Homework Help: Complex number problem

1. Apr 21, 2006

sparsh

using eulers formula express cos5ø in terms of cosø. Hence show that =cos(pie/10) is a root of the equation 16x^4 - 20x^2 +5 =0 ..

2. Apr 21, 2006

HallsofIvy

Looks like homework to me so I'm moving this to the homework section. Do you know what Euler's formula is? If so, go ahead and apply it!

3. Apr 21, 2006

sparsh

Yup i know it

But the problem is that when i convert cos5ø in terms of (cosø + isinø )
then how do i remove the sin term ......

like this complex number has only got its real part. And using eulers formula we cant have angles different for the cos and sine terms..

Moreover we have to use the obtained result in the 2nd part of the question

4. Apr 21, 2006

matt grime

If you know it then use it, or at least write here what you've obtained by using it, so that you can get help (not just the answer).

You know how to get rid of sin's and turn them into cos's from trig (cos^2+sin^2=1) and the real part only has even powers of sin in it.

5. Apr 21, 2006

Reshma

I think you will have to consider only the real part here since you have only 'x' terms.
Use the De Moivre's formula for powers and roots of complex numbers.
$$\left[r(\cos \phi + i\sin \phi)\right]^n = r^n(\cos n\phi + i\sin n\phi)$$

Hope this helps a bit...

6. Apr 21, 2006

sparsh

I converted it into something like this ::

cos 5ø = (cosø)^5 + i cos(90-ø)^5 - icos(90-5ø)

Then what should i do

7. Apr 21, 2006

matt grime

That is not correct. In fact it looks as though you think that (x+y)^5=x^5+y^5.

Do not write sin(x)=cos(90-x), by the way. There is no need to do that.

8. Apr 21, 2006

sparsh

@ matt

By the eulers formula or De Movier's Theorum (cosø+isinø)^5 is cos 5ø+isin5ø
Is there anything wrong with that.

Actually I gotta give the exam after two days . So please it would be really great if you could just give me the solution to this problem. I have spent too much time on it already. Have to study Physics and Chem too

9. Apr 21, 2006

matt grime

that is right, and nothing to do with what i said: you failed to expand the fifth power as a power properly.

certainly (cosx+isinx)^5=cos5x+isin5x, but you also seem to think that this is the same as cos^5(x)+isin^5(x), which is obviously false.

10. Apr 21, 2006

sparsh

I think i got the solution to this problem...

Just to share it with you. We can break e^i5ø into e^i3ø e^i2ø and then solve .
Thanks anyways for all your help ...

11. Apr 22, 2006

HallsofIvy

sparsh, in your original post you said "use Euler's formula". My first response was "Do you know what Euler's formula is? If so, go ahead and apply it!" Your last response was the first one where it looks like you are actually using Euler's formula!

12. Apr 22, 2006

sparsh

Can i know what is the basic difference between the Eulers Formula and the De moivers theorum. I mean they produce the same result.

13. Apr 23, 2006

Reshma

Euler's formula states:
$$e^{i\phi} = \cos \phi + i\sin \phi$$

De Moivre's Formula states:
$$(\cos \phi + i\sin \phi)^n = \cos(n\phi)+i\sin(n\phi)$$

De Moivre's formula is useful when you are dealing with powers and roots of complex numbers.

14. Apr 23, 2006

Curious3141

I think the basic problem is knowledge and application of Binomial Theorem, not De Moivre's or Euler's formulae.

Sparsh, you can expand second and third powers of (x + y).

Can you do the same for the fifth power ?