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Complex Numbers

  1. Feb 8, 2016 #1
    1. The problem statement, all variables and given/known data
    0f9c29f81d.png
    2. Relevant equations

    De Moivres Theorem/ Eulers formula

    3. The attempt at a solution
    fa49e98d1c.jpg

    Honestly don't know where to go with this now. I already applied De Moivres theorem at the very end. It feels like I have to do something more with either De Moivres theorem or Eulers formula. (only 2 we have done)
     
  2. jcsd
  3. Feb 8, 2016 #2

    Samy_A

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    Start from ##a+bi-\frac{1}{a+bi}=i##, multiply both sides with ##a+bi## and work it out. That will allow you to find the value(s) of a and b. Then use De Moivre's formula to get ##z^6##.
     
  4. Feb 8, 2016 #3
    f3d25612e5.jpg
    Where do I go form here?
     
  5. Feb 8, 2016 #4

    Samy_A

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    On the left you have a complex number, say ##X+iY##, that is equal to 0. What can you then conclude about X and Y?
     
  6. Feb 8, 2016 #5
    then X and Y should be equal to 0. But I don't see where is the complex number on the left? I know b-ai is one but I'm not too sure about the 2abi thing?
     
  7. Feb 8, 2016 #6

    Samy_A

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    Correct.
    The complex number on the left is ##a²+2abi-b²-1-ai+b=(a²-b²-1+b)+(2ab-a)i##. Your computation showed that this complex number is 0. You should be able to get the value(s) of a and b from that.
     
  8. Feb 8, 2016 #7
    Ah. I see now. the entire left side was a complex number. Is this correct now?
    d498220a6f.jpg
    1e3f200bd8.jpg
     
  9. Feb 8, 2016 #8

    Samy_A

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    b is correct.
    But you made a sign error while computing a, so that a is not correct.
    Keep in mind that an equation in the form ##a²=c## can have two real solutions.

    Also, to be complete, you should explain why in ##2ab=a##, you reject the solution ##a=0##.
     
  10. Feb 8, 2016 #9
    Ah I see. Should it be a = + or - sqrt(1/4). therefore giving me two solutions for a. a = +1/4 and a = -1/4. I entered both into 2ab-a=0 and both give me true statements.
    Also would the reason for rejecting a=0 be that if a=0 it would give me no value for b either.
     
  11. Feb 8, 2016 #10

    Samy_A

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    Any value for a will give you a true statement if b=1/2.
    No, that is still wrong. As I wrote, somewhere in the calculation you make a sign error: a minus for no reason becomes a plus.
    Correct.
     
  12. Feb 8, 2016 #11

    Ray Vickson

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    There is no need for DeMoivre or Euler. Once you have ##z = a + ib## you can quite quickly and easily get ##z^2##, then get ##z^4 = z^2 \cdot z^2## and finally ##z^6 = z^4 \cdot z^2##.

    However, if you do care to use DeMoivre/Euler you should first examine very carefully the actual form of ##z##.
     
  13. Feb 8, 2016 #12

    ehild

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    Multiplying the equation z+1/z=i by z, you get a quadratic equation for. Get the solution with the quadratic formula. Write z in the exponential/ trigonometric form.
     
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