1. ## complex numbers, help

Let $\displaystyle \alpha = z+z^{-1}.$

show that $\displaystyle \alpha^{2} + \alpha -1 = 0$

I have tried to square the first equation and get,

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

and i can't seem to go any further from here, any help appreciated.

2. ## Re: complex numbers, help

It's true for all z.

3. ## 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 $\displaystyle \alpha^{2} + \alpha -1 = 0$

5. ## 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)

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

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

6. ## Re: complex numbers, help

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.

7. ## Re: complex numbers, help

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

Sorry.

8. ## Re: complex numbers, help

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

9. ## Re: complex numbers, help

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 $\displaystyle \alpha^2+\alpha-1=0~?$

10. ## Re: complex numbers, help

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 ?

11. ## Re: complex numbers, help

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.

Hello!