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anonimnystefy
2013-06-03 03:40:49

Oh, sorry. I didn't understand what he wanted to say. And, anyway, using that would be very hard for a hand method.

bob bundy
2013-06-03 03:38:19

hi Stefy

Why isn't it correct?

Bob

anonimnystefy
2013-06-03 03:29:14

#### fm_hlp wrote:

its a polynomial of power 4 as it simplifies to

That is not correct.

#### SteveB wrote:

Thanks anonimnystefy. I thought there might be a division by zero somewhere !!

You're welcome!

fm_hlp
2013-06-03 03:12:37

its a polynomial of power 4 as it simplifies to

SteveB
2013-06-03 03:10:48

Thanks anonimnystefy. I thought there might be a division by zero somewhere !!

anonimnystefy
2013-06-03 03:04:35

Well, actually, when k=0, z is not defined (cot(0)=not defined). But, the first troubles with k=0 begin when we take the reciprocal of both sides of "1/z=..." .

fm_hlp
2013-06-03 03:03:50

wow
thanks a lot

SteveB
2013-06-03 03:00:46

Yes. I agree. If you let k=0 then z = 1
Obviously (1 + 1)/1 = 2
So this cannot give us 1 when raised to the power of 5.
Why does my algebraic argument not work for k = 0
Usually a fifth root gives us five solutions. (????)

I can see your point about the negative sign as well .....

anonimnystefy
2013-06-03 02:55:14

In this steps you lose s minus sign:

There are 4 solutions. k=0 is not a solution.

SteveB
2013-06-03 02:44:51

I was working on those formulas as you were writing that post.
I think it works and produces the right result.
There are 5 solutions.

anonimnystefy
2013-06-03 02:17:50

Hi SteveB

Have you tried using the half-angle formulas for sine and cosine? (the answer provided has k*pi/5, not 2*k*pi/5 )

SteveB
2013-06-03 01:20:00

I am not sure about this because I cannot do the very last bit of trigonometry but this is what I have got so far:

Using a right angled triangle in the Argand diagram:

Use the rule of multiplying by the complex conjugate pair:

Using standard trigonometric results we need this in terms of cot(A/2):

I have put in the minus sign in now. Well spotted anonimnystefy.
k = 0  is not a solution I agree Stefy, but why does my answer not work for k = 0 ?

EDIT: I now can see that cot(0) is not possible. Given that cot X = 1/(tan X) if X=0 then we have a division by zero.
Therefore the argument is not valid for this case.

fm_hlp
2013-06-02 22:32:17

By writing the equation

in the form

show that its roots are