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Mathematics Q & A
complex number
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bskaterina
December 2008
COULD YOU HELP ME SOLVING THE FOLLOWING EQUATION.
(1+z)^5+i(1-z)^5=0
ps. (use the equation w^5=-1)[/i]
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krrispes
May 2012
(1+z)^5 = -i(1-z)^5
{(1+z)/(1-z)}^5 = -i = e^i(3pi/2)
(1+z)/(1-z) = e^i(3pi/10)
z = {e^(3pi/10)-1}/{e^(3pi/10)+1}
z= {cos(3pi/10)+isin(3pi/10)-1}/{cos(3pi/10)+isin(3pi/10)}
solve this equation...
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availchet
June 2012
(1+z)^5+i(1-z)^5=0
(1+z)^5=-i{(1-z)^5}
[(1+z)/(1-z)]^5= -1(i)
[(1+z)/(1-z)]^5= (w^5)(i^5) {since w^5=-1 and i^4=1}
[(1+z)/(1-z)]^5= (wi)^5
[(1+z)/(1-z)]= (wi)
solving to get
z= (wi-1)/(wi+1) {applying componendo and dividendo}
***Hope that Helps***
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Aditi
December 2008
Hey , Katerina ... really good question ... I remember coming across this some time ago ... but couldnt get a solution ....ppl cmon solve this !!!
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ranjangupta_1
December 2008
nice one....alas!, dont know the solution :-( HELP!!
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Friction
December 2008
Hi .... It is a good question ...
I think I have a solution
Substitute : Z = re^(iθ)
Now , expand using binomial
(1 + Z)^5 = 1 + 5 Z + 10 Z^2 + 10 Z^3 + 5Z^4 + Z^5.
Use : De Moivre's theorem : {exp(iθ)}^n = exp(inθ)
Now , seperate out the real and imaginary parts .
The 2 unknowns : r & θ
and will have 2 eqautions.
This approach should work . But very time taking and lengthy. Try it ... and Lets think about a more elegant approach !!!
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Ayush
November 2012
1+z=1.47 * e^(i*pi/4 )
1-z= 1.47 * e^ (- i pi/4)
If Z is = i
your funtion = (1.47)^5 * [ e^(i 5pi/4) + e^( - i 5pi/4)]
which is equal to = 2^(5/2) * [2cos 5pi/4] = 2^(5/2 +1/2) = 2^(3)= 8
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In this Discussion
Aditi
December 2008
availchet
June 2012
Ayush
November 2012
Friction
December 2008
krrispes
May 2012
ranjangupta_1
December 2008
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