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Equation with complex numbers

  1. Nov 12, 2015 #1
    I have to solve the following equation:

    z4=i*(z-2i)4

    Now, i tried to move everything but i (imaginary number) to the left side and then find the 4-th root of i, when i did that, i had four solutions, with one of them being eiπ/8. But i don't know what to do with the left side, since i get way to complicated expression. So there must be less complicated way to solve this.
     
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  3. Nov 12, 2015 #2

    SteamKing

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    It's not clear what you did here. Can you share the other steps you made to get everything but i to one side of the equation?

    After doing the algebra, it doesn't seem like you would wind up with a simple equation like z4 - i = 0, which has as its solution of the fourth roots of i.
     
  4. Nov 12, 2015 #3

    HallsofIvy

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    If I understand you, you divided both sides by [itex]z- 2i[/itex] to get [itex]\frac{z^4}{(z- 2i)^4}= \sqrt[4]{i}[/itex]. The right side has, of course, four different values. What did you get for the four values of [itex]\sqrt[4]{i}[/itex]?
     
  5. Nov 12, 2015 #4
    Actually, i divided both sides by (z-2i)4 but i assume that you meant that.

    Anyway, I got the following:

    i= 0+i = cos(π/2) + isin(π/2)

    then, fourth root of i is:

    cos[(π/2 + 2kπ)/4) + isin[(π/2 + 2kπ)/4)

    so i have the following 4 solutions:

    i0=cos(π/8) + isin(π/8)
    i1=cos(5π/8) + isin(5π/8)
    i2=cos(9π/8) + isin(9π/8)
    i3=cos(13π/8) + isin(13π/8)

    but i still don't know what can i do next?


    @SteamKing I did exactly what HallsofIvy said.
     
  6. Nov 12, 2015 #5

    SteamKing

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    It's still not clear how calculating i1/4 gets you any closer to finding the value of z which solves the original equation.

    After all,

    ##\frac{z^4}{(z-2i)^4}=i^4##

    but you are looking for the value of z which satisfies the equation ##z^4 = i (z - 2i)^4##.

    Why didn't you expand the RHS, collect all the resulting terms on the LHS, and then solve the resulting polynomial ##f(z) = 0##?
     
  7. Nov 12, 2015 #6
    Well, i thought, if i find the exact value on the rhs (that is eiπ/8 and three other values) and if i have ##\frac{z}{z-2i}## on the lhs then if i write a+ib instead of z (which means that z=a +ib) i could get the solution, but it turns out it isn't working. By expansion, do you mean using binomial theorem?
     
  8. Nov 12, 2015 #7

    SteamKing

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    Yes.
     
  9. Nov 12, 2015 #8

    HallsofIvy

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    Yes, it does. Once you have arrived at [itex]\frac{z}{z- 2i}= a[/itex] (where a is one of the four fourth roots of i) it follows that [itex]z= az- 2ai[/itex] so that [itex]z- az= (1- a)z= -ai[/itex] and then [itex]z= -\frac{ai}{1- a}[/itex].
     
  10. Nov 12, 2015 #9
    Use these values for the right hand side:

    ##i^{\frac{1}{4}}=\left ( e^{i \frac{\pi}{2}}\right )^{\frac{1}{4}}=e^{i \frac{\pi}{8}}=\cos{\left (\frac{\pi}{8}\right)}+i \sin{\left (\frac{\pi}{8} \right )}##
     
  11. Nov 12, 2015 #10

    jbriggs444

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    I think you dropped a factor of 2 in there. ##z = az - 2ai## leads to ##z = -\frac{2ai}{1-a}##
     
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