Algebra question help!

[Destiny of Pi]

Hey guys, I have questions that I failed to solve. Could any of you help me?

1) if (z+1)/(z-1) = cosx + i sinx, show that z= -i cot(x/2)

• I found that z is purely imaginary, and tried to replace z with bi and then solve for b, but it didn't work that well

2) if z is a root of the equation [(z+1)/(z-1)]^5 = 1, show that z^4 + 2z^2 + 0.2 = 0

• I believe I have to solve the prior question in order to do this question.

3) Hence show that [tan^2 (pi/5)] [tan^2 (2pi/5)] = 5 and [tan^2 (pi/5)] [tan^2 (2pi/5)] = 10

• I don't even know where to start off with... help!

It would be great if you can show me your steps! Thanks

Edited March 22, 2017 by Destiny of Pi

[kw0573]

The question is supposed to say (z+1)/(z-1) = cos x + i sin x.
Using identities cot (x) = (cot² (x/2) - 1) / (2 cot (x/2)), and cos θ = 1/2 * (e^iθ + e^-iθ) and i sin θ = 1/2 * (e^iθ - e^-iθ)
I used z = re^iθ. where z* = re^-iθ.

1) Multiply (z+1)/(z-1) by (z*-1) / (z*- 1), and using above identities I get

(r² - 2r i sin θ - 1) / (r² - 2 r cos θ + 1) = cos x + i sin x

2) Match imaginary part to imaginary part to get sin x = (-2r sin θ) / (r² - 2 r cos θ + 1)

3) now divide both sides by sin x, we get (r² - 2r i sin θ - 1) / (-2r sin θ) = cot x + i, which simplifies to (r² - 1) / (-2r sin θ) = cot x
4) Match again with the cot (x) identity above, we see that -2r sin θ = cot (x / 2) ==> r sin θ = - cot (x / 2) and from numerator matching that r² = cot² (x / 2) which implies Re(z) = 0. Hence z =  i r sin θ = -i cot (x/2).
An alternative solution can be found here, which is a more or less equivalent using z = cos θ + i sin θ instead. 

Question 2 is a direct binomial expansion, without any substitution, just (z+1)^2 - (z-1)^2 = 0, odd power terms cancel. Divide by 10 you get the answer.

Question 3 you made a typo somewhere.

Edited March 22, 2017 by kw0573

[Destiny of Pi]

20 minutes ago, kw0573 said:

The question is supposed to say (z+1)/(z-1) = cos x + i sin x.
Using identities cot (x) = (cot² (x/2) - 1) / (2 cot (x/2)), and cos θ = 1/2 * (e^iθ + e^-iθ) and i sin θ = 1/2 * (e^iθ - e^-iθ)
I used z = re^iθ. where z* = re^-iθ.

1) Multiply (z+1)/(z-1) by (z*-1) / (z*- 1), and using above identities I get

(r² - 2r i sin θ - 1) / (r² - 2 r cos θ + 1) = cos x + i sin x

2) Match imaginary part to imaginary part to get sin x = (-2r sin θ) / (r² - 2 r cos θ + 1)

3) now divide both sides by sin x, we get (r² - 2r i sin θ - 1) / (-2r sin θ) = cot x + i, which simplifies to (r² - 1) / (-2r sin θ) = cot x
4) Match again with the cot (x) identity above, we see that -2r sin θ = cot (x / 2) ==> r sin θ = - cot (x / 2) and from numerator matching that r² = cot² (x / 2) which implies Re(z) = 0. Hence z =  i r sin θ = -i cot (x/2).
An alternative solution can be found here, which is a more or less equivalent using z = cos θ + i sin θ instead. 

Question 2 is a direct binomial expansion, without any substitution, just (z+1)^2 - (z-1)^2 = 0, odd power terms cancel. Divide by 10 you get the answer.

Question 3 you made a typo somewhere.

Thanks for your help! 

For question 3, I don't think there's a typo but I did add a few brackets to make it easier to see, does this help?

[kw0573]

1 hour ago, Destiny of Pi said:

Thanks for your help! 

For question 3, I don't think there's a typo but I did add a few brackets to make it easier to see, does this help?

You essentially said the expression is equal to both 5, and 10. In any case you should try your best at such "hence" questions because they appear quite often in exams. Just try something using both results in a) and b). Good luck you can do it

[kw0573]

Key to solve these hence problems is to first find an expression in the form.Note that (cos x + isin x)^5 =1 yields x = n pi / 5, where n is 0, 2, 4, 6, 8. Sketch cot(x/2) to see that it (and hence cot^2(x/2)) is positive and decreasing from 0 to pi.

z = -i cot (x/2) ==> z^2 = -cot^2(x/2)

iz = cot(x/2)

1/(-z^2) = tan^2(x/2)

From quartic you get z^2 = (-2 +/- sqrt(4-0.8))/2 which also = -cot^2(x/2). For tan^2(pi/5) ==> x=2pi/5, ==> z^2 is more negative and it's z^2 = (-2 - sqrt(4-0.8))/2. Do that for the other angles, simplify, you should get expected answers.