Re: complex numbers, help

Re: complex numbers, help

Okay, thank you,

I know $\displaystyle \alpha = z+ z^{-1} = z + z^{10} $

$\displaystyle \alpha^{2} = z^{2} + z^{-2} +2 = z^{2} + z^{9} +2 $

but i am not sure how to put this together to get $\displaystyle \alpha^{2} + \alpha -1 = 0 $

Re: complex numbers, help

I hope your teacher set your task in writing. If so please post it exactly as it was given to you.

Re: complex numbers, help

Okay,

1) Z is the complex number $\displaystyle e^{\frac{2\pi i }{5} $

a) State the value of $\displaystyle z^{5} $ , hence show that $\displaystyle z^{-r} = z^{5-r} $ for all $\displaystyle r \in Z $

b) use the formula for the sum of a geometric series to show that $\displaystyle \sum_{r=0}^{4} z^{r} = 0 $

c) Let $\displaystyle \alpha = z+z^{-1}. $ show that $\displaystyle \alpha^{2} + \alpha -1 = 0 $. Also show that $\displaystyle \alpha = 2cos\frac{2\pi}{5} $ and hence find an expression for $\displaystyle cos\frac{2\pi}{5} $ in surd form.

i have done all parts except part c, am stuck on that bit

my solutons for a and b

$\displaystyle z^{5} = 1 $ recognising z as an 5th root of unity?

$\displaystyle z^{-r} = 1 * z^{-r} = z^{5} z^{-r} = z^{5-r} $

b)

$\displaystyle S_{n} = \frac{a(r^{n}-1)}{r-1} $

$\displaystyle \frac{ 1(z^{5}-1}{z-1} = 0 $

Re: complex numbers, help

Quote:

Originally Posted by

**Tweety** Okay,

1) Z is the complex number $\displaystyle e^{\frac{2\pi i }{5} $

c) Let $\displaystyle \alpha = z+z^{-1}. $ show that $\displaystyle \alpha^{2} + \alpha -1 = 0 $. Also show that $\displaystyle \alpha = 2cos\frac{2\pi}{5} $ and hence find an expression for $\displaystyle cos\frac{2\pi}{5} $ in surd form.

I think that I have written this before. But again, posting the complete problem to begin with saves a lot of time.

Let us use a more convenient notation.

$\displaystyle z=\exp \left( {\frac{{2\pi i}}{5}} \right)$.

Here are useful facts for any nonzero complex number $\displaystyle w$.

If $\displaystyle w=\exp(i\theta)$ then $\displaystyle w^{-1}=\exp(-i\theta)$ and$\displaystyle w+w^{-1}=2\cos(\theta)$

Now try the question.

Re: complex numbers, help

In post 2 I meant it's NOT true for all z.

Sorry.

Re: complex numbers, help

i am actually stil not able to compute part 'c'

Re: complex numbers, help

Quote:

Originally Posted by

**Tweety** i am actually stil not able to compute part 'c'

If $\displaystyle \alpha = 2\cos \left( {\frac{{2\pi }}{5}} \right)$ can't you show that $\displaystyle \alpha^2+\alpha-1=0~?$

Re: complex numbers, help

Quote:

Originally Posted by

**Plato** If $\displaystyle \alpha = 2\cos \left( {\frac{{2\pi }}{5}} \right)$ can't you show that $\displaystyle \alpha^2+\alpha-1=0~?$

BUt I dont understand how alpha equals that ?

Re: complex numbers, help

Quote:

Originally Posted by

**Tweety** BUt I dont understand how alpha equals that ?

I posted it above.

If $\displaystyle w=\exp(i\theta)$ then $\displaystyle w^{-1}=\exp(-i\theta)$ and$\displaystyle w+w^{-1}=2\cos(\theta)$

$\displaystyle \cos(\theta)+i\sin(\theta)+\cos(-\theta)+i\sin(-\theta)=$

$\displaystyle [\cos(\theta)+\cos(-\theta)]+i[\sin(\theta)+\sin(-\theta)]=$

$\displaystyle [\cos(\theta)+\cos(\theta)]+i[\sin(\theta)-\sin(\theta)]=$

$\displaystyle 2\cos(\theta)$

Recall that $\displaystyle \cos$ is an even function and $\displaystyle \sin$ is an odd function.

Re: complex numbers, help

Hello!

Would try to help you!

I'm not sure, but what i got is

Rectangualr form a^2*z^2+1/z^2+2

Polar form a^2*z^2+1/z^2+2

check it here Online Complex Numbers Calculator

I used Complex Numbers Calculator

Good luck