Complex Numbers Questions (2 Viewers)

Jonothancroller · Nov 5, 2015

Hey, I need help with two questions:

http://imgur.com/YmMw5kd
http://imgur.com/hVWn0OS

I only need help with a) ii) and c) iii) B), but the whole question is needed to solve it so I posted the whole thing. Thanks.

Jonothancroller · Nov 9, 2015

InteGrand · Nov 9, 2015

Jonothancroller said:
Hey, I need help with two questions:

http://imgur.com/YmMw5kd
http://imgur.com/hVWn0OS

I only need help with a) ii) and c) iii) B), but the whole question is needed to solve it so I posted the whole thing. Thanks.

$Here's how to do (c) (iii) $(\beta )$.$

$Note that the roots to the equation$

$$z^3 + z^2 + z + 1 + z^{-1} + z^{-2} + z^{-3} = 0\ \ (*)$$

$are precisely the six non-real seventh roots of unity, namely $\mathrm{cis}\left(\pm\frac{2\pi}{7} \right),\mathrm{cis}\left(\pm\frac{4\pi}{7} \right),\mathrm{cis}\left(\pm\frac{6\pi}{7} \right)$.$

$Now observe the that $z^3 + z^2 + z + 1 + z^{-1} + z^{-2} + z^{-3} = \left(z+\frac{1}{z} \right)^3 + \left(z+\frac{1}{z} \right)^2 -2\left(z+\frac{1}{z} \right) -1$ \Big{(}to derive this, consider the expansions of $\left(z+\frac{1}{z} \right)^3$ and $\left(z+\frac{1}{z} \right)^2$ and play around with them; basically our aim is to get Equation (*) into a polynomial equation in $z+\frac{1}{z}$\Big{)}.$

$So we can rewrite Equation (*) as a third degree polynomial in $\zeta$, where $\zeta = z+\frac{1}{z}$:$

$$\zeta ^3 + \zeta ^2 -2\zeta - 1 = 0\ \ (**)$.$

$Since $\zeta = z+\frac{1}{z} = 2\cos \theta$ (where $\theta = \arg (z)$), and there are exactly three different values for $\cos \theta$, namely $\cos \left(\frac{2\pi}{7} \right),\cos \left(\frac{4\pi}{7} \right),\cos \left(\frac{6\pi}{7} \right)$ (which is seen by seeing what the roots of (*) are and recalling that $\cos (-\theta) = \cos \theta$), we have that the roots of (**) are: $2\cos \left(\frac{2\pi}{7} \right),2\cos \left(\frac{4\pi}{7} \right),2\cos \left(\frac{6\pi}{7} \right)$ (notice that there are three roots, as expected of a third degree polynomial equation).$

$Using the identity $\cos \left( \pi - \theta\right) = -\cos \theta$, we can write the three roots of (**) as: $2\cos \left(\frac{2\pi}{7} \right),-2\cos \left(\frac{3\pi}{7} \right),-2\cos \left(\frac{\pi}{7} \right)$.$

$Now via product of roots of (**), we have$

$\begin{align*} 2\cos \left(\frac{2\pi}{7} \right)\times \left(-2\cos \left(\frac{3\pi}{7} \right)\right)\times \left(-2\cos \left(\frac{\pi}{7} \right)\right) &= - (-1) \\ \Rightarrow 8\cos \left(\frac{\pi}{7} \right) \cos \left(\frac{2\pi}{7} \right) \cos \left(\frac{3\pi}{7} \right)&= 1 \\ \Rightarrow \cos \left(\frac{\pi}{7} \right) \cos \left(\frac{2\pi}{7} \right) \cos \left(\frac{3\pi}{7} \right)&= \frac{1}{8}. \quad \blacksquare\end{align*}$

kawaiipotato · Nov 9, 2015

$arg(z^2 - a)$

$= arg((z-a)(z+a))$

$= arg(z-a) + arg(z+a)$

$= arg(z-a) - arg(z+a) + 2arg(z+a) ($I thought of doing this because it's easier to work with vectors if there's a negative sign$)$

$Note that $ arg(z) = 2arg(z+a) $ by constructing vectors z-a and z and using the circle geometry theorem: Angle at the centre is twice the angle at the circumference, subtended by the same arc.$

$= arg(z-a) - arg(z+a) + \theta$

$Draw up the vectors z-a and z+a and use the circle geometry theorem: angle at circumference subtended by diameter is $ \frac{\pi}{2} $radians$

$= \frac{\pi}{2} + \theta$
(only holds true when z is above the x-axis or else the direction of the angle will be the opposite)

InteGrand · Nov 9, 2015

kawaiipotato said:
$arg(z^2 - a)$

$= arg((z-a)(z+a))$

$= arg(z-a) + arg(z+a)$

$= arg(z-a) - arg(z+a) + 2arg(z+a) ($I thought of doing this because it's easier to work with vectors if there's a negative sign$)$

$Note that $ arg(z) = 2arg(z+a) $ by constructing vectors z-a and z and using the circle geometry theorem: Angle at the centre is twice the angle at the circumference, subtended by the same arc.$

$= arg(z-a) - arg(z+a) + \theta$

$Draw up the vectors z-a and z+a and use the circle geometry theorem: angle at circumference subtended by diameter is $ \frac{\pi}{2} $radians$

$= \frac{\pi}{2} + \theta$
(only holds true when z is above the x-axis or else the direction of the angle will be the opposite)

$I haven't looked at this proof closely so I don't know whether this affects the proof, but at the start, $z^2 - a$ is not $(z+a)(z-a)$, it is $\left( z + \sqrt{a}\right)\left( z-\sqrt{a}\right)$.$

kawaiipotato · Nov 9, 2015

InteGrand said:
$I haven't looked at this proof closely so I don't know whether this affects the proof, but at the start, $z^2 - a$ is not $(z+a)(z-a)$, it is $\left( z + \sqrt{a}\right)\left( z-\sqrt{a}\right)$.$

Oops. You're right and it does affect the proof. Mine only works when it's actually "z^2 - a^2" but having "a^2" proved the result so I think it might be a typo.

Carrotsticks · Nov 9, 2015

Sent from my SM-G928I using Tapatalk

Complex Numbers Questions (2 Viewers)

Jonothancroller

Member

Jonothancroller

Member

InteGrand

Well-Known Member

kawaiipotato

Well-Known Member

InteGrand

Well-Known Member

kawaiipotato

Well-Known Member

Carrotsticks

Retired

Users Who Are Viewing This Thread (Users: 0, Guests: 2)