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Solving a complex equation

  1. Jan 31, 2010 #1
    The problem says: Find all solutions z of the equation: z^6 - 4z^3 + 4 = i

    First I factored the equation into (z^3 -2)^2 = i, set w= z^3 -2 and solved w^2 = i for w_1 = sqrt(i) and w_2 = -sqrt(i). I tried setting z^3 - 2 = sqrt(i) and solving but I get stuck there. I really have no idea how to go about this.
     
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  3. Jan 31, 2010 #2

    Dick

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    You want to figure out how to use deMoivre to extract roots. For example to solve w^2=i, write i=e^(i*(pi/2)), e^(i*(pi/2+2pi)), ... Now if w^n=e^(i*theta) a root is w=e^(i*theta/2). Can you write both of your roots +/-sqrt(i) in the rectangular form a+bi?
     
  4. Jan 31, 2010 #3
    hmm...I guess I don't really see what you mean. Is it possible to write +/-sqrt(i) in the form a+bi? I can't figure out how.
     
  5. Jan 31, 2010 #4

    Dick

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    DeMoivre! e^(i*theta)=cos(theta)+i*sin(theta). You must have seen this somewhere. So e^(i*pi/2)=cos(pi/2)+i*sin(pi/2)=0+i*1=i. Now (e^(i*theta))^2=e^(i*2*theta). So w=e^(i*pi/4) solves w^2=e^(i*pi/2)=i. So w=cos(pi/4)+i*sin(pi/4). That's one of the square roots of i. What is it in rectangular form?
     
  6. Jan 31, 2010 #5
    oh yeah, yeah, now I see. That would be w=sqrt(2)/2 + i*sqrt(2)/2. So now I need to solve w=z^3 -2 = sqrt(2)/2 + i*sqrt(2)/2? Since the original equation is of degree six shouldn't I have six solutions?
     
  7. Jan 31, 2010 #6

    Dick

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    Sure. There are two square roots of i. So z^3=2+sqrt(2)/2+sqrt(2)*i/2 or z^3=2+(-sqrt(2)/2-sqrt(2)*i/2). Those are two cubic equations. Each one has three roots. Now it gets a little hairy because you have to express the right side of those equations in the form r*e^(i*theta) and find 3 angles phi such that 3*phi is theta. Then your roots are r^(1/3)*e^(i*phi) for each of those angles. But conceptually its the same and finding the square roots of i.
     
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