Question

Prove that \(z=tan(\frac{\pi}{10})\) is a root of \(5z^5-5iz^4-10z^3-10iz^2+z-i=0\). I simplified it to \((5z^4-10z^2+1)(z-i)=0\), but I'm just not sure from here.

Answer 1

So, if we plug it into the quadratic: \[z^2=\frac{10\left(\begin{matrix}+ \\ -\end{matrix}\right) \sqrt{80}}{10}\]

Answer 2

So we get the 4 roots from \[z=\left(\begin{matrix}+ \\ -\end{matrix}\right) \sqrt{\frac{10\left(\begin{matrix}+ \\ -\end{matrix}\right) \sqrt{80}}{10}}\]

Answer 3

@satellite73 , any ideas?

Answer 4

hmm
i am still at your step, solving the 4th degree equation
i wonder if there is some half or double angle identity to use
any clue? as in what the topic for the class is?

Answer 5

oh i have an idea
not saying it is right
write it in polar form and see if it checks

Answer 6

did you try working with
\[5(z^2-1)^2-4=0\]?

Answer 7

not sure that helps
\[(z^2-1)^2=\frac{4}{5}\] \[z^2-1=\pm\frac{2}{\sqrt5}\]

Answer 8

at least that proves that \(z^2\) is real, not sure where to go from there

Answer 9

How do you get that factorization?

Answer 10

I think there is something off about it unless I multiplied and divided wrong.

Answer 11

I'm talking about what fibs did

Answer 12

\[5z^4-10z^2+1=5(z^24-2z^2)+1=5(z^4-2z^2+1)+1-5=5(z^2-1)^2-4\]

Answer 13

ok typo there, but you get the idea
not sure how it help, but is seems to nice to be a coincidence

Answer 14

was it suppose to be
\[5z^5-5iz^4-10z^3+10iz^2+z-i=0\]?

Answer 15

ooh are you saying the initial factorization is wrong? i didn't check ti

Answer 16

yep i said fibs
maybe i should have spelled out Fibonacci

Answer 17

I think he meant that one part to be + instead -
with it being - there is actually no real solutions (i cheated and used wolfram)

Answer 18

oh i see
lets wolfram

Answer 19

http://www.wolframalpha.com/input/?i=+5z^5%E2%88%925iz^4%E2%88%9210z^3%E2%88%9210iz^2%2Bz%E2%88%92i%3D0.

Answer 20

btw, i assume that even though fibonacci was a guy, fibonacci chick is probably a woman

Answer 21

oops
i don't check last names
my bad

Answer 22

anyway i am stuck

Answer 23

so I guess we need to somehow prove \[\tan(\pi/10)=\sqrt{1-\frac{2}{\sqrt{5}}}\]

Answer 24

if we can prove this then maybe we can work backwards

Answer 25

it is right, according to the wolf

Answer 26

lol want to work through this?
http://math.stackexchange.com/questions/196067/how-to-calculate-the-exact-value-of-tan-frac-pi10

Answer 27

so sorry guys, i was afk

Answer 28

i bet there is a snappier way, but i sure don't know it
probably write the solution as \(z^2=1-\frac{2}{\sqrt5}\) and then make some argument (pun intended) about what \(z^2\) has to be in polar form

Answer 29

topic is complex variables

Answer 30

lol, got that

Answer 31

also there must be a typo in the question
before the factorization
but no matter
we got to
\[(z^2-1)^2=\frac{4}{5}\] making
\[z^2=1\pm\frac{2}{\sqrt5}\] which does in fact make \(z=\tan(\frac{\pi}{10})\)

Answer 32

Sorry again, I was thinking about applying Cauchy Integral Theorem somehow

Answer 33

the issue is pi,

Answer 34

@Zarkon was here a second ago, he teaches this stuff

Answer 35

but so, let me check real quick, the factorization may be wrong, but the initial problem is right

Answer 36

i think the initial is wrong and the factorization is right, because the solution is right to the factored one

Answer 37

That is what I was going to say. Took the words out of my mouth.

Answer 38

nope, the initial is right, I just checked it against the problem

Answer 39

That would lead to all imaginary solutions then

Answer 40

therefore tan(pi/10) could not possibly be a root

Answer 41

it's weird I know
That's what is partially messing with me

Answer 42

\[5z^5-5iz^4-10z^3+10iz^2+z-i=0 \]
This is what I'm thinking what was meant.

Answer 43

not possible
wolf has 5 complex solutions to the original
http://www.wolframalpha.com/input/?i=+5z^5%E2%88%925iz^4%E2%88%9210z^3%E2%88%9210iz^2%2Bz%E2%88%92i%3D0.
but exactly the right solution to the factored one
http://www.wolframalpha.com/input/?i=tan%28pi%2F10%29

Answer 44

hmm if we redistribute the z-i which do we get, then let's go with that and say my teacher wrote the problem wrong

Answer 45

When you multiply it you get the equation I wrote above.

Answer 46

Let's say it's wrong and you are right(because most likely you are)

Answer 47

There is just a one single difference between the questions there is a + in front of the 10iz^2

Answer 48

I agree, what if I wrote the factorization wrong and it's a plus?

Answer 49

well that factorization you wrote is correct for the + version

Answer 50

then I don't get the minus i

Answer 51

ok so i'm convinced,

Answer 52

I was incorrect, thank you for that... it explains so much

Answer 53

But I'm not sure if I can actually be help to this problem. I would have to do some reading.

Answer 54

so far i am at
\[\cos(2\theta)=1\pm\frac{2}{\sqrt5}\]

Answer 55

actually, I think @satellite73 's factorization might do the trick, if I can prove that \(tan(\pi/10)=\sqrt{1(+/-)\frac2 {\sqrt5}}\)

Answer 56

you can
i have provided a link to the proof

Answer 57

http://math.stackexchange.com/questions/196067/how-to-calculate-the-exact-value-of-tan-frac-pi10
but i bet there is another way, using complex numbers
as this method only uses real numbers

Answer 58

then, obviously it's a root right?

Answer 59

well, i was thinking about using CIT and then having a be tan(pi/10)

Answer 60

making it =0

Answer 61

then using squeeze to equal zero

Answer 62

but i just don't know how to apply it

Answer 63

i would go with just using numbers
but i could be wrong
without a similar example worked out, i am not sure how to proceed

Answer 64

i like yours better

Answer 65

a lot better

Answer 66

do you mind best answering @myininaya for me @satellite73 , she deserves a medal too

Answer 67

i bet @Zarkon has a better method

Answer 68

we'll see when he gets back can you check mine in a min?

Answer 69

http://math.stackexchange.com/questions/7695/how-to-prove-cos-frac2-pi-5-frac-1-sqrt54
I want to use the complex way, but I don't understand the series

Answer 70

I don't see a slick way of doing it. What you did is fine

Answer 71

How did they write z=1+z+z^2+z^3+z^4=0?

Answer 72

Thank you @Zarkon

Answer 73

oops \(tan\theta=\frac{z-z^{-1}}{z+z^{-1}}\)

Answer 74

\(z=e^\frac{i\pi}{10}\)

Answer 75

@satellite73 , do you know how they did the series?

Answer 76

ok i got the expansion, but why can i make the expansion?