complex number help!! (1 Viewer)

poptarts12345 · Feb 19, 2021

Ahhh I'm so bad at complex numbers!! I feel like no matter how many times I practice, I am not getting better!!

Could someone tell me how to get 11d? I feel like I'm pretty close, but can't quite get it.

quickoats · Feb 19, 2021

Try finding the root $\alpha$ and note that it is also equal to $\rho + \rho^2 + \rho^4$ . Sub some stuff in and do some trig shuffling and de moivres and equate real/imaginary terms and you should be on your way! The clues are in the earlier part of the question.

Paradoxica · Feb 19, 2021

poptarts12345 said:
Ahhh I'm so bad at complex numbers!! I feel like no matter how many times I practice, I am not getting better!!

Could someone tell me how to get 11d? I feel like I'm pretty close, but can't quite get it.
View attachment 30272

Consider the trigonometric form of α-β by rewriting it in terms of powers of ρ

CM_Tutor · Feb 20, 2021

You know that $\rho^7 = 1 \implies \rho^7 - 1 = (\rho - 1)\left(1 + \rho + \rho^2 + \rho^3 + \rho^4 + \rho^5 + \rho^6\right)=0$ , which solves (a) as $\rho \neq 1$

With $\alpha = \rho + \rho^2 + \rho^4$ and wanting a quadratic with real coefficients, our second root must be the conjugate of $\alpha$ , which is $\beta = \rho^3 + \rho^5 + \rho^6$ . We then have

$\alpha + \beta = -1$

and

$\begin{align*} \alpha\beta &= \left(\rho + \rho^2 + \rho^4\right)\left(\rho^3 + \rho^5 + \rho^6\right) \\ &= \rho^4\left(1 + \rho + \rho^3\right)\left(1 + \rho^2 + \rho^3\right) \\ &= \rho^4\left(1 + \rho^2 + \rho^3 + \rho + \rho^3 + \rho^4 + \rho^3 + \rho^5 + \rho^6\right) \\ &= \rho^4\left(1 + \rho + \rho^2 + 3\rho^3 + \rho^4 + \rho^5 + \rho^6\right) \\ &= \rho^4\left(0 + 2\rho^3\right) \\ &= 2\rho^7 \\ &= 2 \end{align*}$

and thus our quadratic is

$x^2 - (\alpha + \beta)x + \alpha\beta = x^2 + x + 2$

which has solutions

$x= \frac{-1\pm\sqrt{1^2-4(1)(2)}}{2(1)}=\frac{-1\pm i\sqrt{7}}{2}$

Looking at the result in the last part, it appears we need the imaginary part of one of the roots.

$\begin{align*} \Im{\left(\alpha\right)}&=\Im{\left(\rho+\rho^2+\rho^4\right)} \\ &=\Im{\left(\rho\right)}+\Im{\left(\rho^2\right)}+\Im{\left(\rho^4\right)} \\ &=\sin{\frac{2\pi}{7}}+\sin{\frac{4\pi}{7}}+\sin{\frac{8\pi}{7}} \\ &=\sin{\frac{2\pi}{7}}+\sin{\left(\pi-\frac{4\pi}{7}\right)}+\sin{\left(\pi-\frac{8\pi}{7}\right)} \qquad \text{as $\sin{\left(\pi-\theta\right)}=\sin\theta$} \\ &=\sin{\frac{2\pi}{7}}+\sin{\frac{3\pi}{7}}+\sin{\frac{-\pi}{7}} \\ &=\sin{\frac{2\pi}{7}}+\sin{\frac{3\pi}{7}}-\sin{\frac{\pi}{7}} \qquad \text{as $\sin\theta$ is an odd function and so $\sin{-\theta}=-\sin\theta$} \\ &= -\sin{\frac{\pi}{7}}+\sin{\frac{2\pi}{7}}+\sin{\frac{3\pi}{7}} \end{align*}$

$\begin{align*} \Im{\left(\beta\right)}&=\Im{\left(\rho^3+\rho^5+\rho^6\right)} \\ &=\Im{\left(\rho^3\right)}+\Im{\left(\rho^5\right)}+\Im{\left(\rho^6\right)} \\ &=\sin{\frac{6\pi}{7}}+\sin{\frac{10\pi}{7}}+\sin{\frac{12\pi}{7}} \\ &=\sin{\left(\pi-\frac{6\pi}{7}\right)}+\sin{\left(\pi-\frac{10\pi}{7}\right)}+\sin{\left(\pi-\frac{12\pi}{7}\right)} \qquad \text{as $\sin{\left(\pi-\theta\right)}=\sin\theta$} \\ &=\sin{\frac{\pi}{7}}+\sin{\frac{-3\pi}{7}}+\sin{\frac{-5\pi}{7}} \\ &=\sin{\frac{\pi}{7}}-\sin{\frac{3\pi}{7}}-\sin{\frac{5\pi}{7}} \qquad \text{as $\sin\theta$ is an odd function and so $\sin{-\theta}=-\sin\theta$} \\ &=\sin{\frac{\pi}{7}}-\sin{\frac{3\pi}{7}}-\sin{\left(\pi-\frac{5\pi}{7}\right)} \\ &=-\left(-\sin{\frac{\pi}{7}}+\sin{\frac{2\pi}{7}}+\sin{\frac{3\pi}{7}}\right) \end{align*}$

Now, we know that, on $x\in[0, \pi]$ , the function $\sin{x}$ increases from 0 to a maximum of 1 (occurring when $x=\frac{\pi}{2}$ ) and then decreases back to 0. We therefore know that:

$0 < \sin{\frac{\pi}{7} < \sin{\frac{2\pi}{7}} < \sin{\frac{3\pi}{7}} < 1$

and thus that

$\Im{\left(\alpha\right)}=-\sin{\frac{\pi}{7}}+\sin{\frac{2\pi}{7}}+\sin{\frac{3\pi}{7}}> 0$

and that

$\Im{\left(\beta\right)}=\sin{\frac{\pi}{7}}-\sin{\frac{2\pi}{7}}-\sin{\frac{3\pi}{7}}< 0$

and so we can conclude that

$\alpha = \frac{-1+i\sqrt{7}}{2}$

and

$\beta = \frac{-1-i\sqrt{7}}{2}$

and by equating the imaginary parts of either $\alpha$ or $\beta$ , we get

$-\sin{\frac{\pi}{7}}+\sin{\frac{2\pi}{7}}+\sin{\frac{3\pi}{7}}=\frac{\sqrt{7}}{2}$

---

Note, we can get other identities by means such as putting $\rho$ into the expression from (a), and taking the real part, to give:

$\begin{align*} 1 + \cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{8\pi}{7}} + \cos{\frac{10\pi}{7}} + \cos{\frac{12\pi}{7}} &= 0 \\ 1 + \cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{6\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{2\pi}{7}} &= 0 \quad \text{since $\cos(2\pi-\theta) = \cos\theta$} \\ \cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} &= -\frac{1}{2} \\ \cos{\frac{2\pi}{7}} - \cos{\frac{3\pi}{7}} - \cos{\frac{\pi}{7}} &= -\frac{1}{2} \quad \text{since $\cos(\pi-\theta) = -\cos\theta$} \\ \cos{\frac{\pi}{7}} - \cos{\frac{2\pi}{7}} + \cos{\frac{3\pi}{7}} &= \frac{1}{2} \end{align*}$

Paradoxica · Feb 20, 2021

Here is a decidedly slicker way to decide which root is α (IMO)

in this image, α = t

complex number help!! (1 Viewer)

poptarts12345

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