac{1}{2}n(n+1)Arg(z)$

Triangular numbers are the simplest method.

We note that since $21$ is the first triangular number that is divisible by $7$, that is the value for which $Arg(z) | π$, and is thus real.

Because we have $z^{21}$ and $|z|=2$, it follows that $|z^{21}|=2^{21}$.

Good job!

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Your reasoning is correct.

A simpler method would be to add

$$\dfrac{\pi}{7}(1+2+3+...+k) $$

Until the value in the parentheses first hits a multiple of 7.

That happens at $k =6$ and the total phase angle is

$$\dfrac {\pi}{7}\cdot 21 =3\pi $$

which will yield a negative real number.

The absolute value is

$$2^{21}$$

Since the complex exponential has an absolute value of 1, and $r =2$ for every term.

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