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**fm_hlp****Guest**

By writing the equation

in the form

show that its roots are

please provide a step to step answer to teach me

i cant seem to solve

**SteveB****Member**- Registered: 2013-03-07
- Posts: 557

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.

*Last edited by SteveB (2013-06-02 18:28:32)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

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 )

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**SteveB****Member**- Registered: 2013-03-07
- Posts: 557

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

In this steps you lose s minus sign:

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

*Last edited by anonimnystefy (2013-06-02 04:59:15)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**SteveB****Member**- Registered: 2013-03-07
- Posts: 557

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 .....

*Last edited by SteveB (2013-06-02 05:03:40)*

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**fm_hlp****Guest**

wow

thanks a lot

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

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=..." .

*Last edited by anonimnystefy (2013-06-02 05:05:28)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**SteveB****Member**- Registered: 2013-03-07
- Posts: 557

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

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**fm_hlp****Guest**

its a polynomial of power 4 as it simplifies to

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

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!

*Last edited by anonimnystefy (2013-06-02 05:29:43)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,466

hi Stefy

Why isn't it correct?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,600

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

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