# complex numbers, help

• Nov 14th 2012, 03:51 AM
Tweety
complex numbers, help
Let $\alpha = z+z^{-1}.$

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

I have tried to square the first equation and get,

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

and i can't seem to go any further from here, any help appreciated.
• Nov 14th 2012, 04:13 AM
a tutor
Re: complex numbers, help
It's true for all z.
• Nov 14th 2012, 04:38 AM
Tweety
Re: complex numbers, help
Okay, thank you,

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

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

but i am not sure how to put this together to get $\alpha^{2} + \alpha -1 = 0$
• Nov 14th 2012, 06:49 AM
a tutor
Re: complex numbers, help
• Nov 14th 2012, 07:22 AM
Tweety
Re: complex numbers, help
Okay,

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

a) State the value of $z^{5}$ , hence show that $z^{-r} = z^{5-r}$ for all $r \in Z$
b) use the formula for the sum of a geometric series to show that $\sum_{r=0}^{4} z^{r} = 0$
c) Let $\alpha = z+z^{-1}.$ show that $\alpha^{2} + \alpha -1 = 0$. Also show that $\alpha = 2cos\frac{2\pi}{5}$ and hence find an expression for $cos\frac{2\pi}{5}$ in surd form.

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

my solutons for a and b

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

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

b)

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

$\frac{ 1(z^{5}-1}{z-1} = 0$
• Nov 14th 2012, 07:48 AM
Plato
Re: complex numbers, help
Quote:

Originally Posted by Tweety
Okay,
1) Z is the complex number $e^{\frac{2\pi i }{5}$
c) Let $\alpha = z+z^{-1}.$ show that $\alpha^{2} + \alpha -1 = 0$. Also show that $\alpha = 2cos\frac{2\pi}{5}$ and hence find an expression for $cos\frac{2\pi}{5}$ 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.
$z=\exp \left( {\frac{{2\pi i}}{5}} \right)$.

Here are useful facts for any nonzero complex number $w$.
If $w=\exp(i\theta)$ then $w^{-1}=\exp(-i\theta)$ and $w+w^{-1}=2\cos(\theta)$

Now try the question.
• Nov 14th 2012, 10:06 AM
a tutor
Re: complex numbers, help
In post 2 I meant it's NOT true for all z.

Sorry.
• Nov 14th 2012, 10:36 AM
Tweety
Re: complex numbers, help
i am actually stil not able to compute part 'c'
• Nov 14th 2012, 10:59 AM
Plato
Re: complex numbers, help
Quote:

Originally Posted by Tweety
i am actually stil not able to compute part 'c'

If $\alpha = 2\cos \left( {\frac{{2\pi }}{5}} \right)$ can't you show that $\alpha^2+\alpha-1=0~?$
• Nov 14th 2012, 11:34 AM
Tweety
Re: complex numbers, help
Quote:

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

BUt I dont understand how alpha equals that ?
• Nov 14th 2012, 11:50 AM
Plato
Re: complex numbers, help
Quote:

Originally Posted by Tweety
BUt I dont understand how alpha equals that ?

I posted it above.
If $w=\exp(i\theta)$ then $w^{-1}=\exp(-i\theta)$ and $w+w^{-1}=2\cos(\theta)$

$\cos(\theta)+i\sin(\theta)+\cos(-\theta)+i\sin(-\theta)=$
$[\cos(\theta)+\cos(-\theta)]+i[\sin(\theta)+\sin(-\theta)]=$
$[\cos(\theta)+\cos(\theta)]+i[\sin(\theta)-\sin(\theta)]=$
$2\cos(\theta)$

Recall that $\cos$ is an even function and $\sin$ is an odd function.
• Nov 14th 2012, 12:56 PM
mathteacher067
Re: complex numbers, help
Hello!