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Expanding Euler's Identity

  1. Aug 5, 2015 #1
    1. The problem statement, all variables and given/known data

    ## y^({9}) + y''' = 6 ##

    2. Relevant equations


    3. The attempt at a solution

    ## y^({9}) + y''' = 6 ##

    ## r^9 + r^3 = r^{3}(r^{6}+1)=0 ##

    ## r = 0, m = 3 ##

    ## r^6 + 1 = 0 = e^{(i(\pi + 2k\pi)} ##

    ## r = -1 = e^{i(\frac {\pi +2k\pi} {6})} ##

    ## k = 0 , r = e^{i(\frac {\pi} {6})} ##

    ## k = 1 , r = e^{i(\frac {\pi} {2})} ##

    ## k = 2 , r = e^{i(\frac {5\pi} {6})} ##

    My question is, when doing the ##k = 0,1,2, ... ## How does the ##e^{i\pi}## expand? My teacher has the answer for ## k = 0 ## as:

    ## r = e^{i(\frac {\pi} {6})} = \frac {\sqrt{3}} {2} + \frac {i} {2}##

    I don't understand how the exponential works out to the ##\frac {\sqrt{3}} {2} + \frac {i} {2}##
     
  2. jcsd
  3. Aug 5, 2015 #2

    SteamKing

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    Remember, eix = cos (x) + i sin (x). If x = π/6, then eix = ?

    https://en.wikipedia.org/wiki/Euler's_identity
     
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