Hard complex number question (1 Viewer)

cutemouse · Jul 22, 2009

Hi,

Could someone please help me out here?

Thanks

kaz1 · Jul 22, 2009

For the first one try taking out 2ⁿsecⁿ@

Trebla · Jul 22, 2009

For (i), write tan θ = sin θ / cos θ and factor out everything until you end up with cis θ so that De Moivre's theorem can be applied.

For (ii), from (i) let θ = π/10 and n = 5, which gives:
(2 + 2i tan π/10)⁵ + (2 - 2i tan π/10)⁵ = (2⁶cos π/2) / (cos π/10)⁵
The RHS = 0, hence
(2 + 2i tan π/10)⁵ + (2 - 2i tan π/10)⁵ = 0
=> z = 2i tan π/10 satisfies the equation:
(2 + z)⁵ + (2 - z)⁵ = 0

For (iii), the roots take the form 2i tan θ, so we solve cos 5θ = 0 from the RHS of (i), choose the solutions of θ that satisfy this trigonometric equation and since your roots are of the form, 2i tan θ you can then list all 5 roots explicitly given the values of θ found (NB: that you cannot choose 5π/10 since this would make 2i tan θ undefined).

For (iv), expand (ii) using binomial theorem and equate the real parts and let z = 2i tan π/10. This should give a quadratic equation which can be solved for (tan π/10)²

jet · Jul 22, 2009

$LHS = \left (2 + 2itan\, \theta \right )^n + \left (2 - 2itan\, \theta \right )^n \\ = 2^n\left (\frac{cos\theta + isin\theta}{cos\theta} \right )^n +2^n\left (\frac{cos\theta - isin\theta}{cos\theta} \right )^n \\ = \frac{2^n}{cos^n\theta} \left [\left (cos\theta + isin\theta \right )^n + \left (cos\theta - isin\theta \right )^n \right ] \\ = \frac{2^n}{cos^n\theta} \left [\left (cos \, n\theta + isin \, n\theta \right ) + \left (cos \, n\theta - isin \, n\theta \right ) \right ] \text{ noting that } cos \, n\theta - isin \, n\theta = cos \left (-\theta \right ) + isin \left (-\theta \right ) \\ = \frac{2^n}{cos^n\theta} \left (2cos \, n\theta \right ) \\ = \frac{2^{n+1}cos \, n\theta}{cos^n \, \theta} \\ = RHS$

$\text{Letting } z = 2i tan \, \theta, \\ \left (2 + z \right )^n + \left (2 - z \right )^n = \frac{2^{n+1}cos \, n\theta}{cos^n \, \theta} \\ \text{Thus, to solve } \left (2 + z \right )^5 + \left (2 - z \right )^5 = 0, \text{ we can solve:} \\ \frac{2^{6}cos \, 5\theta}{cos^5 \, \theta} = 0 \\ cos \, 5\theta = 0 \\ 5\theta = \frac{\pi}{2}, \frac{-\pi}{2}, \frac{3\pi}{2}, \frac{-3\pi}{2}, \text{ etc.} \\ \theta = \frac{\pi}{10}, \text{ etc.} \\ \text{Thus, one solution is } \theta = \frac{\pi}{10}, \text{ which corresponds to the root } z = 2itan \, \frac{\pi}{10} \\ \text{Hence, } 2itan \, \frac{\pi}{10} \text{ is a root of the equation } \left (2 + z \right )^5 + \left (2 - z \right )^5 = 0 \\$

$\text{Taking } 0 \leq \theta \leq \pi \text{ since the period of tan is } \pi, \text{the solutions of the equation are } \theta = \frac{\pi}{10}, \frac{3\pi}{10},\frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}\\ \text{Hence the other roots are } z = 2itan \, \frac{m\pi}{10}, \text{where} m = 1,3,5,7,9$

Will do last part after dinner.

jet · Jul 22, 2009

cutemouse · Jul 22, 2009

Thanks jetblack, it makes alot of sense to me now.

But 1 question though... Where did you get the tan(pi/10) < tan(pi/4) ?

Thanks

untouchablecuz · Jul 22, 2009

jetblack2007 said:
$\left (2 + z \right )^5 + \left (2 - z \right )^5 = 0 \\ 32 + 80z + 80z^2 + 40z^3 + 10z^4 + z^5 + 32 - 80z + 80z^2 - 40z^3 + 10z^4 - z^5 = 0 \\ 64 + 160z^2 + 20z^4 = 0 \\ \text{Since } z = 2itan \, \frac{\pi}{10} \text{ is a root} \\ 320tan^4 \frac{\pi}{10} - 640tan^2 \frac{\pi}{10} + 64 = 0 \\ 5tan^4 \frac{\pi}{10} - 10tan^2 \frac{\pi}{10} + 1 = 0 \\ \therefore tan^2 \frac{\pi}{10} = \frac{10 \pm \sqrt{100 - 4(1)(5)}}{10} \\ = \frac{10 \pm \sqrt{80}}{10} \\ = 1 + \frac{2\sqrt{5}}{5} \text{ or } 1 - \frac{2\sqrt{5}}{5} \\ \text{But, since } tan\frac{\pi}{10} < tan \frac{\pi}{4} = 1, \\ tan^2 \frac{\pi}{10} = 1 - \frac{2\sqrt{5}}{5}$

just one Q, with this question, did you know to expand the binomials (evetnually leading to a quadratic in x^2) or just tried it in an attempt to answer the Q?

untouchablecuz · Jul 22, 2009

jm01 said:
Thanks jetblack, it makes alot of sense to me now.

But 1 question though... Where did you get the tan(pi/10) < tan(pi/4) ?

Thanks

think of the graph of y = tan x

Aerath · Jul 22, 2009

jm01 said:
But 1 question though... Where did you get the tan(pi/10) < tan(pi/4) ?

I believe you can just say: "by inspection". Or just say: "using calculator".

gurmies · Jul 22, 2009

1/10 < 1/4

For both sinx and tanx as x ===> pi/2, sinx and tanx ===> 1... for cosx it approaches zero. [0<=x<=pi/2]

Trebla · Jul 22, 2009

jetblack2007 said:
$LHS = \left (2 + 2itan\, \theta \right )^n + \left (2 - 2itan\, \theta \right )^n \\ = 2^n\left (\frac{cos\theta + isin\theta}{cos\theta} \right )^n +2^n\left (\frac{cos\theta - isin\theta}{cos\theta} \right )^n \\ = \frac{2^n}{cos^n\theta} \left [\left (cos\theta + isin\theta \right )^n + \left (cos\theta - isin\theta \right )^n \right ] \\ = \frac{2^n}{cos^n\theta} \left [\left (cos \, n\theta + isin \, n\theta \right ) + \left (cos \, n\theta - isin \, n\theta \right ) \right ] \text{ noting that } cos \, n\theta - isin \, n\theta = cos \left (-\theta \right ) + isin \left (-\theta \right ) \\ = \frac{2^n}{cos^n\theta} \left (2cos \, n\theta \right ) \\ = \frac{2^{n+1}cos \, n\theta}{cos^n \, \theta} \\ = RHS$

$\text{Letting } z = 2i tan \, \theta, \\ \left (2 + z \right )^n + \left (2 - z \right )^n = \frac{2^{n+1}cos \, n\theta}{cos^n \, \theta} \\ \text{Thus, to solve } \left (2 + z \right )^5 + \left (2 - z \right )^5 = 0, \text{ we can solve:} \\ \frac{2^{6}cos \, 5\theta}{cos^5 \, \theta} = 0 \\ cos \, 5\theta = 0 \\ 5\theta = \frac{\pi}{2}, \frac{-\pi}{2}, \frac{3\pi}{2}, \frac{-3\pi}{2}, \text{ etc.} \\ \theta = \frac{\pi}{10}, \text{ etc.} \\ \text{Thus, one solution is } \theta = \frac{\pi}{10}, \text{ which corresponds to the root } z = 2itan \, \frac{\pi}{10} \\ \text{Hence, } 2itan \, \frac{\pi}{10} \text{ is a root of the equation } \left (2 + z \right )^5 + \left (2 - z \right )^5 = 0 \\$

$\text{Taking } 0 \leq \theta \leq \pi \text{ since the period of tan is } \pi, \text{the solutions of the equation are } \theta = \frac{\pi}{10}, \frac{3\pi}{10},\frac{5\pi}{10}, \frac{7\pi}{10}, \frac{9\pi}{10}\\ \text{Hence the other roots are } z = 2itan \, \frac{m\pi}{10}, \text{where} m = 1,3,5,7,9$

Will do last part after dinner.

Just to make note, there is a problem at m = 5 since z = 2i tan 5π/10 is undefined.

jet · Jul 22, 2009

Trebla said:
Just to make note, there is a problem at m = 5 since z = 2i tan 5π/10 is undefined.

That's interesting. I'm just unsure how to rectify it....

EDIT: Thinking over it, you can see that z^5 cancels out in the equations, meaning that it is actually a polynomial of degree 4 = four roots. I'm thinking just skip m=5.

jet · Jul 22, 2009

untouchablecuz said:
just one Q, with this question, did you know to expand the binomials (evetnually leading to a quadratic in x^2) or just tried it in an attempt to answer the Q?

Well from the fact that it contained tan pi/10, it was obvious that the root from one of the previous parts came into it. I just worked from there.

untouchablecuz · Jul 22, 2009

jetblack2007 said:
Well from the fact that it contained tan pi/10, it was obvious that the root from one of the previous parts came into it. I just worked from there.

well i realised that part, but did u know that um er

i dunno how to explain myself, dw

jet · Jul 22, 2009

untouchablecuz said:
well i realised that part, but did u know that um er

i dunno how to explain myself, dw

Well there was really nothing else to do.

cutemouse · Jul 22, 2009

Thanks alot Jetblack. I really understand it now =D

cutemouse · Jul 22, 2009

Hi again,

Just 1 last question. For part (ii) couldn't I just sub z=2itan(pi/10) to show that it satisfies the equation, which proves that it's a root?

Thanks

yibbon · Jul 23, 2009

jm01 said:
Hi again,

Just 1 last question. For part (ii) couldn't I just sub z=2itan(pi/10) to show that it satisfies the equation, which proves that it's a root?

Thanks

jetblack does this, however by saying z=2iTan(pi/10) you assume a value for theta, hence can simply sub the theta value into the RHS and find when Cos(5@) = 0. I think doing the expansion would take too long but would show it's a root yes

cutemouse · Jul 23, 2009

Well, if I sub z=2itanpi/10, then I can just use the result from part (i), which yields zero. Nothing too hard about that lol.

yibbon · Jul 23, 2009

jm01 said:
Well, if I sub z=2itanpi/10, then I can just use the result from part (i), which yields zero. Nothing too hard about that lol.

Same thing

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Hard complex number question (1 Viewer)

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