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complex number
  • COULD YOU HELP ME SOLVING THE FOLLOWING EQUATION.
    (1+z)^5+i(1-z)^5=0
    ps. (use the equation w^5=-1)[/i]
  • 6 Comments sorted by
  • Hey , Katerina ... really good question ... I remember coming across this some time ago ... but couldnt get a solution ....ppl cmon solve this !!!
  • nice one....alas!, dont know the solution :-( HELP!!
  • 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 !!!
  • (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...
  • (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***
  • 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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