complex numbers, help

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November 14th 2012, 02:51 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.
November 14th 2012, 03:13 AM
a tutor

Re: complex numbers, help

It's true for all z.
November 14th 2012, 03:38 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 $\alpha^{2} + \alpha -1 = 0$
November 14th 2012, 05:49 AM
a tutor

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.
November 14th 2012, 06:22 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)

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

$\frac{ 1(z^{5}-1}{z-1} = 0$
November 14th 2012, 06:48 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.
November 14th 2012, 09:06 AM
a tutor

Re: complex numbers, help

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

Sorry.
November 14th 2012, 09:36 AM
Tweety

Re: complex numbers, help

i am actually stil not able to compute part 'c'
November 14th 2012, 09:59 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 $\alpha^2+\alpha-1=0~?$
November 14th 2012, 10:34 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 ?
November 14th 2012, 10:50 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.
November 14th 2012, 11:56 AM
mathteacher067

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