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Algebra

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23 Jan 2009 22:42:58 IST

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507

Complex Summation

None

If , then evaluate

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rohit

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24 Jan 2009 00:28:07 IST

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is the answer 4

Hari Shankar

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24 Jan 2009 07:47:08 IST

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no. its not 4

Anant Kumar

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24 Jan 2009 22:52:48 IST

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Out of the many possible ways, we can use the following.
Since $\omega^5=1$ , so $1+\omega + \omega^2+\omega^3+\omega^4=0$
The expression
$E=\dfrac{\omega}{1+\omega^2}+\dfrac{\omega^2}{1+\omega^4}+ \dfrac{\omega^3}{1+\omega}+\dfrac{\omega^4}{1+\omega^3}$
$=\dfrac{1}{\omega+\omega^4}+\dfrac{1}{\omega^2+\omega^3}+\dfrac{1}{\omega^2+\omega^3}+\dfrac{1}{\omega+\omega^4}$ (I used $\omega^k = \dfrac{1}{\omega^{5-k}}$ for $k=1$ to $4$ )
$=2\left(\dfrac{\omega+\omega^2+\omega^3+\omega^4}{\omega^3(1+\omega^3)(1+\omega)}\right)$
$=2\left(\dfrac{-1}{\omega^3(1+\omega+\omega^3+\omega^4)}\right)$
$=2\,\dfrac{-1}{-\omega^5}=2$
So the given expression $E=2$

Hari Shankar

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25 Jan 2009 06:59:11 IST

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My startting point too was that $\omega$ is a solution to

$z^4+z^3+z^2+z+1 = 0 \equiv z^2+\frac{1}{z^2} +z+ \frac{1}{z} +1 = 0 \equiv t^2 - t + 1 = 0$

where $t = {z}+\frac{1}{z}$

Now using we can write the given expression as

$\frac{1}{\omega + \frac{1}{\omega}} + \frac{1}{\omega^2 + \frac{1}{\omega^2}}+ \frac{1}{\omega^2 + \frac{1}{\omega^2}}+\frac{1}{\omega + \frac{1}{\omega}} = 2 \left(\frac{1}{\omega + \frac{1}{\omega}} +\frac{1}{\omega^2 + \frac{1}{\omega^2}} \right)$

Also $t = \omega + \frac{1}{\omega}, t =\omega^2 + \frac{1}{\omega^2}$ are the roots of the quadratic in t. So the sum of reciprocals of the roots is 1.

Hence $2 \left(\frac{1}{\omega + \frac{1}{\omega}}+ \frac{1} {\omega^2 + \frac{1}{\omega^2}} \right) = 2$

sarang

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27 Jan 2009 16:47:27 IST

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Anant sir here its given w=1 but in sum of roots u used that u can explain me why

Gaurav |spideyunlimited| Ragtah

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27 Jan 2009 17:52:50 IST

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sum of nth roots of unity is zero. That's what.

PATHFINDER

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27 Jan 2009 18:00:57 IST

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1+w+w^2=1/..................i dun nu if i am correct

PATHFINDER

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27 Jan 2009 18:01:00 IST

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1+w+w^2=1/..................i dun nu if i am correct

Anshul Rai

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27 Jan 2009 18:18:22 IST

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w/(1+w²) + w²/(1+w⁴) + w³/(1+w) + w⁴/(1+w³)

= [w(1+w⁴) + w²(1+w²)]/(1+w²+w⁴+w⁶) + [w³(1+w³) + w⁴(1+w)]/( 1+w+w³+w⁴)

= [w+w⁵ + w² +w⁴]/(1+w²+w⁴+w) + [w³+w⁶ + w⁴+w⁵]/( 1+w+w³+w⁴)

= [w+1+ w² +w⁴]/(1+w²+w⁴+w) + [w³+w + w⁴+1]/( 1+w+w³+w⁴)

= 1+1 = 2

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